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

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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)

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

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

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

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

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

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

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

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

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

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

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]

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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]

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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]