Solution of differential equation dy dx = 1 xy( x^2 y^2 + 1) (where C is integral constant)

A. e^ y^2 ( 1 x^2 - y^2 2 + y^2 2 ) = C

Solution of differential equation dy dx = 1 xy( x^2 y^2 + 1) (whe…

MCQ

Solution of differential equation $\frac{{dy}}{{dx}} = \frac{1}{{xy({x^2}\sin {y^2} + 1)}}$ (where $C$ is integral constant)

✓
${e^{{y^2}}}\left( {\frac{1}{{{x^2}}} - \frac{{\cos {y^2}}}{2} + \frac{{\sin {y^2}}}{2}} \right) = C$
B
${e^{{y^2}}}\left( {\frac{1}{{{x^2}}} + \frac{{\cos {y^2}}}{2} - \frac{{\sin {y^2}}}{2}} \right) = C$
C
${e^{{y^2}}}\left( {\frac{1}{{{x^2}}} - \frac{{\cos {y^2}}}{2} + \frac{{{{\sin }^2}y}}{2}} \right) = C$
D
${e^{{y^2}}}\left( {\frac{1}{{{x^2}}} - \frac{{\cos y}}{2} + \frac{{\sin y}}{2}} \right) = C$

✓

Correct option: A.

${e^{{y^2}}}\left( {\frac{1}{{{x^2}}} - \frac{{\cos {y^2}}}{2} + \frac{{\sin {y^2}}}{2}} \right) = C$

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

$2 \frac{d t}{d x}=\frac{1}{x\left(x^{2} \sin t+1\right)}$

$\frac{d x}{d t}=\frac{x^{3} \sin t}{2}+\frac{x}{2}$

$x^{-3} \frac{d x}{d t}=\frac{\sin t}{2}+\frac{x^{-2}}{2}$

Let $x^{-2}=z \Rightarrow x^{-3} \frac{d x}{d t}=-\frac{d z}{2 d t}$

$-\frac{1}{2} \frac{d z}{d t}=\frac{\sin t}{2}+\frac{z}{2}$

$\frac{d z}{d t}=-\sin t-z$

$\frac{d z}{d t}+z=-\sin t$

$\mathrm{ze}^{\mathrm{t}}=-\int \mathrm{e}^{\mathrm{t}} \sin \mathrm{t} d \mathrm{t}$

$\mathrm{ze}^{\mathrm{t}}=\frac{\left(-\mathrm{e}^{\mathrm{t}} \cos \mathrm{t}+\mathrm{e}^{\mathrm{t}} \sin \mathrm{t}\right)}{2}+\mathrm{C}$

$\frac{e^{y^{2}}}{x^{2}}=e^{t} \frac{(\cos t-\sin t)}{2}+C$

$e^{y^{2}}\left(\frac{1}{x^{2}}-\frac{\cos t}{2}+\frac{\sin t}{2}\right) \pm C$

$e^{y^{2}}\left(\frac{1}{x^{2}}-\frac{\cos y^{2}}{2}+\frac{\sin y^{2}}{2}\right)=C$

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.