Differential Equations jee mains questions | JEE Main Maths Differential Equations Previous Year Question

Q 21 :

The slope of tangent at any point (x, y) on a curve $y = y (x) is \frac{x^{2} + y^{2}}{2 x y}, x > 0 .$ If $y (2) = 0$ , then a value of $y (8)$ is [2023]

$- 2 \sqrt{3}$
$2 \sqrt{3}$
$4 \sqrt{3}$
$- 4 \sqrt{2}$

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(3)

$We have, \frac{d y}{d x} = \frac{x^{2} + y^{2}}{2 x y}$

Putting $y = v x \Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x}$ , we get

$v + x \frac{d v}{d x} = \frac{1}{2} (\frac{1 + v^{2}}{v})$

$\Rightarrow x \frac{d v}{d x} = \frac{1}{2} (\frac{1}{v} - v) = \frac{1}{2} (\frac{1 - v^{2}}{v})$ $∴$ $\int \frac{2 v}{1 - v^{2}} d v = \int \frac{d x}{x}$

$\Rightarrow \log c - \log$ $| 1 - v^{2} |$ $= \log$ $| x |$

$c = x (1 - \frac{y^{2}}{x^{2}}) = \frac{x^{2} - y^{2}}{x}$

At $y (2) = 0, c = 2$

$∴$ $\frac{x^{2} - y^{2}}{x} = 2$

Now, at $x = 8,$ $2 = \frac{8^{2} - y^{2}}{8} \Rightarrow y^{2} = 64 - 16 = 48 \Rightarrow y = 4 \sqrt{3}$

Q 22 :

Let $f$ be a differentiable function such that $x^{2} f (x) - x = 4 \int_{0}^{x} t f (t) d t, f (1) = \frac{2}{3} .$ Then $18 f (3)$ is equal to [2023]

180
210
160
150

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(3)

We have, $x^{2} f (x) - x = 4 \int_{0}^{x} t f (t) d t$

$\Rightarrow 2 x f (x) + x^{2} f' (x) - 1 = 4 x f (x)$

$\Rightarrow x^{2} \frac{d y}{d x} - 2 x y = 1 [f (x) = y, f' (x) = \frac{d y}{d x}]$

$\Rightarrow \frac{d y}{d x} - \frac{2 y}{x} = \frac{1}{x^{2}}, which is a linear differential equation.$

$I.F. = e^{\int - \frac{2}{x} d x} = e^{- 2 \log x} = \frac{1}{x^{2}}$

$∴$ Solution is given by

$\frac{y}{x^{2}} = \int \frac{1}{x^{4}} d x + C \Rightarrow \frac{y}{x^{2}} = - \frac{1}{3 x^{3}} + C$ ...(i)

$At x = 1, y = \frac{2}{3}$ $[∵ f (1) = \frac{2}{3}]$

$\frac{2}{3} = \frac{- 1}{3} + C \Rightarrow C = 1$

$∴$ $y = \frac{- 1}{3 x} + x^{2} \Rightarrow f (x) = \frac{- 1}{3 x} + x^{2}$

$Now, 18 f (3) = 18 [\frac{- 1}{9} + 9] = 160$

Q 23 :

Let $y = y (x)$ be a solution curve of the differential equation $(1 - x^{2} y^{2}) d x = y d x + x d y$ . If the line $x$ = 1 intersects the curve $y = y (x)$ at $y = 2$ and the line $x = 2$ intersects the curve $y = y (x)$ at $y = α$ , then a value of $α$ is [2023]

$\frac{3 e^{2}}{2 (3 e^{2} - 1)}$
$\frac{3 e^{2}}{2 (3 e^{2} + 1)}$
$\frac{1 - 3 e^{2}}{2 (3 e^{2} + 1)}$
$\frac{1 + 3 e^{2}}{2 (3 e^{2} - 1)}$

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(4)

$(1 - x^{2} y^{2}) d x = y d x + x d y$

$\Rightarrow d x = \frac{d (x y)}{1 - {(x y)}^{2}}$

$\int d x = \int \frac{d (x y)}{1 - {(x y)}^{2}}$ $x = \frac{1}{2} \ln$ $| \frac{1 + x y}{1 - x y} |$ $+ C$

$We are given y (1) = 2 and y (2) = α$

$Put x = 1, y = 2 : 1 = \frac{1}{2} \ln$ $| \frac{1 + 2}{1 - 2} |$ $+ C$

$C = 1 - \frac{1}{2} \ln 3$

$Now put x = 2, y = α$

$2 = \frac{1}{2} \ln$ $| \frac{1 + 2 α}{1 - 2 α} |$ $+ 1 - \frac{1}{2} \ln 3$ $;$ $1 + \frac{1}{2} \ln 3 = \frac{1}{2} \ln$ $| \frac{1 + 2 α}{1 - 2 α} |$

$2 + \ln 3 = \ln (\frac{1 + 2 α}{1 - 2 α})$

$\Rightarrow$ $| \frac{1 + 2 α}{1 - 2 α} |$ $= 3 e^{2}$ $\Rightarrow \frac{1 + 2 α}{1 - 2 α} = 3 e^{2}, \frac{1 + 2 α}{1 - 2 α} = - 3 e^{2}$

$\frac{1 + 2 α}{1 - 2 α} = 3 e^{2} \Rightarrow α = \frac{3 e^{2} - 1}{2 (3 e^{2} + 1)}$

and $\frac{1 + 2 α}{1 - 2 α} = - 3 e^{2} \Rightarrow α = \frac{3 e^{2} + 1}{2 (3 e^{2} - 1)}$

Q 24 :

Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} + \frac{5}{x (x^{5} + 1)} y = \frac{{(x^{5} + 1)}^{2}}{x^{7}}, x > 0 .$ If $y (1) = 2$ , then $y (2)$ is equal to [2023]

$\frac{637}{128}$
$\frac{679}{128}$
$\frac{693}{128}$
$\frac{697}{128}$

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(3)

$I.F. = e^{\int \frac{5 d x}{x (x^{5} + 1)}} = e^{\int \frac{5 x^{- 6}}{x^{- 5} + 1} d x}$

Put $1 + x^{- 5} = t \Rightarrow - 5 x^{- 6} d x = d t$

$∴$ $I.F. = e^{\int \frac{- d t}{t}} = \frac{1}{t} = \frac{x^{5}}{1 + x^{5}}$

$Solution is given by \frac{y x^{5}}{1 + x^{5}} = \int \frac{x^{5}}{(1 + x^{5})} \times \frac{{(1 + x^{5})}^{2}}{x^{7}} d x + C$

$= \int x^{3} d x + \int x^{- 2} d x + C$ $∴$ $\frac{y x^{5}}{1 + x^{5}} = \frac{x^{4}}{4} - \frac{1}{x} + C$

At $x = 1, y = 2$

$\frac{2}{2} = \frac{1}{4} - 1 + C \Rightarrow C = \frac{7}{4}$ $∴$ $\frac{y x^{5}}{1 + x^{5}} = \frac{x^{4}}{4} - \frac{1}{x} + \frac{7}{4}$

$At x = 2$

$y (\frac{32}{33}) = \frac{21}{4} = \frac{693}{128}$

Q 25 :

Let $y = y (x), y > 0$ be a solution curve of the differential equation $(1 + x^{2}) d y = y (x - y) d x .$ If $y (0) = 1$ and $y (2 \sqrt{2}) = β$ , then [2023]

$e^{3 β^{- 1}} = e (5 + \sqrt{2})$
$e^{3 β^{- 1}} = e (3 + 2 \sqrt{2})$
$e^{β^{- 1}} = e^{- 2} (3 + 2 \sqrt{2})$
$e^{β^{- 1}} = e^{- 2} (5 + \sqrt{2})$

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(2)

$Given differential equation is$

$(1 + x^{2}) d y = y (x - y) d x$

$\Rightarrow \frac{d y}{d x} = \frac{x y - y^{2}}{1 + x^{2}} \Rightarrow \frac{d y}{d x} = \frac{x y}{1 + x^{2}} - \frac{y^{2}}{1 + x^{2}}$

$\frac{d y}{d x} - \frac{x y}{1 + x^{2}} = \frac{- y^{2}}{1 + x^{2}}$

$\Rightarrow \frac{1}{y^{2}} \frac{d y}{d x} - \frac{1}{y} \frac{x}{1 + x^{2}} = \frac{- 1}{1 + x^{2}}$ ...(i)

Put $\frac{1}{y} = z \Rightarrow \frac{- 1}{y^{2}} \frac{d y}{d x} = \frac{d z}{d x}$

So, (i) becomes

$\frac{d z}{d x} + \frac{x z}{1 + x^{2}} = \frac{1}{1 + x^{2}}$

Which is a linear differential equation.

$I.F. = e^{\int \frac{x}{1 + x^{2}} d x} = e^{\int \frac{x}{1 + x^{2}} d x} = e^{\ln \sqrt{1 + x^{2}}} = \sqrt{1 + x^{2}}$

$∴$ $Solution is z \sqrt{1 + x^{2}} = \int \frac{\sqrt{1 + x^{2}}}{1 + x^{2}} d x$

$z \sqrt{1 + x^{2}} = \int \frac{d x}{\sqrt{1 + x^{2}}}$

$z \sqrt{1 + x^{2}} = \log$ $| x + \sqrt{1 + x^{2}} |$ $+ C$

$\frac{\sqrt{1 + x^{2}}}{y} = \log$ $| x + \sqrt{1 + x^{2}} |$ $+ C$ ...(ii)

$We have, y (0) = 1$

So, from (ii), $1 = \log 1 + C \Rightarrow C = 1$

$∴$ $(ii) becomes$

$\Rightarrow \frac{\sqrt{1 + x^{2}}}{y} = \log$ $| x + \sqrt{1 + x^{2}} |$ $+ 1$

$\Rightarrow$ $\frac{\sqrt{1 + x^{2}}}{y} = \log$ $| x + \sqrt{1 + x^{2}} |$ $+ \log e$

$\Rightarrow$ $\frac{\sqrt{1 + x^{2}}}{y} = \log (e (x + \sqrt{1 + x^{2}}))$

$\Rightarrow$ Also, $y (2 \sqrt{2}) = β$

$\frac{3}{β} = \log (e (2 \sqrt{2} + 3)) \Rightarrow e^{\frac{3}{β}} = (e (3 + 2 \sqrt{2}))$

$\Rightarrow e^{3 β^{- 1}} = (e (3 + 2 \sqrt{2}))$

Q 26 :

Let $y = y_{1} (x)$ and $y = y_{2} (x)$ be the solution curves of the differential equation $\frac{d y}{d x} = y + 7$ with initial conditions $y_{1} (0) = 0$ and $y_{2} (0) = 1$ respectively. Then the curves $y = y_{1} (x)$ and $y = y_{2} (x)$ intersect at [2023]

infinite number of points
two points
no point
one point

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(3)

We have, $\frac{d y}{d x} = y + 7, y_{1} (0) = 0, y_{2} (0) = 1$

$\int \frac{d y}{y + 7} = \int d x$

$\ln$ $| y + 7 |$ $= x + k \Rightarrow y + 7 = e^{x + k} = e^{x} \cdot e^{k} = C e^{x} [∵ e^{k} = C]$

$\Rightarrow y = - 7 + C e^{x}$

$y_{1} (0) = 0 \Rightarrow 0 = - 7 + C \Rightarrow C = 7$

$y_{2} (0) = 1 \Rightarrow 1 = - 7 + C \Rightarrow C = 8$

$∴$ $y_{1} (x) = - 7 + 7 e^{x} and y_{2} (x) = - 7 + 8 e^{x}$

If $y_{1} (x)$ and $y_{2} (x)$ intersect at any point, then the values of both curves will be the same at that point.

$∴$ $- 7 + 7 e^{x} = - 7 + 8 e^{x} \Rightarrow e^{x} = 0 \Rightarrow Not possible$

$∴$ $The curves y_{1} (x) and y_{2} (x) will not intersect at any point.$

Q 27 :

Let $x = x (y)$ be the solution of the differential equation $2 (y + 2) \log_{e} (y + 2) d x + (x + 4 - 2 \log_{e} (y + 2)) d y = 0,$ $y > - 1$ with $x (e^{4} - 2) = 1$ .Then $x (e^{9} - 2)$ is equal to [2023]

$\frac{10}{3}$
$\frac{32}{9}$
$3$
$\frac{4}{9}$

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(2)

$2 (y + 2) \ln (y + 2) d x + (x + 4 - 2 \ln (y + 2)) d y = 0$

$\Rightarrow 2 \ln (y + 2) + (x + 4 - 2 \ln (y + 2)) \frac{1}{y + 2} \cdot \frac{d y}{d x} = 0$ ...(i)

Let $\ln (y + 2) = t$ $\Rightarrow$ $\frac{1}{y + 2} \cdot \frac{d y}{d x} = \frac{d t}{d x}$

Equation (i) becomes,

$2 t + (x + 4 - 2 t) \cdot \frac{d t}{d x} = 0 \Rightarrow (x + 4 - 2 t) \frac{d t}{d x} = - 2 t$

$\Rightarrow \frac{d x}{d t} = \frac{2 t - 4 - x}{2 t} \Rightarrow \frac{d x}{d t} + \frac{x}{2 t} = \frac{2 t - 4}{2 t}$

which is a linear differential equation in $x$ .

$I . F . = e^{\int \frac{1}{2 t} d t} = e^{\frac{1}{2} \ln t} = t^{1 / 2}$

Now, $x t^{1 / 2} = \int \frac{2 t - 4}{2 t} t^{1 / 2} d t + C$

$\Rightarrow x t^{1 / 2} = \int (t^{1 / 2} - \frac{2}{t^{1 / 2}}) d t + C$

$\Rightarrow x \cdot t^{1 / 2} = \frac{t^{3 / 2}}{3 / 2} - 2 \cdot \frac{t^{1 / 2}}{1 / 2} + C$

$\Rightarrow x \cdot t^{1 / 2} = \frac{2 t^{3 / 2}}{3} - 4 t^{1 / 2} + C \Rightarrow x = \frac{2}{3} \cdot t - 4 + C \cdot t^{- 1 / 2}$

Put $t = \ln (y + 2)$

$x = \frac{2}{3} \ln (y + 2) - 4 + C \cdot (\ln {(y + 2))}^{- 1 / 2}$ ...(ii)

Put $y = e^{4} - 2$ and $x = 1$

$1 = \frac{2}{3} \ln (e^{4} - 2 + 2) - 4 + C (\ln {(e^{4} - 2 + 2))}^{- 1 / 2}$

$\Rightarrow 1 = \frac{2}{3} \times 4 - 4 + C \times \frac{1}{2} \Rightarrow \frac{C}{2} = 5 - \frac{8}{3} = \frac{7}{3} \Rightarrow C = \frac{14}{3}$

Equation (ii) becomes,

$x = \frac{2}{3} \ln (y + 2) - 4 + \frac{14}{3} (\ln {(y + 2))}^{- 1 / 2}$

Put $y = e^{9} - 2$

$x (e^{9} - 2) = \frac{2}{3} \ln (e^{9} - 2 + 2) - 4 + \frac{14}{3} (\ln {(e^{9} - 2 + 2))}^{- 1 / 2}$

$\Rightarrow x (e^{9} - 2) = \frac{2}{3} \times 9 - 4 + \frac{14}{3} \times \frac{1}{3} = 2 + \frac{14}{9} = \frac{32}{9}$

Q 28 :

If $y = y (x)$ is the solution curve of the differential equation $\frac{d y}{d x} + y \tan x = x \sec x, 0 \leq x \leq \frac{π}{3}, y (0) = 1$ , then $y (\frac{π}{6})$ is equal to [2023]

$\frac{π}{12} - \frac{\sqrt{3}}{2} \log_{e} (\frac{2}{e \sqrt{3}})$
$\frac{π}{12} + \frac{\sqrt{3}}{2} \log_{e} (\frac{2}{e \sqrt{3}})$
$\frac{π}{12} + \frac{\sqrt{3}}{2} \log_{e} (\frac{2 \sqrt{3}}{e})$
$\frac{π}{12} - \frac{\sqrt{3}}{2} \log_{e} (\frac{2 \sqrt{3}}{e})$

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(1)

$\frac{d y}{d x} + y \tan x = x \sec x, 0 \leq x \leq \frac{π}{3}, y (0) = 1$

$I.F. = e^{\int \tan x d x} = e^{\log_{e} \sec x} = \sec x$ $\Rightarrow y \cdot \sec x = \int x \sec^{2} x d x$

$\Rightarrow y \sec x = x \tan x - \int \tan x d x$

$\Rightarrow y \cdot \sec x = x \tan x - \log_{e} (\sec x) + c$

$Now y (0) = 1 \Rightarrow 1 = c$

$Now, at x = \frac{π}{6}, we have$

$\frac{2 y}{\sqrt{3}} = \frac{π}{6} \cdot \frac{1}{\sqrt{3}} - \log_{e} \frac{2}{\sqrt{3}} + 1;$ $y = \frac{π}{12} - \frac{\sqrt{3}}{2} \log_{e} \frac{2}{\sqrt{3}} + \frac{\sqrt{3}}{2}$

$= \frac{π}{12} - \frac{\sqrt{3}}{2} [\log_{e} \frac{2}{\sqrt{3}} - \log_{e} e]$ $= \frac{π}{12} - \frac{\sqrt{3}}{2} \log_{e} (\frac{2}{e \sqrt{3}})$

Q 29 :

Let $α x = \exp (x^{β} y^{γ})$ be the solution of the differential equation $2 x^{2} y d y - (1 - x y^{2}) d x = 0, x > 0, y (2) = \sqrt{\log_{e} 2} .$ Then $α + β - γ$ equals [2023]

0
- 1
1
3

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(3)

$α x = e^{x^{β} \cdot y^{γ}}$ and $2 x^{2} y \frac{d y}{d x} = 1 - x \cdot y^{2}$

Let $y^{2} = t \Rightarrow 2 y \frac{d y}{d x} = \frac{d t}{d x}$ $\Rightarrow x^{2} \frac{d t}{d x} = 1 - x t$

$\Rightarrow \frac{d t}{d x} + \frac{t}{x} = \frac{1}{x^{2}}$

$I.F. = e^{\log_{e} x} = x$

$t (x) = \int \frac{1}{x^{2}} \cdot x d x \Rightarrow y^{2} \cdot x = \log_{e} x + C$

$\Rightarrow 2 \log_{e} 2 = \log_{e} 2 + C \Rightarrow C = \log_{e} 2$

$Hence, x y^{2} = \log_{e} 2 x$ $∴$ $2 x = e^{x \cdot y^{2}}$

$Hence, α = 2, β = 1, γ = 2$ $∴$ $α + β - γ = 1$

Q 30 :

$Let y = y (x) be the solution of the differential equation x^{3} d y + (x y - 1) d x = 0, x > 0, y (\frac{1}{2}) = 3 - e . 1$ $Then y (1) is equal to$ [2023]

$2 - e$
$e$
$1$
$3$

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(3)

$Given: x^{3} d y + (x y - 1) d x = 0$

$\Rightarrow x^{3} d y = (1 - x y) d x$

$\Rightarrow \frac{d y}{d x} = \frac{1 - x y}{x^{3}} \Rightarrow \frac{d y}{d x} + \frac{y}{x^{2}} = \frac{1}{x^{3}},$ which is a linear differential equation.

$Now, (I.F.) = e^{\int P d x} = e^{\int \frac{1}{x^{2}} d x} = e^{- 1 / x}$

Solution is given by, $y e^{- 1 / x} = \int \frac{1}{x^{3}} e^{- 1 / x} d x + C$

Putting $u = \frac{- 1}{x} \Rightarrow d u = \frac{1}{x^{2}} d x,$ we get

$y e^{- 1 / x} = \int - e^{u} u d u + c = - [u e^{u} - \int e^{u} d u] + c$

$= - [u e^{u} - e^{u}] + c = (1 - u) e^{u} + c$

$\Rightarrow y e^{- 1 / x} = (1 + \frac{1}{x}) e^{- 1 / x} + c$ ...(i)

Put $x = \frac{1}{2}$ and $y = 3 - e$

$3 e^{- 2} - e^{- 1} = 3 e^{- 2} + c \Rightarrow c = - e^{- 1}$

Equation (i) becomes, $y e^{- 1 / x} = e^{- 1 / x} + \frac{e^{- 1 / x}}{x} - e^{- 1}$

Put $x = 1$ to get $y (1)$ ,

$\Rightarrow y e^{- 1} = e^{- 1} + e^{- 1} - e^{- 1} = e^{- 1} \Rightarrow y = 1$

Q 31 :

Let $y = y (x)$ be the solution of the differential equation $(x^{2} - 3 y^{2}) d x + 3 x y d y = 0, y (1) = 1 .$ Then $6 y^{2} (e)$ is equal to [2023]

$e^{2}$
$\frac{3}{2} e^{2}$
$3 e^{2}$
$2 e^{2}$

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(4)

$(x^{2} - 3 y^{2}) d x + 3 x y d y = 0$

$\Rightarrow x^{2} d x - 3 y^{2} d x + 3 x y d y = 0 \Rightarrow x^{2} - 3 y^{2} + 3 x y \frac{d y}{d x} = 0$

Put $t = y^{2} \Rightarrow 2 y \frac{d y}{d x} = \frac{d t}{d x}$ $\Rightarrow x^{2} - 3 t + \frac{3 x}{2} \frac{d t}{d x} = 0$

$\Rightarrow \frac{2 x}{3} - \frac{2 t}{x} + \frac{d t}{d x} = 0 \Rightarrow \frac{d t}{d x} - \frac{2 t}{x} = - \frac{2 x}{3},$

which is a linear differential equation.

$I.F. = e^{\int - \frac{2}{x} d x} = e^{- 2 \ln | x |} = e^{\ln \frac{1}{x^{2}}} = \frac{1}{x^{2}}$

So, $t \cdot \frac{1}{x^{2}} = \int \frac{1}{x^{2}} (\frac{- 2 x}{3}) d x$ $\Rightarrow \frac{y^{2}}{x^{2}} = - \frac{2}{3} \ln | x | + C$

When, $x = 1, y = 1$

$1 = \frac{- 2}{3} \ln (1) + C \Rightarrow C = 1$ $∴$ $\frac{y^{2}}{x^{2}} = \frac{- 2}{3} \ln | x | + 1$

At $x = e,$ $\frac{y^{2} (e)}{e^{2}} = \frac{- 2}{3} + 1 \Rightarrow y^{2} (e) = \frac{e^{2}}{3} \Rightarrow 6 y^{2} (e) = 2 e^{2}$

Q 32 :

Let $y = y (x)$ be the solution curve of the differential equation $\frac{d y}{d x} = \frac{y}{x} (1 + x y^{2} (1 + \log_{e} x)), x > 0, y (1) = 3 .$ Then, $\frac{y^{2} (x)}{9}$ is equal to [2023]

$\frac{x^{2}}{5 - 2 x^{3} (2 + \log_{e} x^{3})}$
$\frac{x^{2}}{3 x^{3} (1 + \log_{e} x^{2}) - 2}$
$\frac{x^{2}}{7 - 3 x^{3} (2 + \log_{e} x^{2})}$
$\frac{x^{2}}{2 x^{3} (2 + \log_{e} x^{3}) - 3}$

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(1)

$Let y = v x; \frac{d y}{d x} = v + x \frac{d v}{d x}$

$v + x \frac{d v}{d x} = v + v^{3} x^{3} (1 + \log_{e} x)$

$\int v^{- 3} d v = \int (x^{2} + x^{2} \log_{e} x) d x$

$\frac{v^{- 2}}{- 2} = \frac{x^{3}}{3} + \log x \frac{x^{3}}{3} - \int \frac{x^{2}}{3} d x = \frac{x^{3}}{3} + \frac{x^{3}}{3} \log x - \frac{x^{3}}{9} + C$

$- \frac{x^{2}}{2 y^{2}} = \frac{2}{9} x^{3} + \frac{x^{3}}{3} \log x + c; y (1) = 3$

$\frac{- 1}{2 \times 9} = \frac{2}{9} + C \Rightarrow C = \frac{- 5}{18}$

$\frac{- x^{2}}{2 y^{2}} = \frac{2}{9} x^{3} + \frac{x^{3}}{3} \log x - \frac{5}{18} \Rightarrow \frac{9}{y^{2}} = \frac{4 x^{3} + 6 x^{3} \log x - 5}{- x^{2}}$

$\Rightarrow \frac{y^{2} (x)}{9} = \frac{x^{2}}{5 - 2 x^{3} (2 + \log x^{3})}$

Q 33 :

Let $y = y (t)$ be a solution of the differential equation $\frac{d y}{d t} + α y = γ e^{- β t}$ where, $α > 0, β > 0$ and $γ > 0$ . Then $\lim_{t \to \infty} y (t)$ [2023]

is -1
is 0
is 1
does not exist

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(2)

$\frac{d y}{d t} + α y = γ e^{- β t}$

$I.F. = e^{\int α d t} = e^{α t}$

Solution is given by

$y \cdot e^{α t} = \int e^{α t} \cdot γ e^{- β t} d t = γ \int e^{(α - β) t} d t$

$\Rightarrow y \cdot e^{α t} = γ \frac{e^{(α - β) t}}{α - β} + c$ $\Rightarrow y = \frac{γ}{α - β} e^{- β t} + c e^{- α t}$

$So, \lim_{t \to \infty} y (t) = \frac{γ}{α - β} \lim_{t \to \infty} e^{- β t} + c \lim_{t \to \infty} e^{- α t} = 0$

Q 34 :

Let $y = f (x)$ be the solution of the differential equation $y (x + 1) d x - x^{2} d y = 0, y (1) = e .$ Then $\lim_{x \to 0^{+}} f (x)$ is equal to [2023]

$\frac{1}{e^{2}}$
$\frac{1}{e}$
$0$
$e^{2}$

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(3)

$Given differential equation,$ $y (x + 1) d x - x^{2} d y = 0$

$\Rightarrow \frac{d y}{d x} = \frac{y (x + 1)}{x^{2}}$

$On integrating,$ $\int \frac{1}{y} d y = \int \frac{x + 1}{x^{2}} d x \Rightarrow \ln y = \ln x - \frac{1}{x} + C$

$Given y (1) = e, that is at x = 1, y = e$

So, $\ln e = \ln 1 - 1 + C \Rightarrow C = 2$

$So, \ln y = \ln x - \frac{1}{x} + 2$

$\Rightarrow y = x e^{- \frac{1}{x} + 2}$ $\Rightarrow \lim_{x \to 0^{+}} x e^{- \frac{1}{x} + 2} = 0$

Q 35 :

Let $y = y (x)$ be the solution of the differential equation $x \log_{e} x \frac{d y}{d x} + y = x^{2} \log_{e} x, (x > 1) .$ If $y (2) = 2$ , then $y (e)$ is equal to [2023]

$\frac{1 + e^{2}}{2}$
$\frac{4 + e^{2}}{4}$
$\frac{1 + e^{2}}{4}$
$\frac{2 + e^{2}}{2}$

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(2)

$Given: x \log_{e} x \frac{d y}{d x} + y = x^{2} \log_{e} x (x > 1)$

$\Rightarrow \frac{d y}{d x} + \frac{y}{x \ln x} = x$ which is a linear differential equation.

whose Integrating factor is given by

$I.F. = e^{\int \frac{1}{x \ln x} d x} =$ $\ln$ $| x |$

$y \ln$ $| x |$ $= \frac{x^{2}}{2} \ln x - \int \frac{1}{x} \cdot \frac{x^{2}}{2} d x$ ...(i)

$\Rightarrow y \ln$ $| x |$ $= \ln$ $| x |$ $\frac{x^{2}}{2} - \frac{x^{2}}{4} + C$

$Also, y (2) = 2 \Rightarrow C = 1$

On substituting the value of $C$ in (i), we get

$y (x) = \frac{x^{2}}{2} - \frac{x^{2}}{4 | \ln (x) |} + \frac{1}{| \ln (x) |}$

$\Rightarrow y (e) = \frac{e^{2}}{2} - \frac{e^{2}}{4} + 1 = 1 + \frac{e^{2}}{4}$ or $\frac{4 + e^{2}}{4}$

Q 36 :

Let the solution curve $y = y (x)$ of the differential equation $\frac{d y}{d x} - \frac{3 x^{5} \tan^{- 1} (x^{3})}{{(1 + x^{6})}^{3 / 2}} y = 2 x \exp {\frac{x^{3} - \tan^{- 1} x^{3}}{\sqrt{(1 + x^{6})}}}$ pass through the origin. Then $y (1)$ is equal to [2023]

$\exp {\frac{4 + π}{4 \sqrt{2}}}$
$\exp {\frac{4 - π}{4 \sqrt{2}}}$
$\exp {\frac{1 - π}{4 \sqrt{2}}}$
$\exp {\frac{π - 4}{4 \sqrt{2}}}$

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(2)

Given differentiate equation is

$\frac{d y}{d x} - \frac{3 x^{5} \tan^{- 1} (x^{3})}{{(1 + x^{6})}^{3 / 2}} y = 2 x \exp {\frac{x^{3} - \tan^{- 1} x^{3}}{\sqrt{(1 + x^{6})}}}$ ,

Which is in the form $\frac{d y}{d x} + P y = Q$

Here, $P = - \frac{3 x^{5} \tan^{- 1} (x^{3})}{{(1 + x^{6})}^{3 / 2}}, Q = 2 x \exp {\frac{x^{3} - \tan^{- 1} x^{3}}{\sqrt{1 + x^{6}}}}$

Let $I = - \int \frac{3 x^{5} \tan^{- 1} x^{3}}{{(1 + x^{6})}^{3 / 2}} d x$

Put $x^{3} = \tan θ \Rightarrow 3 x^{2} d x = \sec^{2} θ d θ$

$I = - \int \frac{\tan θ \cdot θ}{\sec^{3} θ} \times \sec^{2} θ d θ = - \int θ \cdot \sin θ d θ$

$= - [θ \int \sin θ d θ - \int (\int \sin θ d θ) d θ]$

$= - [- θ \cdot \cos θ + \sin θ] = - [\frac{x^{3}}{\sqrt{1 + x^{6}}} - \tan^{- 1} (x^{3}) \cdot \frac{1}{\sqrt{1 + x^{6}}}]$

$∴$ $I.F. = e^{\int p d x} = e^{I}$

$Solution of the D.E. is$

$y \cdot e^{I} = \int e^{I} \cdot 2 x \cdot \exp {\frac{x^{3} - \tan^{- 1} (x^{3})}{\sqrt{1 + x^{6}}}} d x$

$= \int 2 x \cdot e^{\frac{- x^{3} + \tan^{- 1} (x^{3})}{\sqrt{1 + x^{6}}}} \cdot e^{\frac{x^{3} - \tan^{- 1} (x^{3})}{\sqrt{1 + x^{6}}}} d x$

$= \int 2 x \cdot d x = x^{2} + c$ $∴$ $y e^{I} = x^{2} + c$

$∴$ $y = x^{2} \exp {\frac{- \tan^{- 1} (x^{3}) + x^{3}}{\sqrt{1 + x^{6}}}} + c \exp {\frac{x^{3} - \tan^{- 1} (x^{3})}{\sqrt{1 + x^{6}}}}$

This curve passes through the origin.

So, $0 = 0 + c \Rightarrow c = 0$

The required solution is

$y (x) = x^{2} \exp {\frac{x^{3} - \tan^{- 1} (x^{3})}{\sqrt{1 + x^{6}}}}$

At $x = 1$ $y (1) = \exp (\frac{1 - \frac{π}{4}}{\sqrt{2}}) = \exp (\frac{4 - π}{4 \sqrt{2}})$

Q 37 :

The solution of the differential equation $\frac{d y}{d x} = - (\frac{x^{2} + 3 y^{2}}{3 x^{2} + y^{2}}), y (1) = 0$ is [2023]

$\log_{e}$ $| x + y |$ $- \frac{x y}{{(x + y)}^{2}} = 0$
$\log_{e}$ $| x + y |$ $- \frac{2 x y}{{(x + y)}^{2}} = 0$
$\log_{e}$ $| x + y |$ $+ \frac{x y}{{(x + y)}^{2}} = 0$
$\log_{e}$ $| x + y |$ $+ \frac{2 x y}{{(x + y)}^{2}} = 0$

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(4)

Given, $\frac{d y}{d x} = - [\frac{x^{2} + 3 y^{2}}{3 x^{2} + y^{2}}], y (1) = 0$

$Substitute y = v x \Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x}$

$v + x \frac{d v}{d x} = - [\frac{x^{2} + 3 v^{2} x^{2}}{3 x^{2} + v^{2} x^{2}}] = - [\frac{1 + 3 v^{2}}{3 + v^{2}}]$

$\Rightarrow x \frac{d v}{d x} = - [\frac{1 + 3 v^{2}}{3 + v^{2}}] - v$ $= - [\frac{1 + 3 v^{2} + 3 v + v^{3}}{3 + v^{2}}]$

$\Rightarrow x \frac{d v}{d x} = - \frac{{(1 + v)}^{3}}{3 + v^{2}} \Rightarrow \int \frac{3 + v^{2}}{{(1 + v)}^{3}} d v = - \int \frac{d x}{x}$

$\Rightarrow \int \frac{1 + v^{2} + 2 v - 2 v + 2}{{(1 + v)}^{3}} d v = - \int \frac{d x}{x}$

$\Rightarrow \int \frac{1}{(1 + v)} d v + 2 \int \frac{1 - v}{{(1 + v)}^{3}} d v = - \int \frac{d x}{x}$

$Substitute 1 + v = t \Rightarrow v = t - 1 \Rightarrow d v = d t$

$∴$ $\int \frac{d t}{t} + 2 \int \frac{1 - t + 1}{t^{3}} d t = - \ln x + C$

$\Rightarrow \log_{e} t + 2 \int (\frac{2 - t}{t^{3}}) d t = - \log_{e} x + C$

$\Rightarrow \log_{e} t + 2 \int (2 t^{- 3} - t^{- 2}) d t = - \log_{e} x + C$

$\Rightarrow \log_{e}$ $| \frac{y}{x} + 1 |$ $+ 2 [- t^{- 2} + t^{- 1}] = - \log_{e} x + C$

$\Rightarrow \log_{e}$ $| \frac{y + x}{x} |$ $+ 2$ $[\frac{- 1}{{(1 + \frac{y}{x})}^{2}} + \frac{1}{\frac{y}{x} + 1}] = - \log_{e} x + C$

$\Rightarrow \log_{e}$ $| y + x |$ $+ 2$ $[\frac{- x^{2}}{{(y + x)}^{2}} + \frac{x}{y + x}]$ $= C$

$\Rightarrow \log_{e}$ $| y + x |$ $+ \frac{2 x y}{{(y + x)}^{2}} = C$

$We have y (1) = 0 \Rightarrow C = 0 .$

$\Rightarrow \log_{e}$ $| x + y |$ $+ \frac{2 x y}{{(x + y)}^{2}} = 0$

Q 38 :

Let a differentiable function $f$ satisfy $f (x) + \int_{3}^{x} \frac{f (t)}{t} d t = \sqrt{x + 1}, x \geq 3 .$ Then $12 f (8)$ is equal to: [2023]

19
17
1
34

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(2)

$We have, f (x) + \int_{3}^{x} \frac{f (t)}{t} d t = \sqrt{x + 1}, x \geq 3$

On differentiating, we get

$f' (x) + \frac{f (x)}{x} = \frac{1}{2 \sqrt{x + 1}}$ $\Rightarrow \frac{d y}{d x} + \frac{1}{x} y = \frac{1}{2 \sqrt{x + 1}}$

$I . F . = e^{\int \frac{1}{x} d x} = e^{\ln x} = x$

Solution; $x y = \int \frac{x}{2 \sqrt{x + 1}} d x$

Put $x + 1 = t^{2};$ $d x = 2 t d t$

$x y = \int \frac{(t^{2} - 1) 2 t d t}{2 t} = \int (t^{2} - 1) d t = \frac{t^{3}}{3} - t + C$

$= \frac{{(\sqrt{x + 1})}^{3}}{3} - \sqrt{x + 1} + C$

At $x = 3, y = 2$

$∴$ $3 \times 2 = \frac{{(\sqrt{3 + 1})}^{3}}{3} - \sqrt{3 + 1} + C \Rightarrow$ $6 = \frac{8}{3} - 2 + C \Rightarrow C = \frac{16}{3}$

$∴$ $f (x) = \frac{{(\sqrt{x + 1})}^{3}}{3 x} - \frac{\sqrt{x + 1}}{x} + \frac{16}{3 x}$

Now, $12 f (8) = 12 [\frac{{(\sqrt{8 + 1})}^{3}}{3 \times 8} - \frac{\sqrt{8 + 1}}{8} + \frac{16}{3 \times 8}] = 17$

Q 39 :

Let $y = y (x)$ be the solution of the differential equation $(3 y^{2} - 5 x^{2}) y d x + 2 x (x^{2} - y^{2}) d y = 0$ such that $y (1) = 1$ . Then $| (y {(2))}^{3} - 12 y (2) |$ is equal to [2023]

$64$
$32 \sqrt{2}$
$32$
$16 \sqrt{2}$

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(2)

$Given, (3 y^{2} - 5 x^{2}) y d x + 2 x (x^{2} - y^{2}) d y = 0$

$\Rightarrow \frac{d y}{d x} = \frac{y (5 x^{2} - 3 y^{2})}{2 x (x^{2} - y^{2})}$

It is a homogeneous differential equation.

Put $y = m x \Rightarrow$ $\frac{d y}{d x} = m + x \frac{d m}{d x}$

$\Rightarrow m + x \frac{d m}{d x} = \frac{m (5 - 3 m^{2})}{2 (1 - m^{2})}$ $\Rightarrow \frac{x d m}{d x} = \frac{m (5 - 3 m^{2})}{2 (1 - m^{2})} - m$

$\Rightarrow x \frac{d m}{d x} = \frac{(5 - 3 m^{2}) m - 2 m (1 - m^{2})}{2 (1 - m^{2})}$

$\Rightarrow \frac{d x}{x} = \frac{2 (m^{2} - 1)}{m (m^{2} - 3)} d m$ $\Rightarrow \frac{d x}{x} = [\frac{2}{m} - \frac{\frac{4}{3}}{m} + \frac{\frac{4 m}{3}}{m^{2} - 3}] d m$

Integrating both sides $\int \frac{d x}{x} = \int \frac{(\frac{2}{3})}{m} d m + \int \frac{2}{3} (\frac{2 m}{m^{2} - 3}) d m$

$\Rightarrow \ln$ $| x |$ $= \frac{2}{3} \ln$ $| m |$ $+ \frac{2}{3} \ln$ $| m^{2} - 3 |$ $+ C$

$\Rightarrow \ln$ $| x |$ $= \frac{2}{3} \ln$ $| \frac{y}{x} |$ $+ \frac{2}{3} \ln$ $| {(\frac{y}{x})}^{2} - 3 |$ $+ C$

Put $(x = 1, y = 1)$ , $\ln (1) = \frac{2}{3} \ln (1) + \frac{2}{3} \ln$ $| (1 - 3) |$ $+ C$

$C = - \frac{2}{3} \ln (2) \Rightarrow \ln$ $| x |$ $= \frac{2}{3} \ln$ $| \frac{y}{x} |$ $+ \frac{2}{3} \ln$ $| {(\frac{y}{x})}^{2} - 3 |$ $- \frac{2}{3} \ln (2)$

$\Rightarrow (\frac{y}{x}) [{(\frac{y}{x})}^{2} - 3] = 2 (x^{3 / 2})$

Put $x = 2$ to get $y (2)$ , $\Rightarrow y (y^{2} - 12) = 4 \times 2 \times 2 \times 2 \sqrt{2}$

$\Rightarrow y^{3} - 12 y = 32 \sqrt{2} \Rightarrow$ $| y^{3} (2) - 12 y (2) |$ $= 32 \sqrt{2}$

Q 40 :

Let $y = y (x)$ be a solution of the differential equation $(x \cos x) d y + (x y \sin x + y \cos x - 1) d x = 0, 0 < x < \frac{π}{2} .$ If $\frac{π}{3} y (\frac{π}{3}) = \sqrt{3},$ then $| \frac{π}{6} y'' (\frac{π}{6}) + 2 y' (\frac{π}{6}) |$ is equal to ______ . [2023]

(2)

$(x \cos x) d y + (x y \sin x + y \cos x - 1) d x = 0,$

$0 < x < \frac{π}{2}$ $\Rightarrow \frac{d y}{d x} + \frac{(x \sin x + \cos x)}{x \cos x} y = \frac{1}{x \cos x}$

$I.F. = e^{\int (\frac{x \sin x + \cos x}{x \cos x}) d x} = e^{\int (\tan x + \frac{1}{x}) d x} = e^{\ln | \sec x | + \ln | x |} = e^{\ln | x \sec x |}$

$I.F. = x \sec x$

$y \times x \sec x = \int \frac{x \sec x}{x \cos x} d x \Rightarrow y \times x$ $s e c x = \tan x + C$

Given, $y (\frac{π}{3}) = \frac{3 \sqrt{3}}{π}$

$So, \frac{3 \sqrt{3}}{π} \times \frac{π}{3} \cdot 2 = \sqrt{3} + C$ $\Rightarrow 2 \sqrt{3} = \sqrt{3} + C \Rightarrow C = \sqrt{3}$

$∴$ $y x \sec x = \tan x + \sqrt{3} \Rightarrow x y = \sin x + \sqrt{3} \cos x$

$\Rightarrow x y' + y = \cos x - \sqrt{3} \sin x$

$\Rightarrow x y'' + 2 y' = - \sin x - \sqrt{3} \cos x$

$∴$ $| \frac{π}{6} \cdot y'' (\frac{π}{6}) + 2 y' (\frac{π}{6}) |$ $=$ $| - \frac{1}{2} - \frac{3}{2} |$ $= 2$

Q 41 :

If the solution curve of the differential equation $(y - 2 \log_{e} x) d x + (x \log_{e} x^{2}) d y = 0, x > 1$ passes through the points $(e, \frac{4}{3})$ and $(e^{4}, α),$ then $α$ is equal to ____ . [2023]

(3)

$Given differential equation is$ $(y - 2 \log_{e} x) d x + (x \log_{e} x^{2}) d y = 0$

$\Rightarrow \frac{d y}{d x} = \frac{2 \log_{e} x - y}{x \log_{e} x^{2}}$

$\Rightarrow \frac{d y}{d x} + (\frac{1}{x \log_{e} x^{2}}) y = \frac{2 \log_{e} x}{2 x \log_{e} x} = \frac{1}{x}$

This is a linear differential equation of the type $\frac{d y}{d x} + P y = Q$

$Here, P = \frac{1}{x \log_{e} x^{2}}, Q = \frac{1}{x}$

$I.F. = e^{\int P d x} = e^{\int \frac{1}{2 x \log_{e} x} d x} = \sqrt{\log_{e} x}$

Solution of the differential equation is

$y \sqrt{\log_{e} x} = \int \frac{1}{x} \cdot \sqrt{\log_{e} x} d x$

$\Rightarrow y \sqrt{\log_{e} x} = \frac{2}{3} {(\log_{e} x)}^{3 / 2} + c \dots (i)$

$Now, put (e, \frac{4}{3}) in (i), we get$ $\frac{4}{3} = \frac{2}{3} + c \Rightarrow c = \frac{2}{3}$

$Now, put x = e^{4}, y = α, c = \frac{2}{3} in (i), we get$

$α \sqrt{\log_{e} e^{4}} = \frac{2}{3} {(\log_{e} e^{4})}^{3 / 2} + \frac{2}{3}$

$\Rightarrow α \times 2 = \frac{2}{3} \times 8 + \frac{2}{3} \Rightarrow α = \frac{8}{3} + \frac{1}{3} = 3$

Q 42 :

Let the solution curve $x = x (y), 0 < y < \frac{π}{2}$ , of the differential equation $(\log_{e} {(\cos y))}^{2} \cos y d x - (1 + 3 x \log_{e} (\cos y)) \sin y d y = 0$ satisfy $x (\frac{π}{3}) = \frac{1}{2 \log_{e} 2} .$ If $x (\frac{π}{6}) = \frac{1}{\log_{e} m - \log_{e} n},$ where $m$ and $n$ are coprime, then $m n$ is equal to _______ . [2023]

(12)

Given differential equation is, $\cos y {(logcos y)}^{2} d x = (1 + 3 x (logcos y)) \sin y d y$

$\Rightarrow \frac{d x}{d y} = \tan y (\frac{3 x}{logcos y} + \frac{1}{{(logcos y)}^{2}})$

$\Rightarrow \frac{d x}{d y} - (\frac{3 \tan y}{logcos y}) x = \frac{\tan y}{{(logcos y)}^{2}}$

$I.F. = e^{\int - \frac{3 \tan y}{logcos y} d y} = (\log {(\cos y))}^{3}$

Solution is given by,

$x \cdot {(logcos y)}^{3} = \int \frac{\tan y}{{(logcos y)}^{2}} \times {(logcos y)}^{3} d y + C$

$= - \frac{{(logcos y)}^{2}}{2} + C$

$Given, x (\frac{π}{3}) = \frac{1}{2 \log 2}$

So, $\frac{1}{2 \log 2} (\log {(\frac{1}{2}))}^{3} = - \frac{(\log {(\frac{1}{2}))}^{2}}{2} + C \Rightarrow C = 0$

For $y = \frac{π}{6}$ , we have $x {(\log \frac{\sqrt{3}}{2})}^{3} = - \frac{1}{2} {(\log \frac{\sqrt{3}}{2})}^{2} + 0$

$\Rightarrow x = - \frac{1}{2 \log (\frac{\sqrt{3}}{2})} = \frac{1}{\log (\frac{4}{3})}$

$= \frac{1}{\log 4 - \log 3} = \frac{1}{\log_{e} m - \log_{e} n} \Rightarrow m = 4, n = 3$

$∴$ $m n = 4 \times 3 = 12$

Q 43 :

Let the tangent at any point P on a curve passing through the points (1,1) and $(\frac{1}{10}, 100),$ intersect the positive $x$ -axis and $y$ -axis at the points A and B respectively. If $P A : P B = 1 : k$ and $y = y (x)$ is the solution of the differential equation $e^{\frac{d y}{d x}} = k x + \frac{k}{2},$ $y (0) = k$ , then $4 y (1) - 5 \log_{e} 3$ is equal to ________ . [2023]

(5)

Equation of tangent at $P (x, y)$

$Y - y = \frac{d y}{d x} (X - x)$

$Y = 0 at point A, then X = \frac{- y d x}{d y} + x$

Point $P (x, y)$ divides $A B$ in $k : 1$ ratio,

$∴$ $\frac{k α + 0}{k + 1} = x \Rightarrow α = \frac{k + 1}{k} x$

$x + \frac{x}{k} = - y \frac{d x}{d y} + x$

$\Rightarrow \frac{d y}{d x} + \frac{k}{x} y = 0$

$\Rightarrow y \cdot x^{k} = C$

$\Rightarrow C = 1 [∵ Tangent passing through (1, 1)]$

From point $(\frac{1}{10}, 100),$

$100 {(\frac{1}{10})}^{k} = 1 \Rightarrow k = 2$

$\frac{d y}{d x} = \ln (2 x + 1) [Given, e^{\frac{d y}{d x}} = k x + \frac{k}{2}]$

$\Rightarrow y = \frac{(2 x + 1)}{2} (\ln (2 x + 1) - 1) + C$

Now, $2 = \frac{1}{2} (0 - 1) + C [∵ y (0) = k and k = 2]$

$C = 2 + \frac{1}{2} = \frac{5}{2}$

Now, $y (1) = \frac{3}{2} (\ln 3 - 1) + \frac{5}{2} = \frac{3}{2} \ln 3 + 1 \Rightarrow 4 y (1) = 6 \ln 3 + 4$

$∴$ $4 y (1) - 5 \ln 3 = 6 \ln 3 + 4 - 5 \ln 3 = \ln 3 + 4 \approx 5.09 \approx 5$ (approx.)

Q 44 :

If $y = y (x)$ is the solution of the differential equation $\frac{d y}{d x} + \frac{4 x}{(x^{2} - 1)} y = \frac{x + 2}{{(x^{2} - 1)}^{5 / 2}}, x > 1$ such that $y (2) = \frac{2}{9} \log_{e} (2 + \sqrt{3})$ and $y (\sqrt{2}) = α \log_{e} (\sqrt{α} + β) + β - \sqrt{γ}, α, β, γ \in ℕ,$ then $α β γ$ is equal to ______ . [2023]

(6)

$\frac{d y}{d x} + \frac{4 x}{(x^{2} - 1)} y = \frac{x + 2}{{(x^{2} - 1)}^{5 / 2}}, x > 1$

$I.F. = e^{2 \ln (x^{2} - 1)} = {(x^{2} - 1)}^{2}$

$\Rightarrow y \times {(x^{2} - 1)}^{2} = \int \frac{x + 2}{{(x^{2} - 1)}^{5 / 2}} \times {(x^{2} - 1)}^{2} d x$

$\Rightarrow$ $y \times {(x^{2} - 1)}^{2} = \int \frac{x + 2}{\sqrt{x^{2} - 1}} d x$

$\Rightarrow y \times {(x^{2} - 1)}^{2} = \int \frac{x d x}{\sqrt{x^{2} - 1}} + 2 \int \frac{d x}{\sqrt{x^{2} - 1}}$

$\Rightarrow y \times {(x^{2} - 1)}^{2} = \sqrt{x^{2} - 1} + 2 \ln$ $| x + \sqrt{x^{2} - 1} |$ $+ C$

$\Rightarrow y (2) = {(3)}^{- 3 / 2} + \frac{2 \ln | 2 + \sqrt{3} |}{9} + \frac{C}{9}$

$\Rightarrow \frac{2}{9} \log_{e} (2 + \sqrt{3}) = {(3)}^{- 3 / 2} + \frac{2}{9} \log_{e}$ $| 2 + \sqrt{3} |$ $+ \frac{C}{9}$

$\Rightarrow \frac{C}{9} = - {(3)}^{- 3 / 2} \Rightarrow C = - \sqrt{3}$

$∴$ $y = {(x^{2} - 1)}^{- \frac{3}{2}} + \frac{2 \log | x + \sqrt{x^{2} - 1} |}{{(x^{2} - 1)}^{2}} - \frac{\sqrt{3}}{{(x^{2} - 1)}^{2}}$

$Now, y (\sqrt{2}) = {(2 - 1)}^{- \frac{3}{2}} + \frac{2 \log | \sqrt{2} + \sqrt{1} |}{1} - \frac{\sqrt{3}}{1}$

$\Rightarrow α \log_{e} (\sqrt{α} + β) + β - \sqrt{γ} = 1 + \log (\sqrt{2} + 1) - \sqrt{3}$

$∴$ $α = 2, β = 1, γ = 3 \Rightarrow α β γ = 6$

Q 45 :

If $f : R \to R$ is a differentiable function such that $f' (x) + f (x) = \int_{0}^{2} f (t) d t .$ If $f (0) = e^{- 2},$ then $2 f (0) - f (2)$ is equal to _____ . [2023]

(1)

$f' (x) + f (x) = \int_{0}^{2} f (t) d t and f (0) = e^{- 2}$

$Let \int_{0}^{2} f (t) d t = K$

$\Rightarrow \frac{d y}{d x} + y = K [Where \frac{d y}{d x} = f' (x), y = f (x)]$

$\Rightarrow y e^{x} = K e^{x} + C;$ $Now f (0) = e^{- 2}$

$\Rightarrow e^{- 2} = K + C \Rightarrow C = e^{- 2} - K$ $\Rightarrow y e^{x} = K e^{x} + e^{- 2} - K$

$\Rightarrow y = K + (e^{- 2} - K) e^{- x}$

$Now, K = \int_{0}^{2} f (x) d x = \int_{0}^{2} (K + (e^{- 2} - K) e^{- x}) d x$

$= 2 K + {[(K - e^{- 2}) e^{- x}]}_{0}^{2}$ $= 2 K + (K - e^{- 2}) (e^{- 2} - 1)$

$\Rightarrow K = 2 K + K e^{- 2} - K - e^{- 4} + e^{- 2}$ $\Rightarrow K = \frac{e^{- 4} - e^{- 2}}{e^{- 2}}$

$\Rightarrow K = e^{- 2} - 1$

$Now, 2 f (0) - f (2) = 2 e^{- 2} - 2 e^{- 2} + 1 = 1$

Q 46 :

Let $f$ be a differentiable function defined on $[0, \frac{π}{2}]$ such that $f (x) > 0$ and $f (x) + \int_{0}^{x} f (t) \sqrt{1 - (\log_{e} f {(t))}^{2}} d t = e, \forall x \in [0, \frac{π}{2}] .$ Then $(6 \log_{e} f {(\frac{π}{6}))}^{2}$ is equal to _____ . [2023]

(27)

$Given, f (x) + \int_{0}^{x} f (t) \sqrt{1 - (\log_{e} f {(t))}^{2}} d t = e$ ...(i)

$\Rightarrow f (0) = e$

$Differentiating (i) both sides, we get$

$f' (x) + f (x) \sqrt{1 - (\ln f {(x))}^{2}} = 0$

$Put f (x) = y$

$\Rightarrow \frac{d y}{d x} = - y \sqrt{1 - {(\ln y)}^{2}}$ $\Rightarrow \int \frac{d y}{y \sqrt{1 - {(\ln y)}^{2}}} = - \int d x$

$Put \ln y = t$

$\Rightarrow \int \frac{d t}{\sqrt{1 - t^{2}}} = - x + c \Rightarrow \sin^{- 1} t = - x + c$

$\Rightarrow \sin^{- 1} (\ln y) = - x + c$

$\Rightarrow \sin^{- 1} (\ln f (x)) = - x + c; f (0) = e$

$\Rightarrow \frac{π}{2} = C \Rightarrow \sin^{- 1} (\ln (f (x))) = - x + \frac{π}{2}$

$\Rightarrow \sin^{- 1} (\ln f (\frac{π}{6})) = \frac{- π}{6} + \frac{π}{2} = \frac{π}{3}$

$∴$ $\ln f (\frac{π}{6}) = \sin (\frac{π}{3}) = \frac{\sqrt{3}}{2}$

$Thus,$ $(6 \log_{e} f {(\frac{π}{6}))}^{2} = {(6 \times \frac{\sqrt{3}}{2})}^{2} = 27$

Q 47 :

Let $y = y (x)$ be the solution of the differential equation $x^{4} d y + (4 x^{3} y + 2 \sin x) d x = 0, x > 0,$ $y (\frac{π}{2}) = 0 .$ Then $π^{4} y (\frac{π}{3})$ is equal to: [2026]

81
72
92
64

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(1)

$(x^{4} d y + 4 x^{3} y d x) = - 2 \sin x d x$

$\Rightarrow \int d (x^{4} y) = \int - 2 \sin x d x$

$\Rightarrow x^{4} y = 2 \cos x + c$

$\Rightarrow x^{4} f (x) = 2 \cos x + c$

$As f (\frac{π}{2}) = 0$

$So, c = 0$

${(\frac{π}{3})}^{4} f (\frac{π}{3}) = 2 \cos \frac{π}{3}$

$π^{4} f (\frac{π}{3}) = 81$

Q 48 :

Let $y = y (x)$ be the solution curve of the differential equation $(1 + x^{2}) d y + (y - \tan^{- 1} x) d x = 0, y (0) = 1 .$ Then the value of $y (1)$ is: [2026]

$\frac{4}{e^{π / 4}} - \frac{π}{2} - 1$
$\frac{2}{e^{π / 4}} + \frac{π}{4} - 1$
$\frac{4}{e^{π / 4}} + \frac{π}{2} - 1$
$\frac{2}{e^{π / 4}} - \frac{π}{4} - 1$

1

2

3

4

(2)

$\frac{d y}{d x} + \frac{y}{x^{2} + 1} = \frac{\tan^{- 1} x}{x^{2} + 1}$

$I.F. = e^{\tan^{- 1} x}$

$y \times e^{\tan^{- 1} x} = \int e^{\tan^{- 1} x} \cdot \frac{\tan^{- 1} x}{1 + x^{2}} d x$

$y \times e^{\tan^{- 1} x} = \tan^{- 1} x (e^{\tan^{- 1} x}) - e^{\tan^{- 1} x} + c$

$y (0) = 1 \Rightarrow c = 2$

$y (1) = \frac{2}{e^{π / 4}} + \frac{π}{4} - 1$

Q 49 :

Let a differentiable function $f$ satisfy the equation $\int_{0}^{36} f (\frac{t x}{36}) d t = 4 α f (x) .$ If $y = f (x)$ is a standard parabola passing through the points (2, 1) and $(- 4, β)$ , then $β^{α}$ is equal to _____. [2026]

(64)

$\int_{0}^{36} f (\frac{t x}{36}) d t = 4 α f (x), Put \frac{t x}{36} = y$

$\frac{d y}{d t} = \frac{x}{36}$

$\int_{0}^{x} \frac{f (y) 36 d y}{x} = 4 α f (x)$

$\int_{0}^{x} f (y) d y = \frac{α f (x) x}{9}$

$f (x) = \frac{α}{9} (f (x) + x f' (x))$

$(1 - \frac{α}{9}) f (x) = \frac{α x}{9} f' (x)$ $\Rightarrow (9 - α) f (x) = α x f' (x)$

$\frac{f' (x)}{f (x)} = (\frac{9}{α} - 1) \frac{1}{x}$

$\log_{e} f (x) = (\frac{9}{α} - 1) \log_{e} x + \log_{e} c$

$f (x) = c x^{(\frac{9}{α} - 1)}$ For standard parabola,

$\frac{9}{α} - 1 = 2$

$α = 3$

$f (x) = c x^{2}$

$Passing through (2, 1)$

$1 = 4 c \Rightarrow c = \frac{1}{4}$

$y = \frac{x^{2}}{4} passing through (- 4, β)$

$β = 4$

$β^{x} = 4^{3} = 64$

Q 50 :

Let y=y(x) be the solution of the differential equation $x \frac{d y}{d x} - y = x^{2} \cot x, x \in (0, π) .$ If $y (\frac{π}{2}) = \frac{π}{2},$ then $6 y (\frac{π}{6}) - 8 y (\frac{π}{4})$ is equal to: [2026]

$π$
$3 π$
$- π$
$- 3 π$

1

2

3

4

(3)

$x d y - y d x = x^{2} \cot x d x$

$x^{2} d (\frac{y}{x}) = x^{2} \cot x d x$

$d (\frac{y}{x}) = \cot x d x$

$\int d (\frac{y}{x}) = \int \cot x d x$

$\frac{y}{x} = \log_{e} (\sin x) + C$

$given y (\frac{π}{2}) = \frac{π}{2} \Rightarrow C = 1$

$y = x (\log_{e} (\sin x) + 1)$

$y (\frac{π}{6}) = \frac{π}{6} [- \log_{e} 2 + 1]$

$y (\frac{π}{4}) = \frac{π}{4} [- \frac{1}{2} \log_{e} 2 + 1]$

$6 y (\frac{π}{6}) - 8 y (\frac{π}{4})$

$= π [(- \log_{e} 2 + 1) + 2 (\frac{1}{2} \log_{e} 2 - 1)]$

$= π [1 - 2] = - π$

Topic Question Set

Q 21 :

The slope of tangent at any point (x, y) on a curve $y = y (x) is \frac{x^{2} + y^{2}}{2 x y}, x > 0 .$ If $y (2) = 0$ , then a value of $y (8)$ is [2023]

Q 22 :

Let $f$ be a differentiable function such that $x^{2} f (x) - x = 4 \int_{0}^{x} t f (t) d t, f (1) = \frac{2}{3} .$ Then $18 f (3)$ is equal to [2023]

Q 23 :

Let $y = y (x)$ be a solution curve of the differential equation $(1 - x^{2} y^{2}) d x = y d x + x d y$ . If the line $x$ = 1 intersects the curve $y = y (x)$ at $y = 2$ and the line $x = 2$ intersects the curve $y = y (x)$ at $y = α$ , then a value of $α$ is [2023]

Q 24 :

Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} + \frac{5}{x (x^{5} + 1)} y = \frac{{(x^{5} + 1)}^{2}}{x^{7}}, x > 0 .$ If $y (1) = 2$ , then $y (2)$ is equal to [2023]

Q 25 :

Let $y = y (x), y > 0$ be a solution curve of the differential equation $(1 + x^{2}) d y = y (x - y) d x .$ If $y (0) = 1$ and $y (2 \sqrt{2}) = β$ , then [2023]

Q 26 :

Let $y = y_{1} (x)$ and $y = y_{2} (x)$ be the solution curves of the differential equation $\frac{d y}{d x} = y + 7$ with initial conditions $y_{1} (0) = 0$ and $y_{2} (0) = 1$ respectively. Then the curves $y = y_{1} (x)$ and $y = y_{2} (x)$ intersect at [2023]

Q 27 :

Let $x = x (y)$ be the solution of the differential equation $2 (y + 2) \log_{e} (y + 2) d x + (x + 4 - 2 \log_{e} (y + 2)) d y = 0,$ $y > - 1$ with $x (e^{4} - 2) = 1$ .Then $x (e^{9} - 2)$ is equal to [2023]

Q 28 :

If $y = y (x)$ is the solution curve of the differential equation $\frac{d y}{d x} + y \tan x = x \sec x, 0 \leq x \leq \frac{π}{3}, y (0) = 1$ , then $y (\frac{π}{6})$ is equal to [2023]

Q 29 :

Let $α x = \exp (x^{β} y^{γ})$ be the solution of the differential equation $2 x^{2} y d y - (1 - x y^{2}) d x = 0, x > 0, y (2) = \sqrt{\log_{e} 2} .$ Then $α + β - γ$ equals [2023]

Q 30 :

$Let y = y (x) be the solution of the differential equation x^{3} d y + (x y - 1) d x = 0, x > 0, y (\frac{1}{2}) = 3 - e . 1$ $Then y (1) is equal to$ [2023]

Q 31 :

Let $y = y (x)$ be the solution of the differential equation $(x^{2} - 3 y^{2}) d x + 3 x y d y = 0, y (1) = 1 .$ Then $6 y^{2} (e)$ is equal to [2023]

Q 32 :

Let $y = y (x)$ be the solution curve of the differential equation $\frac{d y}{d x} = \frac{y}{x} (1 + x y^{2} (1 + \log_{e} x)), x > 0, y (1) = 3 .$ Then, $\frac{y^{2} (x)}{9}$ is equal to [2023]

Q 33 :

Let $y = y (t)$ be a solution of the differential equation $\frac{d y}{d t} + α y = γ e^{- β t}$ where, $α > 0, β > 0$ and $γ > 0$ . Then $\lim_{t \to \infty} y (t)$ [2023]

Q 34 :

Let $y = f (x)$ be the solution of the differential equation $y (x + 1) d x - x^{2} d y = 0, y (1) = e .$ Then $\lim_{x \to 0^{+}} f (x)$ is equal to [2023]

Q 35 :

Let $y = y (x)$ be the solution of the differential equation $x \log_{e} x \frac{d y}{d x} + y = x^{2} \log_{e} x, (x > 1) .$ If $y (2) = 2$ , then $y (e)$ is equal to [2023]

Q 36 :

Let the solution curve $y = y (x)$ of the differential equation $\frac{d y}{d x} - \frac{3 x^{5} \tan^{- 1} (x^{3})}{{(1 + x^{6})}^{3 / 2}} y = 2 x \exp {\frac{x^{3} - \tan^{- 1} x^{3}}{\sqrt{(1 + x^{6})}}}$ pass through the origin. Then $y (1)$ is equal to [2023]

Q 37 :

The solution of the differential equation $\frac{d y}{d x} = - (\frac{x^{2} + 3 y^{2}}{3 x^{2} + y^{2}}), y (1) = 0$ is [2023]

Q 38 :

Let a differentiable function $f$ satisfy $f (x) + \int_{3}^{x} \frac{f (t)}{t} d t = \sqrt{x + 1}, x \geq 3 .$ Then $12 f (8)$ is equal to: [2023]

Q 39 :

Let $y = y (x)$ be the solution of the differential equation $(3 y^{2} - 5 x^{2}) y d x + 2 x (x^{2} - y^{2}) d y = 0$ such that $y (1) = 1$ . Then $| (y {(2))}^{3} - 12 y (2) |$ is equal to [2023]

Q 40 :

Let $y = y (x)$ be a solution of the differential equation $(x \cos x) d y + (x y \sin x + y \cos x - 1) d x = 0, 0 < x < \frac{π}{2} .$ If $\frac{π}{3} y (\frac{π}{3}) = \sqrt{3},$ then $| \frac{π}{6} y'' (\frac{π}{6}) + 2 y' (\frac{π}{6}) |$ is equal to ______ . [2023]

Q 41 :

If the solution curve of the differential equation $(y - 2 \log_{e} x) d x + (x \log_{e} x^{2}) d y = 0, x > 1$ passes through the points $(e, \frac{4}{3})$ and $(e^{4}, α),$ then $α$ is equal to ____ . [2023]

Q 45 :

If $f : R \to R$ is a differentiable function such that $f' (x) + f (x) = \int_{0}^{2} f (t) d t .$ If $f (0) = e^{- 2},$ then $2 f (0) - f (2)$ is equal to _____ . [2023]

Q 46 :

Let $f$ be a differentiable function defined on $[0, \frac{π}{2}]$ such that $f (x) > 0$ and $f (x) + \int_{0}^{x} f (t) \sqrt{1 - (\log_{e} f {(t))}^{2}} d t = e, \forall x \in [0, \frac{π}{2}] .$ Then $(6 \log_{e} f {(\frac{π}{6}))}^{2}$ is equal to _____ . [2023]

Q 47 :

Let $y = y (x)$ be the solution of the differential equation $x^{4} d y + (4 x^{3} y + 2 \sin x) d x = 0, x > 0,$ $y (\frac{π}{2}) = 0 .$ Then $π^{4} y (\frac{π}{3})$ is equal to: [2026]

Q 48 :

Let $y = y (x)$ be the solution curve of the differential equation $(1 + x^{2}) d y + (y - \tan^{- 1} x) d x = 0, y (0) = 1 .$ Then the value of $y (1)$ is: [2026]

Q 49 :

Let a differentiable function $f$ satisfy the equation $\int_{0}^{36} f (\frac{t x}{36}) d t = 4 α f (x) .$ If $y = f (x)$ is a standard parabola passing through the points (2, 1) and $(- 4, β)$ , then $β^{α}$ is equal to _____. [2026]

Q 50 :

Let y=y(x) be the solution of the differential equation $x \frac{d y}{d x} - y = x^{2} \cot x, x \in (0, π) .$ If $y (\frac{π}{2}) = \frac{π}{2},$ then $6 y (\frac{π}{6}) - 8 y (\frac{π}{4})$ is equal to: [2026]

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Topic Question Set

Q 21 : The slope of tangent at any point (x, y) on a curve y=y(x) is x2+y22xy,x>0. If y(2)=0, then a value of y(8) is [2023]

Q 22 : Let f be a differentiable function such that x2f(x)-x=4∫0xtf(t) dt, f(1)=23. Then 18f(3) is equal to [2023]

Q 23 : Let y=y(x) be a solution curve of the differential equation (1-x2y2)dx=ydx+xdy. If the line x = 1 intersects the curve y=y(x) at y=2 and the line x=2 intersects the curve y=y(x) at y=α, then a value of α is [2023]

Q 24 : Let y=y(x) be the solution of the differential equation dydx+5x(x5+1)y=(x5+1)2x7, x>0. If y(1)=2, then y(2) is equal to [2023]

Q 25 : Let y=y(x), y>0 be a solution curve of the differential equation (1+x2) dy=y(x-y)dx. If y(0)=1 and y(22)=β, then [2023]

Q 26 : Let y=y1(x) and y=y2(x) be the solution curves of the differential equation dydx=y+7 with initial conditions y1(0)=0 and y2(0)=1 respectively. Then the curves y=y1(x) and y=y2(x) intersect at [2023]

Q 27 : Let x=x(y) be the solution of the differential equation 2(y+2)loge(y+2) dx+(x+4-2loge(y+2)) dy=0, y>-1 with x(e4-2)=1.Then x(e9-2) is equal to [2023]

Q 28 : If y=y(x) is the solution curve of the differential equation dydx+ytanx=xsecx, 0≤x≤π3, y(0)=1, then y(π6) is equal to [2023]

Q 29 : Let αx=exp(xβyγ) be the solution of the differential equation 2x2y dy-(1-xy2)dx=0, x>0, y(2)=loge2. Then α+β-γ equals [2023]

Q 30 : Let y=y(x) be the solution of the differential equation x3 dy+(xy-1)dx=0, x>0, y(12)=3-e.1 Then y(1) is equal to [2023]

Q 31 : Let y=y(x) be the solution of the differential equation (x2-3y2) dx+3xy dy=0, y(1)=1. Then 6y2(e) is equal to [2023]

Q 32 : Let y=y(x) be the solution curve of the differential equation dydx=yx(1+xy2(1+logex)),x>0, y(1)=3. Then, y2(x)9 is equal to [2023]

Q 33 : Let y=y(t) be a solution of the differential equation dydt+αy=γe-βt where, α>0,β>0 and γ>0. Then limt→∞y(t) [2023]

Q 34 : Let y=f(x) be the solution of the differential equation y(x+1) dx-x2 dy=0,y(1)=e. Then limx→0+f(x) is equal to [2023]

Q 35 : Let y=y(x) be the solution of the differential equation xlogex dydx+y=x2logex,(x>1). If y(2)=2, then y(e) is equal to [2023]

Q 36 : Let the solution curve y=y(x) of the differential equation dydx-3x5tan-1(x3)(1+x6)3/2y=2xexp{x3-tan-1x3(1+x6)} pass through the origin. Then y(1) is equal to [2023]

Q 37 : The solution of the differential equation dydx=-(x2+3y23x2+y2), y(1)=0 is [2023]

Q 38 : Let a differentiable function f satisfy f(x)+∫3xf(t)tdt=x+1, x≥3. Then 12f(8) is equal to: [2023]

Q 39 : Let y=y(x) be the solution of the differential equation (3y2-5x2)ydx+2x(x2-y2)dy=0 such that y(1)=1. Then |(y(2))3-12y(2)| is equal to [2023]

Q 40 : Let y=y(x) be a solution of the differential equation (xcosx)dy+(xysinx+ycosx-1)dx=0, 0<x<π2. If π3y(π3)=3, then |π6 y''(π6)+2 y'(π6)| is equal to ______ . [2023]

Q 41 : If the solution curve of the differential equation (y-2logex)dx+(xlogex2)dy=0, x>1 passes through the points (e,43) and (e4, α), then α is equal to ____ . [2023]

Q 42 : Let the solution curve x=x(y),0<y<π2, of the differential equation (loge(cosy))2cosydx-(1+3xloge(cosy))siny dy=0 satisfy x(π3)=12loge2. If x(π6)=1logem-logen, where m and n are coprime, then mn is equal to _______ . [2023]

Q 44 : If y=y(x) is the solution of the differential equation dydx+4x(x2-1)y=x+2 (x2-1)5/2, x>1 such that y(2)=29loge(2+3) and y(2)=αloge(α+β)+β-γ, α,β,γ∈ℕ, then αβγ is equal to ______ . [2023]

Q 45 : If f:R→R is a differentiable function such that f'(x)+f(x)=∫02f(t) dt. If f(0)=e-2, then 2f(0)-f(2) is equal to _____ . [2023]

Q 46 : Let f be a differentiable function defined on [0,π2] such that f(x)>0 and f(x)+∫0xf(t)1-(logef(t))2 dt=e,∀x∈[0,π2]. Then (6logef(π6))2 is equal to _____ . [2023]

Q 47 : Let y=y(x) be the solution of the differential equation x4 dy+(4x3y+2sinx)dx=0, x>0, y(π2)=0. Then π4y(π3) is equal to: [2026]

Q 48 : Let y=y(x) be the solution curve of the differential equation (1+x2)dy+(y-tan-1x)dx=0, y(0)=1. Then the value of y(1) is: [2026]

Q 49 : Let a differentiable function f satisfy the equation ∫036f(tx36)dt=4αf(x). If y=f(x) is a standard parabola passing through the points (2, 1) and (-4,β), then βα is equal to _____. [2026]

Q 50 : Let y=y(x) be the solution of the differential equation xdydx-y=x2cotx, x∈(0,π). If y(π2)=π2, then 6y(π6)-8y(π4) is equal to: [2026]

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Q 21 :

The slope of tangent at any point (x, y) on a curve $y = y (x) is \frac{x^{2} + y^{2}}{2 x y}, x > 0 .$ If $y (2) = 0$ , then a value of $y (8)$ is [2023]

Q 22 :

Let $f$ be a differentiable function such that $x^{2} f (x) - x = 4 \int_{0}^{x} t f (t) d t, f (1) = \frac{2}{3} .$ Then $18 f (3)$ is equal to [2023]

Q 23 :

Let $y = y (x)$ be a solution curve of the differential equation $(1 - x^{2} y^{2}) d x = y d x + x d y$ . If the line $x$ = 1 intersects the curve $y = y (x)$ at $y = 2$ and the line $x = 2$ intersects the curve $y = y (x)$ at $y = α$ , then a value of $α$ is [2023]

Q 24 :

Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} + \frac{5}{x (x^{5} + 1)} y = \frac{{(x^{5} + 1)}^{2}}{x^{7}}, x > 0 .$ If $y (1) = 2$ , then $y (2)$ is equal to [2023]

Q 25 :

Let $y = y (x), y > 0$ be a solution curve of the differential equation $(1 + x^{2}) d y = y (x - y) d x .$ If $y (0) = 1$ and $y (2 \sqrt{2}) = β$ , then [2023]

Q 26 :

Let $y = y_{1} (x)$ and $y = y_{2} (x)$ be the solution curves of the differential equation $\frac{d y}{d x} = y + 7$ with initial conditions $y_{1} (0) = 0$ and $y_{2} (0) = 1$ respectively. Then the curves $y = y_{1} (x)$ and $y = y_{2} (x)$ intersect at [2023]

Q 27 :

Let $x = x (y)$ be the solution of the differential equation $2 (y + 2) \log_{e} (y + 2) d x + (x + 4 - 2 \log_{e} (y + 2)) d y = 0,$ $y > - 1$ with $x (e^{4} - 2) = 1$ .Then $x (e^{9} - 2)$ is equal to [2023]

Q 28 :

If $y = y (x)$ is the solution curve of the differential equation $\frac{d y}{d x} + y \tan x = x \sec x, 0 \leq x \leq \frac{π}{3}, y (0) = 1$ , then $y (\frac{π}{6})$ is equal to [2023]

Q 29 :

Let $α x = \exp (x^{β} y^{γ})$ be the solution of the differential equation $2 x^{2} y d y - (1 - x y^{2}) d x = 0, x > 0, y (2) = \sqrt{\log_{e} 2} .$ Then $α + β - γ$ equals [2023]

Q 30 :

$Let y = y (x) be the solution of the differential equation x^{3} d y + (x y - 1) d x = 0, x > 0, y (\frac{1}{2}) = 3 - e . 1$ $Then y (1) is equal to$ [2023]

Q 31 :

Let $y = y (x)$ be the solution of the differential equation $(x^{2} - 3 y^{2}) d x + 3 x y d y = 0, y (1) = 1 .$ Then $6 y^{2} (e)$ is equal to [2023]

Q 32 :

Let $y = y (x)$ be the solution curve of the differential equation $\frac{d y}{d x} = \frac{y}{x} (1 + x y^{2} (1 + \log_{e} x)), x > 0, y (1) = 3 .$ Then, $\frac{y^{2} (x)}{9}$ is equal to [2023]

Q 33 :

Let $y = y (t)$ be a solution of the differential equation $\frac{d y}{d t} + α y = γ e^{- β t}$ where, $α > 0, β > 0$ and $γ > 0$ . Then $\lim_{t \to \infty} y (t)$ [2023]

Q 34 :

Let $y = f (x)$ be the solution of the differential equation $y (x + 1) d x - x^{2} d y = 0, y (1) = e .$ Then $\lim_{x \to 0^{+}} f (x)$ is equal to [2023]

Q 35 :

Let $y = y (x)$ be the solution of the differential equation $x \log_{e} x \frac{d y}{d x} + y = x^{2} \log_{e} x, (x > 1) .$ If $y (2) = 2$ , then $y (e)$ is equal to [2023]

Q 36 :

Let the solution curve $y = y (x)$ of the differential equation $\frac{d y}{d x} - \frac{3 x^{5} \tan^{- 1} (x^{3})}{{(1 + x^{6})}^{3 / 2}} y = 2 x \exp {\frac{x^{3} - \tan^{- 1} x^{3}}{\sqrt{(1 + x^{6})}}}$ pass through the origin. Then $y (1)$ is equal to [2023]

Q 37 :

The solution of the differential equation $\frac{d y}{d x} = - (\frac{x^{2} + 3 y^{2}}{3 x^{2} + y^{2}}), y (1) = 0$ is [2023]

Q 38 :

Let a differentiable function $f$ satisfy $f (x) + \int_{3}^{x} \frac{f (t)}{t} d t = \sqrt{x + 1}, x \geq 3 .$ Then $12 f (8)$ is equal to: [2023]

Q 39 :

Let $y = y (x)$ be the solution of the differential equation $(3 y^{2} - 5 x^{2}) y d x + 2 x (x^{2} - y^{2}) d y = 0$ such that $y (1) = 1$ . Then $| (y {(2))}^{3} - 12 y (2) |$ is equal to [2023]

Q 40 :

Let $y = y (x)$ be a solution of the differential equation $(x \cos x) d y + (x y \sin x + y \cos x - 1) d x = 0, 0 < x < \frac{π}{2} .$ If $\frac{π}{3} y (\frac{π}{3}) = \sqrt{3},$ then $| \frac{π}{6} y'' (\frac{π}{6}) + 2 y' (\frac{π}{6}) |$ is equal to ______ . [2023]

Q 41 :

If the solution curve of the differential equation $(y - 2 \log_{e} x) d x + (x \log_{e} x^{2}) d y = 0, x > 1$ passes through the points $(e, \frac{4}{3})$ and $(e^{4}, α),$ then $α$ is equal to ____ . [2023]

Q 45 :

If $f : R \to R$ is a differentiable function such that $f' (x) + f (x) = \int_{0}^{2} f (t) d t .$ If $f (0) = e^{- 2},$ then $2 f (0) - f (2)$ is equal to _____ . [2023]

Q 46 :

Let $f$ be a differentiable function defined on $[0, \frac{π}{2}]$ such that $f (x) > 0$ and $f (x) + \int_{0}^{x} f (t) \sqrt{1 - (\log_{e} f {(t))}^{2}} d t = e, \forall x \in [0, \frac{π}{2}] .$ Then $(6 \log_{e} f {(\frac{π}{6}))}^{2}$ is equal to _____ . [2023]

Q 47 :

Let $y = y (x)$ be the solution of the differential equation $x^{4} d y + (4 x^{3} y + 2 \sin x) d x = 0, x > 0,$ $y (\frac{π}{2}) = 0 .$ Then $π^{4} y (\frac{π}{3})$ is equal to: [2026]

Q 48 :

Let $y = y (x)$ be the solution curve of the differential equation $(1 + x^{2}) d y + (y - \tan^{- 1} x) d x = 0, y (0) = 1 .$ Then the value of $y (1)$ is: [2026]

Q 49 :

Let a differentiable function $f$ satisfy the equation $\int_{0}^{36} f (\frac{t x}{36}) d t = 4 α f (x) .$ If $y = f (x)$ is a standard parabola passing through the points (2, 1) and $(- 4, β)$ , then $β^{α}$ is equal to _____. [2026]

Q 50 :

Let y=y(x) be the solution of the differential equation $x \frac{d y}{d x} - y = x^{2} \cot x, x \in (0, π) .$ If $y (\frac{π}{2}) = \frac{π}{2},$ then $6 y (\frac{π}{6}) - 8 y (\frac{π}{4})$ is equal to: [2026]