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Differential Equations — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations3 Marks
Question
Find the general solution of $y d x+\left(x-y^{2}\right) d y=0$
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Answer

It is given that $\mathrm{ydx}+\left(\mathrm{x}-\mathrm{y}^{2}\right) \mathrm{dy}=0$ 
$\Rightarrow \mathrm{ydx}=\left(\mathrm{y}^{2}-\mathrm{x}\right) \mathrm{dy}$ 
$\Rightarrow \frac{d x}{d y}=\frac{\left(y^{2}-x\right)}{y}=y-\frac{x}{y}$ 
$\Rightarrow \frac{d x}{d y}+\frac{x}{y}=y$ 
This is equation in the form of $\frac{d x}{d y}+p x=Q$ (where, p = $\frac{1}{y}$ and Q = y)
Now, I.F. = $\mathrm{e}^{\int \mathrm{pdy}}=\mathrm{e}^{\int \frac{\mathrm{dy}}{\mathrm{y}}}=\mathrm{e}^{\log \mathrm{y}}=\mathrm{y}$ 
Thus, the solution of the given differential equation is given by the relation:
$x(I . F .)=\int(Q \times I . F .) d y+C$ 
$\Rightarrow \mathrm{x} \cdot \mathrm{y}=\int[\mathrm{y} \cdot \mathrm{y}] \mathrm{d} \mathrm{y}+\mathrm{C}$ 
$\Rightarrow x . y=\int y^{2} d y+C$ 
$\Rightarrow x \cdot y=\frac{y^{3}}{3}+C$ 
$\Rightarrow x =\frac{y^{2}}{3}+\frac{C}{y}$ 
Therefore, the required general solution of the given differential equation is $x =\frac{y^{2}}{3}+\frac{c}{y}$.

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