Solve the following differential equations : x^2 y d x- (x^3+y^3 …

Question

Solve the following differential equations : $x^2 y d x-\left(x^3+y^3\right) d y=0$

✓

Answer

$
\begin{aligned}
& x ^2 y dx -\left( x ^3+ y ^3\right) d y=0 \\
& \therefore\left( x ^3+ y ^3\right) d y= x ^2 y dx \\
& \therefore \frac{d y}{d x}=\frac{x^2 y}{x^3+y^3} \ldots \ldots .(1)
\end{aligned}
$
Put $y = vx$
$
\therefore \frac{d y}{d x}=v+x \frac{d v}{d x}
$
$\therefore$ (1) becomes, $v+x \frac{d v}{d x}=\frac{x^2 \cdot v x}{x^3+v^3 x^3}=\frac{v}{1+v^3}$
$
\therefore x \frac{d v}{d x}=\frac{v}{1+v^3}-v=\frac{v-v-v^4}{1+v^3}
$
$\therefore x \frac{d v}{d x}=\frac{-v^4}{1+v^3}$
$
\therefore \frac{1+v^3}{v^4} d v=-\frac{1}{x} d x
$
Integrating, we get
$
\begin{aligned}
& \int \frac{1+v^3}{v^4} d v=-\int \frac{1}{x} d x \\
\therefore & \int\left(\frac{1}{v^4}+\frac{1}{v}\right) d v=-\int \frac{1}{x} d x \\
\therefore & \int v^{-4} d v+\int \frac{1}{v} d v=-\int \frac{1}{x} d x \\
\therefore & \frac{v^{-3}}{-3}+\log |v|=-\log |x|+c_1 \\
\therefore & -\frac{1}{3 v^3}+\log |v|=-\log |x|+c_1
\end{aligned}
$
$
\begin{aligned}
& \therefore-\frac{1}{3} \cdot \frac{1}{\left(\frac{y}{x}\right)^3}+\log \left|\frac{y}{x}\right|=-\log |x|+c_1 \\
& \therefore-\frac{x^3}{3 y^3}+\log |y|-\log |x|=-\log |x|-\log c \\
& \text{where }c_1=-\log c, \\
& \therefore \frac{x^3}{3 y^3}=\log c+\log y \\
& \therefore \frac{x^3}{3 y^3}=\log |c y|
\end{aligned}
$
This is the general solution.