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

Q 1 :

If the Solution y = y(x) of the differential equation $(x^{4} + 2 x^{3} + 3 x^{2} + 2 x + 2) d y - (2 x^{2} + 2 x + 3) d x = 0$ satisfies $y (- 1) = - \frac{π}{4}$ , then y(0) is equal [2024]

$\frac{π}{4}$
$\frac{π}{2}$
$- \frac{π}{12}$
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(1)

$(x^{4} + 2 x^{3} + 3 x^{2} + 2 x + 2) d y - (2 x^{2} + 2 x + 3) d x = 0$

$\frac{d y}{d x} = \frac{2 x^{2} + 2 x + 3}{x^{4} + 2 x^{3} + 3 x^{2} + 2 x + 2}$

$\Rightarrow \int d y = \int (\frac{2 x^{2} + 2 x + 3}{x^{4} + 2 x^{3} + 3 x^{2} + 2 x + 2}) d x$

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

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

$\Rightarrow y = \tan^{- 1} (x) + \tan^{- 1} (1 + x) + C$ ... (i)

Now, $y (- 1) = \frac{- π}{4}$

$∴$ From (i), we get $\frac{- π}{4} = \tan^{- 1} (- 1) + \tan^{- 1} (1 - 1) + C$

$\Rightarrow \frac{- π}{4} = - \frac{π}{4} + C \Rightarrow C = 0$

So, $y (x) = \tan^{- 1} (x) + \tan^{- 1} (1 + x)$

Now, $y (0) = \tan^{- 1} (0) + \tan^{- 1} (1 + 0) = \frac{π}{4}$ .

Q 2 :

Let y = y(x) be the solution of the differential equation ${(x^{2} + 4)}^{2} d y + (2 x^{3} y + 8 x y - 2) d x = 0$ . If y(0) = 0, then y(2) is equal to [2024]

$2 π$
$\frac{π}{16}$
$\frac{π}{8}$
$\frac{π}{32}$

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(4)

We Have, ${(x^{2} + 4)}^{2} d y + (2 x^{3} y + 8 x y - 2) d x = 0$

$\Rightarrow \frac{d y}{d x} + \frac{y (2 x^{3} + 8 x)}{{(x^{2} + 4)}^{2}} = \frac{2}{{(x^{2} + 4)}^{2}}$

$\Rightarrow \frac{d y}{d x} + \frac{2 x y}{x^{2} + 4} = \frac{2}{{(x^{2} + 4)}^{2}}$

$I F = e^{\int \frac{2 x}{x^{2} + 4} d x} = e^{\log_{e} (x^{2} + 4)} = x^{2} + 4$

$∴$ Solution is $y (x^{2} + 4) = \int \frac{2}{{(x^{2} + 4)}^{2}} \times (x^{2} + 4) d x + C$

$\Rightarrow y (x^{2} + 4) = \int \frac{2}{x^{2} + 4} d x + C$

$\Rightarrow y (x^{2} + 4) = 2 \times \frac{1}{2} \tan^{- 1} (\frac{x}{2}) + C$

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

Now. We have y(0) = 0

$\Rightarrow 0 \cdot (0 + 4) = \tan^{- 1} (0) + C \Rightarrow C = 0$

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

So, at x = 2

$y (4 + 4) = \tan^{- 1} (\frac{2}{2}) \Rightarrow y = \frac{1}{8} \times \tan^{- 1} 1 = \frac{1}{8} \times \frac{π}{4}$

$\Rightarrow y = \frac{π}{32}$

Q 3 :

If y = y(x) is the solution of the differential equation $\frac{d y}{d x} + 2 y = \sin (2 x)$ , $y (0) = \frac{3}{4}$ then $y (\frac{π}{8})$ is equal to [2024]

$e^{\frac{- π}{4}}$
$e^{\frac{π}{4}}$
$e^{\frac{π}{8}}$
$e^{\frac{- π}{8}}$

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(1)

$\frac{d y}{d x} + 2 y = \sin 2 x$ , $y (0) = \frac{3}{4}$

$IF = e^{\int 2 d x} = e^{2 x}$

$∴$ General solution of given differential equation is,

$y \times e^{2 x} = \int e^{2 x} \sin 2 x d x$

$\Rightarrow y e^{2 x} = \frac{e^{2 x}}{8} (2 \sin 2 x - 2 \cos 2 x) + c [∵ \int e^{a x} \sin b x d x = \frac{e^{a x}}{a^{2} + b^{2}} (a \sin b x - b \cos b x) + c]$

$\Rightarrow y e^{2 x} = \frac{e^{2 x}}{4} (\sin 2 x - \cos 2 x) + c$

Now, $y (0) = \frac{3}{4}$

$\Rightarrow \frac{3}{4} = \frac{1}{4} (- 1) + c \Rightarrow c = \frac{3}{4} + \frac{1}{4} = 1$

$∴ y (\frac{π}{8}) = \frac{1}{4} (\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}) + \frac{1}{e^{π / 4}} = e^{- π / 4}$

Q 4 :

Let y = y(x) be the solution of the differential equation $(1 + x^{2}) \frac{d y}{d x} + y = e^{\tan^{- 1} x}$ , y(1) = 0. Then y(0) is [2024]

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

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(2)

We Have, $(1 + x^{2}) \frac{d y}{d x} + y = e^{\tan^{- 1} x}$

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

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

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

Let $\tan^{- 1} x = t$

$\Rightarrow \frac{1}{1 + x^{2}} d x = d t \Rightarrow y e^{t} = \int e^{2 t} d t + c$

$\Rightarrow y e^{t} = \frac{e^{2 t}}{2} + c \Rightarrow y e^{\tan^{- 1} x} = \frac{e^{2 \tan^{- 1} x}}{2} + c$

Since y(1) = 0

$\Rightarrow 0 e^{π / 4} = \frac{e^{π / 2}}{2} + c \Rightarrow c = \frac{- e^{π / 2}}{2}$

$∴ y = \frac{e^{\tan^{- 1} x}}{2} - \frac{e^{π / 2}}{2 e^{\tan^{- 1} x}}$

$\Rightarrow y (0) = \frac{1}{2} - \frac{e^{π / 2}}{2} = \frac{1 - e^{π / 2}}{2}$

Q 5 :

Let y = y(x) be the solution of the differential equation $(2 x \log_{e} x) \frac{d y}{d x} + 2 y = \frac{3}{x} \log_{e} x$ , x > 0 and $y (e^{- 1}) = 0$ . Then y(e) is equal to [2024]

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

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(3)

We Have, $\frac{d y}{d x} + \frac{2 y}{2 x \log_{e} x} = \frac{3 \log_{e} x}{x (2 x \log_{e} x)}$

$\Rightarrow \frac{d y}{d x} + \frac{y}{x \log_{e} x} = \frac{3}{2 x^{2}}$

$IF = e^{\int \frac{1}{x \log_{e} x} d x} = e^{\log (\log_{e} x)} = \log_{e} x$

$\Rightarrow y \log_{e} x = \int \frac{3}{2 x^{2}} \log_{e} x d x$

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

$= \frac{3}{2} \log_{e} x (\frac{- 1}{x}) + \int \frac{3}{2 x^{2}} d x + c = \frac{- 3 \log_{e} x}{2 x} + (\frac{- 3}{2 x}) + c$

Now, $y (e^{- 1}) = 0$

$\Rightarrow 0 = \frac{3 e}{2} - \frac{3 e}{2} + c \Rightarrow c = 0$

$∴ y \log_{e} x = \frac{- 3 \log_{e} x}{2 x} - \frac{3}{2 x} \Rightarrow y = \frac{- 3}{2 x} - \frac{3}{2 x \log_{e} x}$

$\Rightarrow y (e) = \frac{- 3}{2 e} - \frac{3}{2 e} = \frac{- 3}{e}$

Q 6 :

Suppose the solution of the differential equation $\frac{d y}{d x} = \frac{(2 + α) x - β y + 2}{β x - 2 α y - (β γ - 4 α)}$ represents a circle passing through origin. Then the radius of this circle is: [2024]

2
$\frac{1}{2}$
$\sqrt{17}$
$\frac{\sqrt{17}}{2}$

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(4)

We have, $\frac{d y}{d x} = \frac{(2 + α) x - β y + 2}{β x - 2 α y - (β γ - 4 α)}$

$\Rightarrow β x d y - 2 α y d y - (β γ - 4 α) d y = 2 x d x + α x d x - β y d x + 2 d x$

On integrating, we get

$\int β x d y + β y d x - α y^{2} - (β γ - 4 α) y = x^{2} + \frac{α x^{2}}{2} + 2 x + c$

$\Rightarrow β x y - α y^{2} - (β γ - 4 α) y = x^{2} (1 + \frac{α}{2}) + 2 x + c$

$\Rightarrow (1 + \frac{α}{2}) x^{2} + α y^{2} - β x y + 2 x + (β γ - 4 α) y + c = 0$

Since, it represents a circle which is passing through origin, then

$1 + \frac{α}{2} = α, β = 0 and c = 0 [\begin{matrix} ∵ coeff . of x^{2} = coeff . of y^{2} \\ coeff . of x y = 0 \end{matrix}]$

$\Rightarrow α = 2$

$∴$ Equation of circle is given by

$2 x^{2} + 2 y^{2} + 2 x - 8 y = 0 \Rightarrow x^{2} + y^{2} + x - 4 y = 0$

$\Rightarrow Centre \equiv (- \frac{1}{2}, 2)$

$\Rightarrow Radius = \sqrt{{(\frac{1}{2})}^{2} + {(2)}^{2} - 0} = \frac{\sqrt{17}}{2}$

Q 7 :

Let y = y(x) be the solution of the differential equation $(1 + y^{2}) e^{\tan x} d x + \cos^{2} x (1 + e^{2 \tan x}) d y = 0$ , y(0) = 1. Then $y (\frac{π}{4})$ is equal to [2024]

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

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(4)

We Have, $(1 + y^{2}) e^{\tan x} d x + \cos^{2} x (1 + e^{2 \tan x}) d y = 0$

$\Rightarrow \frac{e^{\tan x}}{\cos^{2} x (1 + e^{2 \tan x})} d x + \frac{1}{1 + y^{2}} d y = 0$

Integrating both sides, we get

$\int \frac{\sec^{2} x e^{\tan x}}{1 + e^{2 \tan x}} d x + \int \frac{d y}{1 + y^{2}} = C$

Put $e^{\tan x} = t \Rightarrow \sec^{2} x e^{\tan x} d x = d t$

$\Rightarrow \tan^{- 1} (t) + \tan^{- 1} y = C \Rightarrow \tan^{- 1} (e^{\tan x}) + \tan^{- 1} y = C$

Since, y(0) = 1

$\Rightarrow \frac{π}{4} + \frac{π}{4} = C \Rightarrow C = \frac{π}{2}$

So, $\tan^{- 1} (e^{\tan x}) + \tan^{- 1} y = \frac{π}{2}$

For $x = \frac{π}{4}$ , we have

$\tan^{- 1} y = \frac{π}{2} - \tan^{- 1} e$

$\Rightarrow \tan^{- 1} y = \cot^{- 1} e \Rightarrow \tan^{- 1} y = \tan^{- 1} \frac{1}{e} \Rightarrow y = \frac{1}{e}$

Q 8 :

Let y = y(x) be the solution curve of the differential equation $\sec y \frac{d y}{d x} + 2 x \sin y = x^{3} \cos y$ , y(1) = 0. Then $y (\sqrt{3})$ is equal to: [2024]

$\frac{π}{6}$
$\frac{π}{3}$
$\frac{π}{4}$
$\frac{π}{12}$

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(3)

We have, $\sec y \frac{d y}{d x} + 2 x \sin y = x^{3} \cos y$

$\Rightarrow \sec^{2} y \frac{d y}{d x} + 2 x \tan y = x^{3}$ ... (i)

Let $z = \tan y \Rightarrow \frac{d z}{d x} = \sec^{2} y \frac{d y}{d x}$

$∴ From (i), \frac{d z}{d x} + 2 x z = x^{3}$

$IF = e^{\int 2 x d x} = e^{x^{2}} \Rightarrow z \cdot e^{x^{2}} = \int e^{x^{2}} \cdot x^{3} d x + c$

$\Rightarrow \tan y \cdot e^{x^{2}} = \frac{1}{2} (x^{2} e^{x^{2}} - e^{x^{2}}) + c$

Since, y(1) = 0

$∴ \tan (0) \cdot e = \frac{1}{2} (1 \cdot e - e) + c \Rightarrow c = 0$

So, $\tan y e^{x^{2}} = \frac{1}{2} (x^{2} e^{x^{2}} - e^{x^{2}})$

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

$∴ y (\sqrt{3}) = \tan^{- 1} (\frac{{(\sqrt{3})}^{2} - 1}{2}) = \tan^{- 1} (1) = \frac{π}{4}$

Q 9 :

The Solution curve of the differential equation $2 y = \frac{d y}{d x} + 3 = 5 \frac{d y}{d x}$ , passing through the point (0, 1) is a conic, whose vertex lies on the line: [2024]

2x + 3y = –6
2x + 3y = 6
2x + 3y = –9
2x + 3y = 9

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(4)

We have, $2 y \frac{d y}{d x} + 3 = 5 \frac{d y}{d x}$

$\Rightarrow$ 2ydy + 3dx = 5dy

Integrating both sides, we get

$y^{2} + 3 x = 5 y + c$

Now, at point (0, 1) i.e., at x = 0, y = 1, we have

1 + 0 = 5 + c $\Rightarrow$ c = –4

So, $y^{2} - 5 y = - 3 x - 4$

$\Rightarrow y^{2} - 5 y + \frac{25}{4} - \frac{25}{4} = - 3 x - 4$

$\Rightarrow {(y - \frac{5}{2})}^{2} = - 3 x + \frac{9}{4} \Rightarrow {(y - \frac{5}{2})}^{2} = - 3 (x - \frac{3}{4})$

Vertex = $(\frac{3}{4}, \frac{5}{2})$ , Which satisfies by 2x + 3y = 9.

Q 10 :

The Solution of the differential equation $(x^{2} + y^{2}) d x - 5 x y d y = 0$ , y(1) = 0, is: [2024]

$| x^{2} - 2 y^{2} |^{5} = x^{2}$
$| x^{2} - 2 y^{2} |^{6} = x$
$| x^{2} - 4 y^{2} |^{5} = x^{2}$
$| \begin{matrix} x^{2} - 4 y^{2} \end{matrix} |^{6} = x$

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(3)

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

$\frac{d y}{d x} = \frac{x^{2} + y^{2}}{5 x y}$

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

$∴ v + x \frac{d y}{d x} = \frac{1 + v^{2}}{5 v} \Rightarrow x \frac{d v}{d x} = \frac{1 + v^{2}}{5 v} - v$

$\Rightarrow \frac{5 v d v}{1 - 4 v^{2}} = \frac{d x}{x} \Rightarrow \frac{1}{8} \int \frac{8 \times 5 v d v}{1 - 4 v^{2}} = \int \frac{d x}{x}$

$\Rightarrow \frac{- 5}{8} \ln | 1 - 4 v^{2} | = \ln | x | + \ln c$

$\Rightarrow \frac{5}{8} \ln \frac{| x^{2} - 4 y^{2} |}{x^{2}} + \ln | x | + \ln c = 0$

$\Rightarrow | \frac{x^{2} - 4 y^{2}}{x^{2}} |^{5 / 8} | x | = k$

$\Rightarrow \frac{| x^{2} - 4 y^{2} |^{5 / 8}}{| x |^{\frac{1}{4}}} = k ∵ y (1) = 0 \Rightarrow k = 1$

$\Rightarrow | \begin{matrix} x^{2} - 4 y^{2} \end{matrix} |^{5 / 8} = | x |^{\frac{1}{4}} \Rightarrow | x^{2} - 4 y^{2} |^{5} = x^{2}$

Q 11 :

Let $\int_{0}^{x} \sqrt{1 - (y' {(t))}^{2}} d t = \int_{0}^{x} y (t) d t$ , $0 \leq x \leq 3, y \geq 0$ , y(0) = 0. Then at x = 2, y" + y + 1 is equal to [2024]

2
$\frac{1}{2}$
$\sqrt{2}$
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(4)

We have $\int_{0}^{x} \sqrt{1 - (y' {(t))}^{2}} d t = \int_{0}^{x} y (t) d t$

On differentiating both sides, we get

$\sqrt{1 - (y' {(x))}^{2}} = y (x)$

$\Rightarrow 1 - {(y')}^{2} = y^{2} \Rightarrow y^{2} + {(y')}^{2} = 1$

$\Rightarrow 2 y y' + 2 y " = 0 \Rightarrow y y' + y " = 0$

So, at x = 2, y" + y + 1 = 0 + 1 = 1.

Q 12 :

Let y = y(x) be the solution of the differential equation $\frac{d y}{d x} = 2 x {(x + y)}^{3} - x (x + y) - 1, y (0) = 1$ . Then, $(\frac{1}{\sqrt{2}} + y {(\frac{1}{\sqrt{2}}))}^{2}$ equals: [2024]

$\frac{1}{2 - \sqrt{e}}$
$\frac{4}{4 + \sqrt{e}}$
$\frac{2}{1 + \sqrt{e}}$
$\frac{3}{1 - \sqrt{e}}$

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(1)

We have, $\frac{d y}{d x} = 2 x {(x + y)}^{3} - x (x + y) - 1, y (0) = 1$

Let $x + y = t \Rightarrow 1 + \frac{d y}{d x} = \frac{d t}{d x}$

Now, $\frac{d t}{d x} - 1 = 2 x t^{3} - x t - 1$

$\Rightarrow \frac{d t}{d x} = 2 x t^{3} - x t \Rightarrow \frac{1}{t^{3}} \frac{d t}{d x} + \frac{x}{t^{2}} - 2 x = 0$

Put $\frac{1}{t^{2}} = u \Rightarrow \frac{- 2}{t^{3}} \frac{d t}{d x} = \frac{d u}{d x}$ we get

$\frac{- 1}{2} \frac{d u}{d x} + x u = 2 x \Rightarrow \frac{d u}{d x} - 2 x u = - 4 x$

$∴ IF = e^{\int - 2 x d x = e^{- x^{2}}}$

Required solution = $u \cdot e^{- x^{2}} = \int e^{- x^{2}} \cdot (- 4 x) d x$

$\Rightarrow \frac{e^{- x^{2}}}{t^{2}} = \int e^{- x^{2}} \cdot (- 4 x) d x$

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

At x = 0, y = 1, we get $1 = 2 + C \Rightarrow C = - 1$

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

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

$∴ (y {(\frac{1}{\sqrt{2}}) + \frac{1}{\sqrt{2}})}^{2} = (\frac{1}{2 - \sqrt{e}})$ .

Q 13 :

Let $α$ be a non-zero real number. Suppose $f : R \to R$ is a differentiable function such that f(0) = 2 and $\lim_{x \to - \infty} f (x) = 1$ . If f'(x) = $α$ f(x) + 3, for all $x \in R$ , then $f (- \log_{e} 2)$ is equal to ________. [2024]

7
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3
5

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*

We have, $f' (x) = α f (x) + 3 \Rightarrow f' (x) - α f (x) = 3$

Now, Let $α > 0, IF = e^{- α \int d x} = e^{- α x}$

$∴$ Solution is given by, $e^{- α x} f (x) = 3 \int e^{- α x} d x + C$

$\Rightarrow f (x) = \frac{- 3}{α} + C e^{α x}$

Now, let $α > 0$ , we have $\lim_{x \to - \infty} f (x) = 1$

$\Rightarrow \frac{- 3}{α} + C \lim_{x \to - \infty} e^{α x} = 1$

Let $α < 0, \lim_{x \to \infty} f (x) = - \frac{3}{α} + C \lim_{x \to - \infty} e^{α x} = 1$

$\Rightarrow \frac{- 3}{α} + \infty = 1$ which is not possible

$∴$ Value of $α$ does not exist.

Q 14 :

Let x = x(t) and y = y(t) be solutions of the differential equations $\frac{d x}{d t} + a x = 0$ and $\frac{d y}{d t} + b y = 0$ respectively, $a, b \in R$ . Given that x(0) = 2; y(0)= 1 and 3y(1) = 2x(1), the value of t, for which x(t) = y(t), is: [2024]

$\log_{\frac{4}{3}} 2$
$\log_{\frac{2}{3}} 2$
$\log_{4} 3$
$\log_{3} 4$

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(1)

We have, $\frac{d x}{d t} + a x = 0$

$\int \frac{d x}{a x} = - \int d t \Rightarrow \frac{1}{a} \log x = - t + c_{1}$

$∵$ x(0) = 2

$c_{1} = \frac{1}{a} \log 2 \Rightarrow \frac{1}{a} \log x = - t + \frac{1}{a} \log 2 \Rightarrow x = 2 e^{- a t}$

Now, $\frac{d y}{d t} + b y = 0 \Rightarrow \frac{1}{b} \log y = - t + c_{2}$

$∵ y (0) = 1$

$∴ \frac{1}{b} \log (1) = - 0 + c_{2} \Rightarrow c_{2} = 0 \Rightarrow y = e^{- b t}$

Now, $3 y (1) = 2 x (1) \Rightarrow 3 e^{- b} = 4 e^{- a}$

$\Rightarrow e^{a - b} = \frac{4}{3} \Rightarrow a - b = \log (\frac{4}{3})$ ... (i)

For $x (t) = y (t), 2 e^{- a t} = e^{- b t} \Rightarrow e^{(a - b) t} = 2 \Rightarrow (a - b) t = \log 2$

$\Rightarrow \log (\frac{4}{3}) t = \log 2$ [From (i)]

$\Rightarrow t = \frac{\log 2}{\log \frac{4}{3}} \Rightarrow t = \log_{\frac{4}{3}} 2$

Q 15 :

If y = y(x) is the solution curve of the differential equation $(x^{2} - 4) d y - (y^{2} - 3 y) d x = 0, x > 2, y (4) = \frac{3}{2}$ and the slope of the curve is never zero, then the value of y(10) equals: [2024]

$\frac{3}{1 + {(8)}^{1 / 4}}$
$\frac{3}{1 - {(8)}^{1 / 4}}$
$\frac{3}{1 - 2 \sqrt{2}}$
$\frac{3}{1 + 2 \sqrt{2}}$

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(1)

Given differential equation is

$(x^{2} - 4) d y - (y^{2} - 3 y) d x = 0 \Rightarrow \frac{d y}{y^{2} - 3 y} = \frac{d x}{x^{2} - 4}$

Integrating on both sides, we get

$\int \frac{d y}{y^{2} - 3 y} = \int \frac{d x}{x^{2} - 4}$

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

$\Rightarrow \int \frac{d y}{3 (y - 3)} - \int \frac{d y}{3 y} = \int \frac{d x}{4 (x - 2)} - \int \frac{d x}{4 (x + 2)}$

$\Rightarrow \frac{1}{3} \ln | \frac{y - 3}{y} | = \frac{1}{4} \ln | \frac{x - 2}{x + 2} | + \ln C \Rightarrow \frac{y - 3}{y} = \frac{C_{1} {(x - 2)}^{3 / 4}}{{(x + 2)}^{3 / 4}}$

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

Now, $y (4) = \frac{3}{2} \Rightarrow C_{1} = - 3^{3 / 4}$

$∴ y = \frac{3 {(x + 2)}^{3 / 4}}{{(x + 2)}^{3 / 4} + {(3 x - 6)}^{3 / 4}}$

So, $y (10) = \frac{3 \times 12^{3 / 4}}{12^{3 / 4} + 24^{3 / 4}} = \frac{3}{1 + 2^{3 / 4}} = \frac{3}{1 + {(8)}^{1 / 4}}$

Q 16 :

A function y = f(x) satisfies $f (x) \sin 2 x + \sin x - (1 + \cos^{2} x) f' (x) = 0$ with condition f(0) = 0. Then $f (\frac{π}{2})$ is equal to [2024]

2
0
1
– 1

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(3)

We have, $y \sin 2 x + \sin x - (1 + \cos^{2} x) \frac{d y}{d x} = 0$

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

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

$IF = e^{- \int \frac{\sin 2 x}{1 + \cos^{2} x} d x} = e^{\log (1 + \cos^{2} x)} = 1 + \cos^{2} x$

So, solution is given by

$y \cdot (1 + \cos^{2} x) = \int \frac{\sin x}{(1 + \cos^{2} x)} (1 + \cos^{2} x) d x$

$\Rightarrow y (1 + \cos^{2} x) = \int \sin x d x \Rightarrow y (1 + \cos^{2} x) = - \cos x + c$ ... (i)

Now, y(0) = 0

$\Rightarrow 0 = - \cos (0) + c \Rightarrow c = 1$

From (i), $y (1 + \cos^{2} x) = - \cos x + 1 \Rightarrow y = \frac{1 - \cos x}{1 + \cos^{2} x}$

$∴ y (\frac{π}{2}) = \frac{1 - 0}{1 + 0} = 1$

Q 17 :

If $\sin (\frac{y}{x}) = \log_{e} | x | + \frac{α}{2}$ is the solution of the differential equation $x \cos (\frac{y}{x}) \frac{d y}{d x} = y \cos (\frac{y}{x}) + x$ and $y (1) = \frac{π}{3}$ , then $α^{2}$ is equal to [2024]

4
3
9
12

1

2

3

4

(2)

The given differential equation is,

$x \cos (\frac{y}{x}) \frac{d y}{d x} = y \cos (\frac{y}{x}) + x$ ... (i)

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

$∴$ (i) becomes,

$x \cos v (v + x \frac{d v}{d x}) = v x \cdot \cos v + x \Rightarrow x \frac{d v}{d x} = \sec v$

$\Rightarrow \int \cos v d v = \int \frac{d x}{x} \Rightarrow \sin v = \log | x | + c$

$\Rightarrow \sin \frac{y}{x} = \log | x | + c$

Now, $y (1) = \frac{π}{3} \Rightarrow \sin \frac{π}{3} = c \Rightarrow c = \frac{\sqrt{3}}{2}$

So, the solution of given differential equation is,

$\sin (\frac{y}{x}) = \log | x | + \frac{\sqrt{3}}{2}$

On comparing with given solution, we get

$α = \sqrt{3} \Rightarrow α^{2} = 3$ .

Q 18 :

Let y = y(x) be the solution of the differential equation sec xdy + {2(1 – x) tan x + x(2 – x)} dx = 0 such that y(0) = 2. Then y(2) is equal to [2024]

2{1 – sin (2)}
1
2{sin (2) + 1}
2

1

2

3

4

(4)

sec xdy + {2(1 – x) tan x + x(2 – x)} dx = 0

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

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

$\Rightarrow \int d y = - 2 \int (1 - x) \sin x d x - \int \cos x (2 x - x^{2}) d x$

$\Rightarrow y = - 2 \int (1 - x) \sin x d x - (2 x - x^{2}) \sin x + \int (2 - 2 x) \sin x d x + c$

$\Rightarrow y = - (2 x - x^{2}) \sin x + c$

Now, $y (0) = 2 \Rightarrow 2 = c$

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

Q 19 :

Let y = y(x) be the solution of the differential equation $\frac{d y}{d x} = \frac{(\tan x) + y}{\sin x (\sec x - \sin x \tan x)}, x \in (0, \frac{π}{2})$ satisfying the condition $y (\frac{π}{4}) = 2$ . Then, $y (\frac{π}{3})$ is [2024]

$\sqrt{3} (2 + \log_{e} 3)$
$\sqrt{3} (2 + \log_{e} \sqrt{3})$
$\frac{\sqrt{3}}{2} (2 + \log_{e} 3)$
$\sqrt{3} (1 + 2 \log_{e} 3)$

1

2

3

4

(2)

We have, $\frac{d y}{d x} = \frac{\tan x + y}{\sin x (\sec x - \sin x \cdot \tan x)}$

$= \frac{\frac{\sin x}{\cos x} + y}{\sin x (\frac{1}{\cos x} - \sin x \cdot \frac{\sin x}{\cos x})}$

$= \frac{\sin x + y \cos x}{\sin x (1 - \sin^{2} x)} = \frac{\sin x + y \cos x}{\sin x \cdot \cos^{2} x}$

$\frac{d y}{d x} = \sec^{2} x + y \cdot \frac{1}{\sin x \cdot \cos x}$

Now, $IF = e^{- \int \frac{1}{\sin x \cdot \cos x} d x} = e^{- 2 \int cosec 2 x d x}$

$= e^{- 2} (\frac{\log | cosec 2 x - \cot 2 x |}{2}) = e^{- \log | cosec 2 x - \cot 2 x |}$

$= \frac{1}{| cosec 2 x - \cot 2 x |} = \frac{1}{| \tan x |}$

The solution is

$y \cdot \frac{1}{| \tan x |} = \int \frac{1}{| \tan x |} \cdot \sec^{2} x d x = \log | \tan x | + c$

$y = | \tan x | \cdot \log | \tan x | + c | \tan x |$

At $x = \frac{π}{4}, y = 2 \Rightarrow 2 = 0 + c \Rightarrow c = 2$

At $x = \frac{π}{3}$

$y = \sqrt{3} \cdot \log \sqrt{3} + 2 \sqrt{3} = \sqrt{3} (\log \sqrt{3} + 2)$ .

Q 20 :

The Solution curve of the differential equation $y \frac{d x}{d y} = x (\log_{e} x - \log_{e} y + 1)$ , x > 0, y > 0 passing through the point (e, 1) is [2024]

$| \log_{e} \frac{x}{y} |$ $= y$
$| \log_{e} \frac{y}{x} |$ $= y^{2}$
$2$ $| \log_{e} \frac{x}{y} |$ $= y + 1$
$| \log_{e} \frac{y}{x} |$ $= x$

1

2

3

4

(1)

$y \frac{d y}{d x} = x (\log_{e} x - \log_{e} y + 1)$

$\frac{d x}{d y} = \frac{x}{y} (\log x - \log y + 1)$

$\frac{d x}{d y} = \frac{x}{y} (\log \frac{x}{y} + 1) (∵ \log x - \log y = \log \frac{x}{y})$

Let $\frac{x}{y} = t \Rightarrow x = t y$

Differentiating w.r.t y

$\Rightarrow \frac{d x}{d y} = t + y \frac{d t}{d y} \Rightarrow t + y \frac{d t}{d y} = t \log t + t$

$\Rightarrow \frac{d t}{t \log t} = \frac{d y}{y} \Rightarrow \log (\log t) = \log y + c$

$\Rightarrow \log (\log \frac{x}{y}) = \log y + c$

at x = e and y = 1 (passing point)

$\Rightarrow \log (\log \frac{e}{1}) = \log 1 + c \Rightarrow c = 0$

$\Rightarrow \log (\log \frac{x}{y}) = \log y + 0$

$\Rightarrow \log (\log \frac{x}{y}) = \log y \Rightarrow \log (\frac{x}{y}) = y$

Q 21 :

The temperature T(t) of a body at time t = 0 is 160º F and it decreases continuously as per the differential equation $\frac{d T}{d t} = - K (T - 80)$ , where K is positive constant. If T(15) = 120º F, then T(45) is equal to [2024]

80º F
90º F
85º F
95º F

1

2

3

4

(2)

Given differential equation is

$\frac{d T}{d t} = - K (T - 80)$

Now, $\frac{d T}{T - 80} = - K d t$

On integrating, we get

$\int \frac{d T}{T - 80} = - K \int d t$

$\Rightarrow \log | T - 80 | = - K t + C$ ... (i)

When t = 0, T = 160º F

$∴ \log | 160 - 80 | = - K \times 0 + C$

$\Rightarrow C = \log (80)$

From (i),

log |T – 80| = – Kt + log (80) ...(ii)

Now, when t = 15, T = 120º F

$∴$ From (ii),

$\log (120 - 80) = - K \times 15 + \log (80)$

$\Rightarrow 15 K = \log (\frac{80}{40}) = \log (2) ∴ k = \frac{\log (2)}{15}$

When t = 45, then from (ii),

$\log (T - 80) = \frac{- \log (2)}{15} \times 45 + \log (80)$

$\Rightarrow \log (T - 80) = - 3 \log (2) + \log (80) = \log (\frac{80}{8}) = \log (10)$

$\Rightarrow \log (T - 80) = 1 \Rightarrow T - 80 = 10^{1}$

$\Rightarrow T = 90 ° F$

Q 22 :

Let the solution y = y(x) of the differential equation $\frac{d y}{d x} - y = 1 + 4 \sin x$ satisfy y( $π$ ) = 1. Then $y (\frac{π}{2}) + 10$ is equal to __________. [2024]

(7)

Given, $\frac{d y}{d x} - y = 1 + 4 \sin x$ , which is a linear differential equation.

$∴$ Integrating factor = $= e^{- \int d x} = e^{- x}$

Hence, solution is $y e^{- x} = \int (1 + 4 \sin x) e^{- x} d x$

$= - e^{- x} + 2 e^{- x} (- \sin x - \cos x) + C$ ... (i)

y( $π$ ) = 1 i.e., at x = $π$ , y = 1

$∴$ From (i), we get

$(1) e^{- π} = - e^{- π} + 2 e^{- π} (- \sin π - \cos π) + C$

$\Rightarrow e^{- π} = - e^{- π} + 2 e^{- π} + C \Rightarrow C = 0$

Hence, y(x) = –1 – 2(sin x + cos x)

Now, $y (\frac{π}{2}) + 10 = - 1 - 2 (\sin \frac{π}{2} + \cos \frac{π}{2}) + 10$

= –1 – 2(1) + 10 = 7.

Q 23 :

Let y = y(x) be the solution of the differential equation ${(x + y + 2)}^{2} d x = d y$ , y(0) = –2. Let the maximum and minimum values of the function y = y(x) in $[0, \frac{π}{3}]$ be $α$ and $β$ , respectively. If ${(3 α + π)}^{2} + β^{2} = γ + δ \sqrt{3}, γ, δ \in Z$ , then $γ + δ$ equals __________. [2024]

(31)

We have, $\frac{d y}{d x} = {(x + y + 2)}^{2}$ ... (i)

Put x + y + 2 = t

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

$∴$ From (i) $\frac{d t}{d x} - 1 = t^{2} \Rightarrow \frac{d t}{1 + t^{2}} = d x \Rightarrow \tan^{- 1} t = x + C$

$\Rightarrow x + y + 2 = \tan (x + C) \Rightarrow y = \tan (x + C) - x - 2$

$∵ y (0) = - 2 \Rightarrow - 2 = \tan C - 0 - 2$

$\Rightarrow \tan C = 0 \Rightarrow C = 0 \Rightarrow y = \tan x - x - 2$

$\Rightarrow \frac{d y}{d x} = \sec^{2} x - 1 \geq 0$

$\Rightarrow y (x) is increasing if x \in (0, \frac{π}{3}) \Rightarrow α = y (\frac{π}{3}), β = y (0)$

$\Rightarrow α = \tan \frac{π}{3} - \frac{π}{3} - 2 = - \frac{π}{3} - 2 + \sqrt{3}$

and $β$ = tan 0 – 0 – 2 = –2

Hence, ${(3 α + π)}^{2} + β^{2} = {(- π - 6 + 3 \sqrt{3} + π)}^{2} + (- 2)^{2}$

$= {(- 6 + 3 \sqrt{3})}^{2} + 4 = 36 + 27 - 36 \sqrt{3} + 4$

$= 67 - 36 \sqrt{3} = y + δ \sqrt{3}$ [Given]

On comparing, we get

$γ$ = 67 and $δ$ = –36

Hence, $γ + δ = 67 - 36 = 31$ .

Q 24 :

Let f be a differentiable function in the interval (0, $\infty$ ) such that f(1) =1 and $\lim_{t \to x} \frac{t^{2} f (x) - x^{2} f (t)}{t - x} = 1$ for each x > 0. Then 2f(2) + 3f(3) is equal to __________. [2024]

(24)

We Have $\underset{t \to x}{l t} \frac{t^{2} f (x) - x^{2} f (t)}{t - x} = 1 (\frac{0}{0} form)$

$\Rightarrow \underset{t \to x}{l t} \frac{2 t f (x) - x^{2} f' (t)}{1} = 1 [L' Hospital Rule]$

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

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

$∴ f (x) \frac{1}{x^{2}} = \int \frac{- 1}{x^{2}} \cdot \frac{1}{x^{2}} d x + C \Rightarrow f (x) \frac{1}{x^{2}} = \frac{1}{3 x^{3}} + C$

Now, f(1) = 1

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

Now, $2 f (2) + 3 f (3) = 2 [\frac{1}{6} + \frac{8}{3}] + 3 [\frac{1}{9} + \frac{18}{3}]$

$= \frac{1}{3} + \frac{16}{3} + \frac{1}{3} + 18 = 6 + 18 = 24$ .

Q 25 :

Let y = y(x) be the solution of the differential equation $\frac{d y}{d x} + \frac{2 x}{{(1 + x^{2})}^{2}} y = x e^{\frac{1}{(1 + x^{2})}}$ ; y(0) = 0. Then the area enclosed by the curve $f (x) = y (x) e^{- \frac{1}{(1 + x^{2})}}$ and the line y – x = 4 is __________. [2024]

(18)

We have, $\frac{d y}{d x} + \frac{2 x}{{(1 + x^{2})}^{2}} y = x e^{\frac{1}{(1 + x^{2})}}$

$IF = e^{\int \frac{2 x}{{(1 + x^{2})}^{2}} d x} = e^{\frac{- 1}{1 + x^{2}}}$

$∴$ General solution of given differential equation is,

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

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

Since y(0) = 0

$\Rightarrow 0 = 0 + c$

$\Rightarrow c = 0$

$∴ y (x) = \frac{x^{2}}{2} e^{\frac{1}{1 + x^{2}}}$

$\Rightarrow f (x) = \frac{x^{2}}{2}$

So, intersection of curve $y = f (x) = \frac{x^{2}}{2}$ and y = x + 4 is given by $\frac{x^{2}}{2} = x + 4$

$\Rightarrow x^{2} - 2 x - 8 = 0 \Rightarrow (x - 4) (x + 2) = 0$

$\Rightarrow x = 4, - 2 \Rightarrow y = 8, 2$

$∴$ Required area = $\int_{- 2}^{4} ((x + 4) - \frac{x^{2}}{2}) d x$

= ${[\frac{x^{2}}{2} + 4 x - \frac{x^{3}}{6}]}_{- 2}^{4} = [8 + 16 - \frac{32}{3} - 2 + 8 - \frac{8}{6}]$

= 30 – 12 = 18.

Q 26 :

If the solution y(x) of the given differential equation $(e^{y} + 1) \cos x d x + e^{y} \sin x d y = 0$ passes through the point $(\frac{π}{2}, 0)$ , then the value of $e^{y (\frac{π}{6})}$ is equal to _________. [2024]

(3)

Givem, $(e^{y} + 1) \cos x d x + e^{y} \sin x d y = 0$

$\Rightarrow d (e^{y} \sin x) + \cos x d x = 0 \Rightarrow e^{y} \sin x + \sin x = C$

The curve passes through $(\frac{π}{2}, 0)$

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

$\Rightarrow (e^{y (\frac{π}{6})} \frac{1}{2} + \frac{1}{2}) = 2 \Rightarrow e^{y (\frac{π}{6})} + 1 = 4 \Rightarrow e^{y (\frac{π}{6})} = 3$ .

Q 27 :

Let $α | x | = | y | e^{x y - β}$ , $α, β \in N$ be the solution of the differential equation xdy – ydx + xy (xdy + ydx) = 0, y(1) = 2. Then $α + β$ is equal to __________. [2024]

(4)

We have, $α | x | = | y | e^{x y - β}$

and $\frac{x d y - y d x}{y^{2}} + \frac{x y (x d y + y d x)}{y^{2}} = 0$

$\Rightarrow - d (\frac{x}{y}) + \frac{x}{y} d (x y) = 0 \Rightarrow \int d (x y) = \int \frac{d (\frac{x}{y})}{x / y}$

$\Rightarrow x y = \ln | \frac{x}{y} | + \ln c \Rightarrow x y = \ln (| \frac{x}{y} | \cdot c)$ ... (i)

$∵$ y(1) = 2, i.e., x = 1, y = 2

$∴ From (i), 2 = \ln (| \frac{1}{2} | c) \Rightarrow c = 2 e^{2}$

$∴ x y = \ln (| \frac{x}{y} | \cdot 2 e^{2})$

$\Rightarrow e^{x y} = \frac{| x |}{| y |} \cdot 2 e^{2} \Rightarrow 2 | x | = | y | e^{x y - 2}$

Also, solution of equation is $α | x | = | y | e^{x y - β}$ (Given)

On comparing, $α = 2, β = 2$ . Hence, $α + β = 4$ .

Q 28 :

If x = x(t) is the solution of the differential equation $(t + 1) d x = (2 x + {(t + 1)}^{4}) d t$ , x(0) = 2, then x(1) equals __________. [2024]

(14)

We have, $(t + 1) d x = (2 x + {(t + 1)}^{4}) d t$

$\Rightarrow \frac{d x}{d t} = \frac{2 x}{t + 1} + {(t + 1)}^{3}; P = \frac{- 2}{t + 1}, Q = {(t + 1)}^{3}$

So, $I . F . = e^{\int P d t} = e^{\int \frac{- 2}{t + 1} d t} = e^{- 2 \log (t + 1)} = {(t + 1)}^{- 2}$

The Solution of given differential equation is given by $x (I . F .) = \int Q \cdot (I . F .) d t + C$

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

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

Putting x(0) = 2, we get

$2 = {(1)}^{2} (0 + 0 + C) \Rightarrow C = 2$

$∴ x = {(t + 1)}^{2} (\frac{t^{2}}{2} + t + 2)$

At $t = 1, x = {(2)}^{2} (\frac{1}{2} + 1 + 2) = 4 (\frac{7}{2}) = 14$ .

Q 29 :

If $\frac{d x}{d y} = \frac{1 + x - y^{2}}{y}$ , x(1) = 1, then 5x(2) is equal to __________. [2024]

(5)

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

Now, $I . F . = e^{- \int \frac{1}{y} d y} = e^{- \log y} = \frac{1}{y}$

So, solution is given by,

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

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

$∵ x (1) = 1 \Rightarrow 1 = - 1 - 1 + C \Rightarrow C = 3$

$\Rightarrow x = - 1 - y^{2} + 3 y$

$∴$ 5x(2) = 5(–1 – 4 + 6) = 5.

Q 30 :

If the solution of the differential equation (2x + 3y – 2)dx + (4x + 6y – 7)dy = 0, y(0) = 3, is $α x + β y + 3 \log_{e} | 2 x + 3 y - γ | = 6$ , then $α + 2 β + 3 γ$ is equal to __________. [2024]

(29)

(2x + 3y – 2)dx + (4x + 6y – 7)dy = 0

$\frac{d y}{d x} = \frac{- (2 x + 3 y - 2)}{(4 x + 6 y - 7}$ ... (i)

Let 2x + 3y = t

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

Putting in (i), we get

$\frac{1}{3} [\frac{d t}{d x}] - \frac{2}{3} = \frac{- (t - 2)}{2 t - 7}$

$\Rightarrow \frac{1}{3} [\frac{d t}{d x}] = \frac{2}{3} - \frac{t - 2}{2 t - 7} = \frac{4 t - 14 - 3 t + 6}{3 (2 t - 7)}$

$\Rightarrow \frac{d t}{d x} = \frac{t - 8}{2 t - 7} \Rightarrow d x = \frac{2 t - 7}{t - 8} d t$

$\Rightarrow \int d x + c = \int 2 d t + 9 \int \frac{d t}{t - 8}$

$\Rightarrow x + c = 2 t + 9 \ln | t - 8 |$

$\Rightarrow 2 (2 x + 3 y) + 9 \ln | 2 x + 3 y - 8 | = x + c$

which is the solution of given differential equation.

Now, $y (0) = 3 \Rightarrow 2 (9) + 9 \ln | 1 | = c \Rightarrow c = 18$

Putting the value of c in (i), we get

$4 x + 6 y + 9 \ln | 2 x + 3 y - 8 | = x + 18$

$\Rightarrow 3 x + 6 y + 9 \ln | 2 x + 3 y - 8 | = 18$

$\Rightarrow x + 2 y + 3 \ln | 2 x + 3 y - 8 | = 6$

$∴ α = 1 . β = 2, γ = 8$

Therefore, $α + 2 β + 3 γ = 1 + 2 (2) + 3 (8) = 1 + 4 + 24 = 29$ .

Topic Question Set

Q 1 :

If the Solution y = y(x) of the differential equation $(x^{4} + 2 x^{3} + 3 x^{2} + 2 x + 2) d y - (2 x^{2} + 2 x + 3) d x = 0$ satisfies $y (- 1) = - \frac{π}{4}$ , then y(0) is equal [2024]

Q 2 :

Let y = y(x) be the solution of the differential equation ${(x^{2} + 4)}^{2} d y + (2 x^{3} y + 8 x y - 2) d x = 0$ . If y(0) = 0, then y(2) is equal to [2024]

Q 3 :

If y = y(x) is the solution of the differential equation $\frac{d y}{d x} + 2 y = \sin (2 x)$ , $y (0) = \frac{3}{4}$ then $y (\frac{π}{8})$ is equal to [2024]

Q 4 :

Let y = y(x) be the solution of the differential equation $(1 + x^{2}) \frac{d y}{d x} + y = e^{\tan^{- 1} x}$ , y(1) = 0. Then y(0) is [2024]

Q 5 :

Let y = y(x) be the solution of the differential equation $(2 x \log_{e} x) \frac{d y}{d x} + 2 y = \frac{3}{x} \log_{e} x$ , x > 0 and $y (e^{- 1}) = 0$ . Then y(e) is equal to [2024]

Q 6 :

Suppose the solution of the differential equation $\frac{d y}{d x} = \frac{(2 + α) x - β y + 2}{β x - 2 α y - (β γ - 4 α)}$ represents a circle passing through origin. Then the radius of this circle is: [2024]

Q 7 :

Let y = y(x) be the solution of the differential equation $(1 + y^{2}) e^{\tan x} d x + \cos^{2} x (1 + e^{2 \tan x}) d y = 0$ , y(0) = 1. Then $y (\frac{π}{4})$ is equal to [2024]

Q 8 :

Let y = y(x) be the solution curve of the differential equation $\sec y \frac{d y}{d x} + 2 x \sin y = x^{3} \cos y$ , y(1) = 0. Then $y (\sqrt{3})$ is equal to: [2024]

Q 9 :

The Solution curve of the differential equation $2 y = \frac{d y}{d x} + 3 = 5 \frac{d y}{d x}$ , passing through the point (0, 1) is a conic, whose vertex lies on the line: [2024]

Q 10 :

The Solution of the differential equation $(x^{2} + y^{2}) d x - 5 x y d y = 0$ , y(1) = 0, is: [2024]

Q 11 :

Let $\int_{0}^{x} \sqrt{1 - (y' {(t))}^{2}} d t = \int_{0}^{x} y (t) d t$ , $0 \leq x \leq 3, y \geq 0$ , y(0) = 0. Then at x = 2, y" + y + 1 is equal to [2024]

Q 12 :

Let y = y(x) be the solution of the differential equation $\frac{d y}{d x} = 2 x {(x + y)}^{3} - x (x + y) - 1, y (0) = 1$ . Then, $(\frac{1}{\sqrt{2}} + y {(\frac{1}{\sqrt{2}}))}^{2}$ equals: [2024]

Q 13 :

Let $α$ be a non-zero real number. Suppose $f : R \to R$ is a differentiable function such that f(0) = 2 and $\lim_{x \to - \infty} f (x) = 1$ . If f'(x) = $α$ f(x) + 3, for all $x \in R$ , then $f (- \log_{e} 2)$ is equal to ________. [2024]

Q 14 :

Let x = x(t) and y = y(t) be solutions of the differential equations $\frac{d x}{d t} + a x = 0$ and $\frac{d y}{d t} + b y = 0$ respectively, $a, b \in R$ . Given that x(0) = 2; y(0)= 1 and 3y(1) = 2x(1), the value of t, for which x(t) = y(t), is: [2024]

Q 15 :

If y = y(x) is the solution curve of the differential equation $(x^{2} - 4) d y - (y^{2} - 3 y) d x = 0, x > 2, y (4) = \frac{3}{2}$ and the slope of the curve is never zero, then the value of y(10) equals: [2024]

Q 16 :

A function y = f(x) satisfies $f (x) \sin 2 x + \sin x - (1 + \cos^{2} x) f' (x) = 0$ with condition f(0) = 0. Then $f (\frac{π}{2})$ is equal to [2024]

Q 17 :

If $\sin (\frac{y}{x}) = \log_{e} | x | + \frac{α}{2}$ is the solution of the differential equation $x \cos (\frac{y}{x}) \frac{d y}{d x} = y \cos (\frac{y}{x}) + x$ and $y (1) = \frac{π}{3}$ , then $α^{2}$ is equal to [2024]

Q 18 :

Let y = y(x) be the solution of the differential equation sec xdy + {2(1 – x) tan x + x(2 – x)} dx = 0 such that y(0) = 2. Then y(2) is equal to [2024]

Q 19 :

Let y = y(x) be the solution of the differential equation $\frac{d y}{d x} = \frac{(\tan x) + y}{\sin x (\sec x - \sin x \tan x)}, x \in (0, \frac{π}{2})$ satisfying the condition $y (\frac{π}{4}) = 2$ . Then, $y (\frac{π}{3})$ is [2024]

Q 20 :

The Solution curve of the differential equation $y \frac{d x}{d y} = x (\log_{e} x - \log_{e} y + 1)$ , x > 0, y > 0 passing through the point (e, 1) is [2024]

Q 21 :

The temperature T(t) of a body at time t = 0 is 160º F and it decreases continuously as per the differential equation $\frac{d T}{d t} = - K (T - 80)$ , where K is positive constant. If T(15) = 120º F, then T(45) is equal to [2024]

Q 22 :

Let the solution y = y(x) of the differential equation $\frac{d y}{d x} - y = 1 + 4 \sin x$ satisfy y( $π$ ) = 1. Then $y (\frac{π}{2}) + 10$ is equal to __________. [2024]

Q 24 :

Let f be a differentiable function in the interval (0, $\infty$ ) such that f(1) =1 and $\lim_{t \to x} \frac{t^{2} f (x) - x^{2} f (t)}{t - x} = 1$ for each x > 0. Then 2f(2) + 3f(3) is equal to __________. [2024]

Q 25 :

Let y = y(x) be the solution of the differential equation $\frac{d y}{d x} + \frac{2 x}{{(1 + x^{2})}^{2}} y = x e^{\frac{1}{(1 + x^{2})}}$ ; y(0) = 0. Then the area enclosed by the curve $f (x) = y (x) e^{- \frac{1}{(1 + x^{2})}}$ and the line y – x = 4 is __________. [2024]

Q 26 :

If the solution y(x) of the given differential equation $(e^{y} + 1) \cos x d x + e^{y} \sin x d y = 0$ passes through the point $(\frac{π}{2}, 0)$ , then the value of $e^{y (\frac{π}{6})}$ is equal to _________. [2024]

Q 27 :

Let $α | x | = | y | e^{x y - β}$ , $α, β \in N$ be the solution of the differential equation xdy – ydx + xy (xdy + ydx) = 0, y(1) = 2. Then $α + β$ is equal to __________. [2024]

Q 28 :

If x = x(t) is the solution of the differential equation $(t + 1) d x = (2 x + {(t + 1)}^{4}) d t$ , x(0) = 2, then x(1) equals __________. [2024]

Q 29 :

If $\frac{d x}{d y} = \frac{1 + x - y^{2}}{y}$ , x(1) = 1, then 5x(2) is equal to __________. [2024]

Q 30 :

If the solution of the differential equation (2x + 3y – 2)dx + (4x + 6y – 7)dy = 0, y(0) = 3, is $α x + β y + 3 \log_{e} | 2 x + 3 y - γ | = 6$ , then $α + 2 β + 3 γ$ is equal to __________. [2024]

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Topic Question Set

Q 1 : If the Solution y = y(x) of the differential equation (x4+2x3+3x2+2x+2)dy-(2x2+2x+3)dx=0 satisfies y(-1)=-π4, then y(0) is equal [2024]

Q 2 : Let y = y(x) be the solution of the differential equation (x2+4)2dy+(2x3y+8xy-2)dx=0. If y(0) = 0, then y(2) is equal to [2024]

Q 3 : If y = y(x) is the solution of the differential equation dydx+2y=sin (2x), y(0)=34 then y (π8) is equal to [2024]

Q 4 : Let y = y(x) be the solution of the differential equation (1+x2) dydx+y=etan-1x, y(1) = 0. Then y(0) is [2024]

Q 5 : Let y = y(x) be the solution of the differential equation (2xlogex)dydx+2y=3xlogex, x > 0 and y(e-1)=0. Then y(e) is equal to [2024]

Q 6 : Suppose the solution of the differential equation dydx=(2 + α)x - βy + 2βx - 2αy - (βγ - 4α) represents a circle passing through origin. Then the radius of this circle is: [2024]

Q 7 : Let y = y(x) be the solution of the differential equation (1+y2)etanxdx+cos2x(1+e2tanx)dy=0 , y(0) = 1. Then y(π4) is equal to [2024]

Q 8 : Let y = y(x) be the solution curve of the differential equation secydydx+2x siny=x3cosy, y(1) = 0. Then y(3) is equal to: [2024]

Q 9 : The Solution curve of the differential equation 2y=dydx+3=5dydx, passing through the point (0, 1) is a conic, whose vertex lies on the line: [2024]

Q 10 : The Solution of the differential equation (x2+y2) dx-5xydy=0, y(1) = 0, is: [2024]

Q 11 : Let ∫0x1-(y'(t))2dt=∫0xy(t)dt, 0≤x≤3, y≥0, y(0) = 0. Then at x = 2, y" + y + 1 is equal to [2024]

Q 12 : Let y = y(x) be the solution of the differential equation dydx=2x(x+y)3-x(x+y)-1, y(0)=1. Then, (12+y(12))2 equals: [2024]

Q 13 : Let α be a non-zero real number. Suppose f : R → R is a differentiable function such that f(0) = 2 and limx → -∞f(x) = 1. If f'(x) = αf(x) + 3, for all x∈ R, then f(- loge 2) is equal to ________. [2024]

Q 14 : Let x = x(t) and y = y(t) be solutions of the differential equations dxdt+ax=0 and dydt+by=0 respectively, a,b∈R. Given that x(0) = 2; y(0)= 1 and 3y(1) = 2x(1), the value of t, for which x(t) = y(t), is: [2024]

Q 15 : If y = y(x) is the solution curve of the differential equation (x2-4)dy-(y2-3y)dx=0, x>2, y(4)=32 and the slope of the curve is never zero, then the value of y(10) equals: [2024]

Q 16 : A function y = f(x) satisfies f(x)sin2x+sin x-(1+cos2 x) f'(x)=0 with condition f(0) = 0. Then f(π2) is equal to [2024]

Q 17 : If sin(yx)=loge|x|+α2 is the solution of the differential equation xcos(yx)dydx=ycos(yx)+x and y(1)=π3, then α2 is equal to [2024]

Q 18 : Let y = y(x) be the solution of the differential equation sec xdy + {2(1 – x) tan x + x(2 – x)} dx = 0 such that y(0) = 2. Then y(2) is equal to [2024]

Q 19 : Let y = y(x) be the solution of the differential equation dydx=(tan x)+ysinx(secx-sinx tanx), x∈(0,π2) satisfying the condition y(π4)=2. Then, y(π3) is [2024]

Q 20 : The Solution curve of the differential equation ydxdy=x(loge x-loge y+1), x > 0, y > 0 passing through the point (e, 1) is [2024]

Q 21 : The temperature T(t) of a body at time t = 0 is 160º F and it decreases continuously as per the differential equation dTdt=- K(T-80), where K is positive constant. If T(15) = 120º F, then T(45) is equal to [2024]

Q 22 : Let the solution y = y(x) of the differential equation dydx–y=1+4 sin x satisfy y(π) = 1. Then y(π2)+10 is equal to __________. [2024]

Q 23 : Let y = y(x) be the solution of the differential equation (x+y+2)2dx=dy, y(0) = –2. Let the maximum and minimum values of the function y = y(x) in [0,π3] be α and β, respectively. If (3α+π)2+β2=γ+δ3, γ, δ∈Z, then γ+δ equals __________. [2024]

Q 24 : Let f be a differentiable function in the interval (0, ∞) such that f(1) =1 and limt→xt2f(x)–x2f(t)t–x=1 for each x > 0. Then 2f(2) + 3f(3) is equal to __________. [2024]

Q 25 : Let y = y(x) be the solution of the differential equation dydx+2x(1+x2)2y=xe1(1+x2); y(0) = 0. Then the area enclosed by the curve f(x)=y(x)e–1(1+x2) and the line y – x = 4 is __________. [2024]

Q 26 : If the solution y(x) of the given differential equation (ey+1) cos xdx+ey sin xdy = 0 passes through the point (π2,0), then the value of ey(π6) is equal to _________. [2024]

Q 27 : Let α|x|=|y|exy–β, α, β∈N be the solution of the differential equation xdy – ydx + xy (xdy + ydx) = 0, y(1) = 2. Then α + β is equal to __________. [2024]

Q 28 : If x = x(t) is the solution of the differential equation (t+1)dx=(2x+(t+1)4)dt, x(0) = 2, then x(1) equals __________. [2024]

Q 29 : If dxdy=1+x–y2y, x(1) = 1, then 5x(2) is equal to __________. [2024]

Q 30 : If the solution of the differential equation (2x + 3y – 2)dx + (4x + 6y – 7)dy = 0, y(0) = 3, is αx+βy+3 loge |2x+3y–γ|=6, then α+2β+3γ is equal to __________. [2024]

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Q 1 :

If the Solution y = y(x) of the differential equation $(x^{4} + 2 x^{3} + 3 x^{2} + 2 x + 2) d y - (2 x^{2} + 2 x + 3) d x = 0$ satisfies $y (- 1) = - \frac{π}{4}$ , then y(0) is equal [2024]

Q 2 :

Let y = y(x) be the solution of the differential equation ${(x^{2} + 4)}^{2} d y + (2 x^{3} y + 8 x y - 2) d x = 0$ . If y(0) = 0, then y(2) is equal to [2024]

Q 3 :

If y = y(x) is the solution of the differential equation $\frac{d y}{d x} + 2 y = \sin (2 x)$ , $y (0) = \frac{3}{4}$ then $y (\frac{π}{8})$ is equal to [2024]

Q 4 :

Let y = y(x) be the solution of the differential equation $(1 + x^{2}) \frac{d y}{d x} + y = e^{\tan^{- 1} x}$ , y(1) = 0. Then y(0) is [2024]

Q 5 :

Let y = y(x) be the solution of the differential equation $(2 x \log_{e} x) \frac{d y}{d x} + 2 y = \frac{3}{x} \log_{e} x$ , x > 0 and $y (e^{- 1}) = 0$ . Then y(e) is equal to [2024]

Q 6 :

Suppose the solution of the differential equation $\frac{d y}{d x} = \frac{(2 + α) x - β y + 2}{β x - 2 α y - (β γ - 4 α)}$ represents a circle passing through origin. Then the radius of this circle is: [2024]

Q 7 :

Let y = y(x) be the solution of the differential equation $(1 + y^{2}) e^{\tan x} d x + \cos^{2} x (1 + e^{2 \tan x}) d y = 0$ , y(0) = 1. Then $y (\frac{π}{4})$ is equal to [2024]

Q 8 :

Let y = y(x) be the solution curve of the differential equation $\sec y \frac{d y}{d x} + 2 x \sin y = x^{3} \cos y$ , y(1) = 0. Then $y (\sqrt{3})$ is equal to: [2024]

Q 9 :

The Solution curve of the differential equation $2 y = \frac{d y}{d x} + 3 = 5 \frac{d y}{d x}$ , passing through the point (0, 1) is a conic, whose vertex lies on the line: [2024]

Q 10 :

The Solution of the differential equation $(x^{2} + y^{2}) d x - 5 x y d y = 0$ , y(1) = 0, is: [2024]

Q 11 :

Let $\int_{0}^{x} \sqrt{1 - (y' {(t))}^{2}} d t = \int_{0}^{x} y (t) d t$ , $0 \leq x \leq 3, y \geq 0$ , y(0) = 0. Then at x = 2, y" + y + 1 is equal to [2024]

Q 12 :

Let y = y(x) be the solution of the differential equation $\frac{d y}{d x} = 2 x {(x + y)}^{3} - x (x + y) - 1, y (0) = 1$ . Then, $(\frac{1}{\sqrt{2}} + y {(\frac{1}{\sqrt{2}}))}^{2}$ equals: [2024]

Q 13 :

Let $α$ be a non-zero real number. Suppose $f : R \to R$ is a differentiable function such that f(0) = 2 and $\lim_{x \to - \infty} f (x) = 1$ . If f'(x) = $α$ f(x) + 3, for all $x \in R$ , then $f (- \log_{e} 2)$ is equal to ________. [2024]

Q 14 :

Let x = x(t) and y = y(t) be solutions of the differential equations $\frac{d x}{d t} + a x = 0$ and $\frac{d y}{d t} + b y = 0$ respectively, $a, b \in R$ . Given that x(0) = 2; y(0)= 1 and 3y(1) = 2x(1), the value of t, for which x(t) = y(t), is: [2024]

Q 15 :

If y = y(x) is the solution curve of the differential equation $(x^{2} - 4) d y - (y^{2} - 3 y) d x = 0, x > 2, y (4) = \frac{3}{2}$ and the slope of the curve is never zero, then the value of y(10) equals: [2024]

Q 16 :

A function y = f(x) satisfies $f (x) \sin 2 x + \sin x - (1 + \cos^{2} x) f' (x) = 0$ with condition f(0) = 0. Then $f (\frac{π}{2})$ is equal to [2024]

Q 17 :

If $\sin (\frac{y}{x}) = \log_{e} | x | + \frac{α}{2}$ is the solution of the differential equation $x \cos (\frac{y}{x}) \frac{d y}{d x} = y \cos (\frac{y}{x}) + x$ and $y (1) = \frac{π}{3}$ , then $α^{2}$ is equal to [2024]

Q 18 :

Let y = y(x) be the solution of the differential equation sec xdy + {2(1 – x) tan x + x(2 – x)} dx = 0 such that y(0) = 2. Then y(2) is equal to [2024]

Q 19 :

Let y = y(x) be the solution of the differential equation $\frac{d y}{d x} = \frac{(\tan x) + y}{\sin x (\sec x - \sin x \tan x)}, x \in (0, \frac{π}{2})$ satisfying the condition $y (\frac{π}{4}) = 2$ . Then, $y (\frac{π}{3})$ is [2024]

Q 20 :

The Solution curve of the differential equation $y \frac{d x}{d y} = x (\log_{e} x - \log_{e} y + 1)$ , x > 0, y > 0 passing through the point (e, 1) is [2024]

Q 21 :

The temperature T(t) of a body at time t = 0 is 160º F and it decreases continuously as per the differential equation $\frac{d T}{d t} = - K (T - 80)$ , where K is positive constant. If T(15) = 120º F, then T(45) is equal to [2024]

Q 22 :

Let the solution y = y(x) of the differential equation $\frac{d y}{d x} - y = 1 + 4 \sin x$ satisfy y( $π$ ) = 1. Then $y (\frac{π}{2}) + 10$ is equal to __________. [2024]

Q 24 :

Let f be a differentiable function in the interval (0, $\infty$ ) such that f(1) =1 and $\lim_{t \to x} \frac{t^{2} f (x) - x^{2} f (t)}{t - x} = 1$ for each x > 0. Then 2f(2) + 3f(3) is equal to __________. [2024]

Q 25 :

Let y = y(x) be the solution of the differential equation $\frac{d y}{d x} + \frac{2 x}{{(1 + x^{2})}^{2}} y = x e^{\frac{1}{(1 + x^{2})}}$ ; y(0) = 0. Then the area enclosed by the curve $f (x) = y (x) e^{- \frac{1}{(1 + x^{2})}}$ and the line y – x = 4 is __________. [2024]

Q 26 :

If the solution y(x) of the given differential equation $(e^{y} + 1) \cos x d x + e^{y} \sin x d y = 0$ passes through the point $(\frac{π}{2}, 0)$ , then the value of $e^{y (\frac{π}{6})}$ is equal to _________. [2024]

Q 27 :

Let $α | x | = | y | e^{x y - β}$ , $α, β \in N$ be the solution of the differential equation xdy – ydx + xy (xdy + ydx) = 0, y(1) = 2. Then $α + β$ is equal to __________. [2024]

Q 28 :

If x = x(t) is the solution of the differential equation $(t + 1) d x = (2 x + {(t + 1)}^{4}) d t$ , x(0) = 2, then x(1) equals __________. [2024]

Q 29 :

If $\frac{d x}{d y} = \frac{1 + x - y^{2}}{y}$ , x(1) = 1, then 5x(2) is equal to __________. [2024]

Q 30 :

If the solution of the differential equation (2x + 3y – 2)dx + (4x + 6y – 7)dy = 0, y(0) = 3, is $α x + β y + 3 \log_{e} | 2 x + 3 y - γ | = 6$ , then $α + 2 β + 3 γ$ is equal to __________. [2024]