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Q.
Solution of the differential equation $x^2 y d x=\left(x^3+y^3\right)$ dy with $y(0)=1$ is given by
Differential Equations
Solution:
$ \Theta x^2 y d x=\left(x^3+y^3\right) d y=x^3 d y+y^3 d y$ $\Rightarrow x^2(y d x-x d y)=y^3 d y$
$\Rightarrow \frac{x^2}{y^2} d\left(\frac{x}{y}\right)=\frac{d y}{y}$
$\Rightarrow \frac{ x ^3}{3 y ^3}=\ln y + C$
$\Theta x =0, y =1$
$\therefore 0=0+ c \Rightarrow c =0$
$\therefore x ^3=3 y ^3 \ln y $