QJEEM22Ma2d05: JEE Main 2022 Mathematics - Class 12 JEEM22Ma2d05 Solution: The general solution of the differential equation is :

Question

Verified Solution • 2025–26 Session moderate

The general solution of the differential equation $\left(x-y^{2}\right) \mathrm{d} x+y\left(5 x+y^{2}\right) \mathrm{d} y=0$ is :

Accepted Answer

\left(x-y^{2}\right) d x+y\left(5 x+y^{2}\right) d y=0

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

Let

y^{2}=t

\frac{1}{2} \cdot \frac{d t}{d x}=\frac{t-x}{5 x+t}

Now substitute,

t=v x

\begin{aligned}&amp; \frac{d t}{d x}=v+x \frac{d v}{d x} \\\\&amp; \frac{1}{2}\left\{v+x \frac{d v}{d x}\right\}=\frac{v-1}{5+v} \\\\&amp; x \frac{d v}{d x}=\frac{2 v-2}{5+v}-v=\frac{-3 v-v^{2}-2}{5+v} \\\\&amp; \int \frac{5+v}{v^{2}+3 v+2} d v=\int-\frac{d x}{x} \\\\&amp; \int \frac{4}{v+1} d v-\int \frac{3}{v+2} d v=-\int \frac{d x}{x} \\\\&amp; 4 \ln |v+1|-3 \ln |v+2|=-\ln x+\ln C \\\\&amp; \left|\frac{(v+1)^{4}}{(v+2)^{3}}\right|=\frac{c}{x} \\\\&amp; \left| \frac{\left(\frac{y^{2}}{x}+1\right)^{4}}{\left(\frac{y^{2}}{x}+2\right)^{3}}\right|=\frac{c}{x} \\\\&amp; \left|\left(y^{2}+x\right)^{4}\right|=C\left|\left(y^{2}+2 x\right)^{3}\right|\end{aligned}

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