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

Q 31 :

Let the solution curve of the differential equation $x d y - y d x = \sqrt{x^{2} + y^{2}} d x, x > 0,$ $y (1) = 0$ be $y = y (x)$ . Then $y (3)$ is equal to [2026]

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(4)

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

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

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

$= l n (\frac{y}{x} + \sqrt{1 + \frac{y^{2}}{x^{2}}}) = l n x + l n k = l n k x$

$\Rightarrow y + \sqrt{x^{2} + y^{2}} = k x^{2}$

$0 + 1 = k$

$\Rightarrow y + \sqrt{x^{2} + y^{2}} = x^{2}$

$y + \sqrt{9 + y^{2}} = 9$

$y = 4$

Q 32 :

Let $y = y (x)$ be the solution of the differential equation $\sec x \frac{d y}{d x} - 2 y = 2 + 3 \sin x, x \in (- \frac{π}{2}, \frac{π}{2}),$ $y (0) = - \frac{7}{4} .$ Then $y (\frac{π}{6})$ is equal to: [2026]

$- \frac{5}{2}$
$- \frac{5}{4}$
$- 3 \sqrt{3} - 7$
$- 3 \sqrt{2} - 7$

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(1)

$\frac{d y}{d x} - 2 y \cos x = 2 \cos x + 3 \sin x \cos x$

$I.F. = e^{- 2 \sin x}$

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

$y . e^{- 2 \sin x} = e^{- 2 \sin x} (- \frac{3}{2} \sin x - \frac{7}{4}) + C$

$\Rightarrow y = - \frac{3}{2} \sin x - \frac{7}{4} + C e^{2 \sin x}$

$∵$ $y (0) = - \frac{7}{4} \Rightarrow C = 0$

$y (\frac{π}{6}) = \frac{- 3}{2} \cdot \frac{1}{2} - \frac{7}{4} = \frac{- 5}{2}$

Q 33 :

Let $f (x) = x^{3} + x^{2} f' (1) + 2 x f'' (2) + f''' (3), x \in R .$ Then the value of $f' (5)$ is [2026]

$\frac{117}{5}$
$\frac{2}{5}$
$\frac{62}{5}$
$\frac{657}{5}$

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(1)

$f' (x) = 3 x^{2} + 2 x f' (1) + 2 f'' (2)$

$f'' (x) = 6 x + 2 f' (1)$

$f'' (2) = 12 + 2 f' (1)$

$∴$ $f' (x) = 3 x^{2} + 2 x f' (1) + 2 (12 + 2 f' (1))$

$f' (x) = 3 x^{2} + 2 (x + 2) f' (1) + 24$

Putting $x = 1$

$f' (1) = 3 + 6 f' (1) + 24$

$- 5 f' (1) = 27 \Rightarrow f' (1) = \frac{- 27}{5}$

$∴$ $f'' (2) = 12 + 2 (\frac{- 27}{5}) = 12 - \frac{54}{5} = \frac{6}{5}$

$∴$ $f' (x) = 3 x^{2} - \frac{54}{5} x + \frac{12}{5}$

$∴$ $f' (5) = 75 - 54 + \frac{12}{5} = \frac{117}{5}$

Q 34 :

If $y = y (x)$ satisfies the differential equation $16 (\sqrt{x + 9 \sqrt{x}}) (4 + \sqrt{9 + \sqrt{x}}) \cos y d y = (1 + 2 \sin y) d x, x > 0$ and $y (256) = \frac{π}{2}, y (49) = α$ , then $2 \sin α$ is equal to_____. [2026]

$3 (\sqrt{2} - 1)$
$\sqrt{2} - 1$
$2 (\sqrt{2} - 1)$
$2 \sqrt{2} - 1$

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(4)

$\int \frac{\cos y}{1 + 2 \sin y} d y = \int \frac{d x}{16 (\sqrt{9 \sqrt{x} + x}) (4 + \sqrt{9 + \sqrt{x}})}$

$4 + \sqrt{9 + \sqrt{x}} = t$

$\frac{1}{2 \sqrt{9 + \sqrt{x}}} \times \frac{d x}{2 \sqrt{x}} = 1 d x$

$\frac{1}{2} l n$ $| 1 + 2 \sin y |$ $= \int \frac{4 d t}{16 t} + C$

$\frac{1}{2} l n$ $| 1 + 2 \sin y |$ $= \frac{1}{4} l n | 4 + \sqrt{9 + \sqrt{x}} | + C$

$\frac{1}{2} l n (2 \sin y + 1) = \frac{1}{4} l n | 4 + \sqrt{9 + \sqrt{x}} | + C$

$Substituting (256, \frac{π}{2})$

$\frac{1}{2} l n 3 = \frac{1}{2} l n 3 + C$ $C = 0$

$Substituting (49, α)$

$\frac{1}{2} l n (2 \sin α + 1) = \frac{1}{4} l n 8$

$l n (2 \sin α + 1) = \frac{1}{2} l n 8$

$l n (2 \sin α + 1) = l n (2 \sqrt{2})$

$2 \sin α + 1 = 2 \sqrt{2}$

$2 \sin α = 2 \sqrt{2} - 1$

Q 35 :

Let y = y(x) be the solution of the differential equation $x \frac{d y}{d x} - \sin 2 y = x^{3} (2 - x^{3}) \cos^{2} y, x \neq 0 .$ If y(2) = 0, then $\tan (y (1))$ is equal to [2026]

$\frac{3}{4}$
$- \frac{3}{4}$
$\frac{7}{4}$
$- \frac{7}{4}$

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(3)

$x \frac{d y}{d x} - \sin 2 y = x^{3} (2 - x^{3}) \cos^{2} y$

$\sec^{2} y \frac{d y}{d x} - 2 \tan y \cdot \frac{1}{x} = x^{2} (2 - x^{3})$

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

$\frac{d t}{d x} - \frac{2 t}{x} = x^{2} (2 - x^{3}) (LDE)$

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

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

$\frac{\tan y}{x^{2}} = 2 x - \frac{x^{4}}{4} + C$

$y (2) = 0 \Rightarrow 0 = 4 - 4 + C \Rightarrow C = 0$

$\tan y = 2 x^{3} - \frac{1}{4} x^{6}$

$x = 1 \Rightarrow \tan y = 2 - \frac{1}{4} = \frac{7}{4}$

Q 36 :

The solution of the differential equation $(1 - x y - x^{5} y^{5}) d x - x^{2} (x^{4} y^{4} + 1) d y = 0$ given by (c is arbitrary constant)

$x = c e^{x y + \frac{1}{5} x^{5} y^{5}}$
$x = c e^{x y - \frac{1}{5} x^{5} y^{5}}$
$x = c e^{x^{2} y^{2} + \frac{1}{5} x^{5} y^{5}}$
$x = c e^{x^{2} y^{2} - \frac{1}{5} x^{5} y^{5}}$

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(1)

$The given equation is d x - x (y d x + x d y) = x^{5} y^{4} (y d x + x d y)$

$\Rightarrow \frac{d x}{x} = (1 + x^{4} y^{4}) d (x y)$ $\Rightarrow \ln x = x y + \frac{1}{5} x^{5} y^{5} + \ln c$

$\Rightarrow x = c e^{x y + \frac{1}{5} x^{5} y^{5}}$

Q 37 :

Let $y = f (x)$ satisfy the differential equation $\frac{d y}{d x} = 2 x e^{- y}, \forall x \in R$ . If $y' (1) = 1$ then the number of solutions of the equation $f (x) = f' (x)$ in $(0, \infty)$ is

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zero

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(1)

$\int e^{y} d y = \int 2 x d x \Rightarrow f (x) = \ln (x^{2} + c)$

$f' (1) = 1 \Rightarrow C = 1$

Q 38 :

Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} + 2 y = f (x)$ , where $f (x) =$ ${\begin{matrix} 1, & x \in [0, 1] \\ 0, & otherwise \end{matrix}$ . If $y (0) = 0$ , then $y (\ln 2)$ is

$\frac{e^{2} - 1}{4 e^{2}}$
$\frac{e^{2} - 1}{2 e^{2}}$
$\frac{1}{4}$
$\frac{3}{8}$

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(4)

$When x \in [0, 1], then \frac{d y}{d x} + 2 y = 1 \Rightarrow y = \frac{1}{2} + C_{1} e^{- 2 x}$

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

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

$When x \notin [0, 1] then \frac{d y}{d x} + 2 y = 0 \Rightarrow C_{2} = \frac{e^{2} - 1}{2}$

$∵$ $y (1) = \frac{e^{2} - 1}{2} \Rightarrow \frac{e^{2} - 1}{2} = C_{2} e^{- 2} \Rightarrow C_{2} = \frac{e^{2} - 1}{2}$

$∵$ $y (x) = (\frac{e^{2} - 1}{2}) e^{- 2 x} \Rightarrow y (\frac{3}{2}) = \frac{e^{2} - 1}{2 e^{3}}$

Q 39 :

$The Solution of the differential equation (1 + x^{3}) \frac{d y}{d x} + 3 x^{2} y + \cos^{2} x - 1 = 0$

$y (1 + x^{3}) = \frac{x}{2} - \frac{1}{4} \sin 2 x + C$
$y (1 + x^{3}) = \frac{x}{2} + \frac{1}{4} \sin 2 x + C$
$y (1 + x^{3}) = \frac{x}{4} - \frac{1}{2} \sin 2 x + C$
$y (1 - x^{3}) = \frac{x}{2} - \frac{1}{4} \sin 2 x + C$

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(1)

$I . F = e^{\int (\frac{3 x^{2}}{1 + x^{3}}) d x} = 1 + x^{3}$

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

$y (1 + x^{3}) = \frac{x}{2} - \frac{1}{4} \sin 2 x + C$

Q 40 :

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

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

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(1)

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

$Where, y (0) = 1, y (2 \sqrt{2}) = β$

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

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

$Divide by y^{2} both sides and proceed.$

Q 41 :

Let the solution curve $y = f (x)$ of the differential equation $\frac{d y}{d x} + \frac{x y}{x^{2} - 1} = \frac{15 x^{4} + 4 x^{3} + 4 x + 1}{\sqrt{1 - x^{2}}}, x \in (- 1, 1)$ passes through the origin. Then $\int_{- 1 / 2}^{1 / 2} \frac{f (x) d x}{x^{2} + 2}$ is equal to.

$\frac{π}{6}$
$\frac{π}{6} - \frac{\sqrt{3}}{4}$
$\frac{π}{3} - \frac{\sqrt{3}}{4}$
$\frac{π}{3} - \frac{1}{4}$

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(2)

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

$y = f (x) = \frac{3 x^{5} + x^{4} + 2 x^{2} + x}{\sqrt{1 - x^{2}}}$

$= \int_{- 1 / 2}^{1 / 2} \frac{f (x)}{(x^{2} + 2)} d x = \frac{π}{6} - \frac{1}{4}$

Q 42 :

The function $y = f (x)$ is the solution of the differential equation $\frac{d y}{d x} + \frac{x y}{x^{2} - 1} = \frac{x^{4} + 2}{\sqrt{1 - x^{2}}}$ in $(- 1, 1)$ satisfying $f (0) = 0$ . Then $2 \int_{- \sqrt{3} / 2}^{\sqrt{3} / 2} f (x) d x$ is

$\frac{2 π}{3} + \frac{\sqrt{3}}{2}$
$\frac{2 π}{3} - \frac{\sqrt{3}}{2}$
$\frac{π}{6} - \frac{\sqrt{3}}{4}$
$\frac{π}{6} - \frac{\sqrt{3}}{2}$

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(2)

$\frac{d y}{d x} + \frac{x}{x^{2} - 1} y = \frac{x^{4} + 2 x}{\sqrt{1 - x^{2}}}$

$The I.F. of this differential equation is$ $e^{\int \frac{x}{x^{2} - 1} d x} = e^{- \int \frac{x}{1 - x^{2}} d x} = e^{\frac{1}{2} \log (1 - x^{2})} = \sqrt{1 - x^{2}}$

$The solution is given by$ $y \sqrt{1 - x^{2}} = \int \frac{x^{3} + 2}{\sqrt{1 - x^{2}}} \sqrt{1 - x^{2}} d x + λ$

$= \int (x^{4} + 2 x) d x + λ$ $= \frac{x^{5}}{5} + x^{2} + λ$

$At y (0) = 0 \Rightarrow λ = 0$ $\Rightarrow y \sqrt{1 - x^{2}} = \frac{x^{5}}{5} + x^{2}$

$\int_{- \sqrt{3} / 2}^{\sqrt{3} / 2} \frac{\frac{x^{5}}{5} + x^{2}}{\sqrt{1 - x^{2}}} d x = \int_{- \sqrt{3} / 2}^{\sqrt{3} / 2} \frac{x^{2}}{\sqrt{1 - x^{2}}} d x$

$(The other part is odd)$ $= 2 \int_{0}^{\sqrt{3} / 2} \frac{x^{2}}{\sqrt{1 - x^{2}}} d x$

$Let x = \sin θ, we get$

$I = 2 \int_{0}^{π / 3} \frac{\sin^{2} θ}{\cos θ} \cos θ d θ$

$= 2 \int_{0}^{π / 3} \sin^{2} θ d θ$

$= \int_{0}^{π / 3} (1 - \cos 2 θ) d θ$

$= {[θ - \frac{\sin 2 θ}{2}]}_{0}^{π / 3}$

$= \frac{π}{3} - \frac{\sqrt{3}}{4}$

$∴$ $2 I = \frac{2 π}{3} - \frac{\sqrt{3}}{2}$

Q 43 :

If $\frac{d y}{d x} = \frac{2^{x + y} - 2^{x}}{2^{y}},$ $y (0) = 1$ , then $y (1) =$

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

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(2)

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

$\Rightarrow \frac{\log (2^{y} - 1)}{\log 2} = \frac{2^{x}}{\log 2} + c, y (0) = 1 \Rightarrow 0 = \frac{1}{\log 2} + c \Rightarrow c = - \frac{1}{\log 2}$

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

If $x = 1$ than $\log (2^{y} - 1) = 2 - 1 = 1 \Rightarrow 2^{y} - 1 = e \Rightarrow 2^{y} = e + 1 \Rightarrow y = \log_{2} (e + 1)$

$\Rightarrow y (1) = \log_{2} (e + 1)$

Q 44 :

In column I differential equations are given in column II their solutions are given then match them

Column I Column II

A) $y \frac{d y}{d x} + x = 0$ p) $2 y = x^{2} + c$

B) $x \frac{d y}{d x} + y = 0$ q) $x^{2} + y^{2} = c$

C) $\frac{d y}{d x} = x$ r) $x y = c$

D) $\frac{d x}{d y} = y$ s) $y^{2} = 2 x + c$

A – r, B – q, C – p, D – s
A – q, B – r, C – p, D – s
A – s, B – q, C – p, D – r
A – p, B – q, C – r, D – s

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Use variable seperable

Topic Question Set

Q 31 :

Let the solution curve of the differential equation $x d y - y d x = \sqrt{x^{2} + y^{2}} d x, x > 0,$ $y (1) = 0$ be $y = y (x)$ . Then $y (3)$ is equal to [2026]

Q 32 :

Let $y = y (x)$ be the solution of the differential equation $\sec x \frac{d y}{d x} - 2 y = 2 + 3 \sin x, x \in (- \frac{π}{2}, \frac{π}{2}),$ $y (0) = - \frac{7}{4} .$ Then $y (\frac{π}{6})$ is equal to: [2026]

Q 33 :

Let $f (x) = x^{3} + x^{2} f' (1) + 2 x f'' (2) + f''' (3), x \in R .$ Then the value of $f' (5)$ is [2026]

Q 34 :

If $y = y (x)$ satisfies the differential equation $16 (\sqrt{x + 9 \sqrt{x}}) (4 + \sqrt{9 + \sqrt{x}}) \cos y d y = (1 + 2 \sin y) d x, x > 0$ and $y (256) = \frac{π}{2}, y (49) = α$ , then $2 \sin α$ is equal to_____. [2026]

Q 35 :

Let y = y(x) be the solution of the differential equation $x \frac{d y}{d x} - \sin 2 y = x^{3} (2 - x^{3}) \cos^{2} y, x \neq 0 .$ If y(2) = 0, then $\tan (y (1))$ is equal to [2026]

Q 36 :

The solution of the differential equation $(1 - x y - x^{5} y^{5}) d x - x^{2} (x^{4} y^{4} + 1) d y = 0$ given by (c is arbitrary constant)

Q 37 :

Let $y = f (x)$ satisfy the differential equation $\frac{d y}{d x} = 2 x e^{- y}, \forall x \in R$ . If $y' (1) = 1$ then the number of solutions of the equation $f (x) = f' (x)$ in $(0, \infty)$ is

Q 38 :

Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} + 2 y = f (x)$ , where $f (x) =$ ${\begin{matrix} 1, & x \in [0, 1] \\ 0, & otherwise \end{matrix}$ . If $y (0) = 0$ , then $y (\ln 2)$ is

Q 39 :

$The Solution of the differential equation (1 + x^{3}) \frac{d y}{d x} + 3 x^{2} y + \cos^{2} x - 1 = 0$

Q 40 :

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

Q 41 :

Let the solution curve $y = f (x)$ of the differential equation $\frac{d y}{d x} + \frac{x y}{x^{2} - 1} = \frac{15 x^{4} + 4 x^{3} + 4 x + 1}{\sqrt{1 - x^{2}}}, x \in (- 1, 1)$ passes through the origin. Then $\int_{- 1 / 2}^{1 / 2} \frac{f (x) d x}{x^{2} + 2}$ is equal to.

Q 42 :

The function $y = f (x)$ is the solution of the differential equation $\frac{d y}{d x} + \frac{x y}{x^{2} - 1} = \frac{x^{4} + 2}{\sqrt{1 - x^{2}}}$ in $(- 1, 1)$ satisfying $f (0) = 0$ . Then $2 \int_{- \sqrt{3} / 2}^{\sqrt{3} / 2} f (x) d x$ is

Q 43 :

If $\frac{d y}{d x} = \frac{2^{x + y} - 2^{x}}{2^{y}},$ $y (0) = 1$ , then $y (1) =$

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	Column I		Column II
A)	$y \frac{d y}{d x} + x = 0$	p)	$2 y = x^{2} + c$
B)	$x \frac{d y}{d x} + y = 0$	q)	$x^{2} + y^{2} = c$
C)	$\frac{d y}{d x} = x$	r)	$x y = c$
D)	$\frac{d x}{d y} = y$	s)	$y^{2} = 2 x + c$

Topic Question Set

Q 31 : Let the solution curve of the differential equation x dy-y dx=x2+y2 dx, x>0, y(1)=0 be y=y(x). Then y(3) is equal to [2026]

Q 32 : Let y=y(x) be the solution of the differential equation secxdydx-2y=2+3sinx, x∈(-π2,π2), y(0)=-74.Then y(π6) is equal to: [2026]

Q 33 : Let f(x)=x3+x2f'(1)+2xf''(2)+f'''(3), x∈R. Then the value of f'(5) is [2026]

Q 34 : If y=y(x) satisfies the differential equation 16(x+9x)(4+9+x)cosy dy=(1+2siny) dx, x>0 and y(256)=π2, y(49)=α, then 2 sinα is equal to_____. [2026]

Q 35 : Let y = y(x) be the solution of the differential equation xdydx-sin2y=x3(2-x3)cos2y, x≠0. If y(2) = 0, then tan(y(1)) is equal to [2026]

Q 36 : The solution of the differential equation (1-xy-x5y5)dx-x2(x4y4+1)dy=0 given by (c is arbitrary constant)

Q 37 : Let y=f(x) satisfy the differential equation dydx=2xe-y, ∀x∈R. If y'(1)=1 then the number of solutions of the equation f(x)=f'(x) in (0,∞) is

Q 38 : Let y=y(x) be the solution of the differential equation dydx+2y=f(x), where f(x)={1,x∈[0,1]0,otherwise. If y(0)=0, then y(ln2) is

Q 39 : The Solution of the differential equation (1+x3)dydx+3x2y+cos2x-1=0

Q 40 : 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

Q 41 : Let the solution curve y=f(x) of the differential equation dydx+xyx2-1=15x4+4x3+4x+11-x2, x∈(-1,1) passes through the origin. Then ∫-1/21/2f(x) dxx2+2 is equal to.

Q 42 : The function y=f(x) is the solution of the differential equation dydx+xyx2-1=x4+21-x2 in (-1,1) satisfying f(0)=0. Then 2∫-3/23/2f(x) dx is

Q 43 : If dydx=2x+y-2x2y, y(0)=1, then y(1)=

Q 44 : In column I differential equations are given in column II their solutions are given then match them Column I Column II A) ydydx+x=0 p) 2y=x2+c B) xdydx+y=0 q) x2+y2=c C) dydx=x r) xy=c D) dxdy=y s) y2=2x+c

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Q 31 :

Let the solution curve of the differential equation $x d y - y d x = \sqrt{x^{2} + y^{2}} d x, x > 0,$ $y (1) = 0$ be $y = y (x)$ . Then $y (3)$ is equal to [2026]

Q 32 :

Let $y = y (x)$ be the solution of the differential equation $\sec x \frac{d y}{d x} - 2 y = 2 + 3 \sin x, x \in (- \frac{π}{2}, \frac{π}{2}),$ $y (0) = - \frac{7}{4} .$ Then $y (\frac{π}{6})$ is equal to: [2026]

Q 33 :

Let $f (x) = x^{3} + x^{2} f' (1) + 2 x f'' (2) + f''' (3), x \in R .$ Then the value of $f' (5)$ is [2026]

Q 34 :

If $y = y (x)$ satisfies the differential equation $16 (\sqrt{x + 9 \sqrt{x}}) (4 + \sqrt{9 + \sqrt{x}}) \cos y d y = (1 + 2 \sin y) d x, x > 0$ and $y (256) = \frac{π}{2}, y (49) = α$ , then $2 \sin α$ is equal to_____. [2026]

Q 35 :

Let y = y(x) be the solution of the differential equation $x \frac{d y}{d x} - \sin 2 y = x^{3} (2 - x^{3}) \cos^{2} y, x \neq 0 .$ If y(2) = 0, then $\tan (y (1))$ is equal to [2026]

Q 36 :

The solution of the differential equation $(1 - x y - x^{5} y^{5}) d x - x^{2} (x^{4} y^{4} + 1) d y = 0$ given by (c is arbitrary constant)

Q 37 :

Let $y = f (x)$ satisfy the differential equation $\frac{d y}{d x} = 2 x e^{- y}, \forall x \in R$ . If $y' (1) = 1$ then the number of solutions of the equation $f (x) = f' (x)$ in $(0, \infty)$ is

Q 38 :

Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} + 2 y = f (x)$ , where $f (x) =$ ${\begin{matrix} 1, & x \in [0, 1] \\ 0, & otherwise \end{matrix}$ . If $y (0) = 0$ , then $y (\ln 2)$ is

Q 39 :

$The Solution of the differential equation (1 + x^{3}) \frac{d y}{d x} + 3 x^{2} y + \cos^{2} x - 1 = 0$

Q 40 :

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

Q 41 :

Let the solution curve $y = f (x)$ of the differential equation $\frac{d y}{d x} + \frac{x y}{x^{2} - 1} = \frac{15 x^{4} + 4 x^{3} + 4 x + 1}{\sqrt{1 - x^{2}}}, x \in (- 1, 1)$ passes through the origin. Then $\int_{- 1 / 2}^{1 / 2} \frac{f (x) d x}{x^{2} + 2}$ is equal to.

Q 42 :

The function $y = f (x)$ is the solution of the differential equation $\frac{d y}{d x} + \frac{x y}{x^{2} - 1} = \frac{x^{4} + 2}{\sqrt{1 - x^{2}}}$ in $(- 1, 1)$ satisfying $f (0) = 0$ . Then $2 \int_{- \sqrt{3} / 2}^{\sqrt{3} / 2} f (x) d x$ is

Q 43 :

If $\frac{d y}{d x} = \frac{2^{x + y} - 2^{x}}{2^{y}},$ $y (0) = 1$ , then $y (1) =$