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Differential Equation and Applications (p-1) — Maths (commerce) STD 12 Commerce / Arts — Question

Maharashtra BoardEnglish MediumSTD 12 Commerce / ArtsMaths (commerce)Differential Equation and Applications (p-1)3 Marks
Question
Solve the following differential equations : $\left(x^2-y x^2\right) d y+\left(y^2+x y^2\right) d x=0$
✓

Answer

$
\begin{aligned}
& \left( x ^2- yx ^2\right) d y +\left( y ^2+ xy ^2\right) dx =0 \\
& \therefore x ^2(1- y ) d y + y ^2(1+ x ) d x =0 \\
& \therefore \frac{1-y}{y^2} d y+\frac{1+x}{x^2} d x=0
\end{aligned}
$
Integrating, we get
$
\begin{aligned}
& \int \frac{1-y}{y^2} d y+\int \frac{1+x}{x^2} d x=c \\
& \therefore \int\left(\frac{1}{y^2}-\frac{1}{y}\right) d y+\int\left(\frac{1}{x^2}+\frac{1}{x}\right) d x=c \\
& \therefore \int y^{-2} d y-\int \frac{1}{y} d y+\int x^{-2} d x+\int \frac{1}{x} d x=c \\
& \therefore \frac{y^{-1}}{-1}-\log |y|+\frac{x^{-1}}{-1}+\log |x|=c \\
& \therefore-\frac{1}{y}-\log |y|-\frac{1}{x}+\log |x|=c
\end{aligned}
$
$
\therefore \log |x|-\log |y|=\frac{1}{x}+\frac{1}{y}+c
$
This is the general solution.

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