Solve the following differential equations : y d x+ (x-y^2 ) d y=… - Vidyadip

Question

Solve the following differential equations : $y d x+\left(x-y^2\right) d y=0$

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Answer

$
\begin{aligned}
& yd x +\left( x - y ^2\right) dy =0 \\
& \therefore y dx =-\left( x - y ^2\right) d y \\
& \therefore \frac{d x}{d y}=-\frac{\left(x-y^2\right)}{y}=-\frac{x}{y}+y \\
& \therefore \frac{d x}{d y}+\left(\frac{1}{y}\right) \cdot x=y \ldots \ldots .(1)
\end{aligned}
$
This is the linear differential equation of the form
$
\begin{aligned}
& \frac{d x}{d y}+P \cdot x=Q, \text { where } P=\frac{1}{y} \text { and } Q=y \\
& \therefore \text { I.F. }=e^{\int P d y}=e^{\int \frac{1}{y} d y}=e^{\log y}=y_1
\end{aligned}
$
$\therefore$ the solution of (1) is given by
$
\begin{aligned}
& x \text {.(I.F.) }=\int Q \cdot\left(\text { I.F.) } d y+c_1\right. \\
& \therefore x y=\int y \cdot y d y+c_1 \\
& \therefore x y=\int y^2 d y+c_1 \\
& \therefore x y=\frac{y^3}{3}+c_1 \\
& \therefore 3 x y=y^3+3 c_1 \\
& \therefore 3 x y=y^2+c, \text { where } c=3 c_1
\end{aligned}
$
This is the general solution.

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