solve
$\displaystyle y'=\frac{y}{x} +2x^{2} $, y(1)=0
really helpful if someone could show me full work through i know they're not hard but need gd understanding thanks
$\displaystyle \begin{aligned}
y'&=\frac{y}{x}+2x^{2} \\
y'-\frac{y}{x}&=2x^{2} \\
\frac{y'}{x}-\frac{y}{x^{2}}&=2x \\
\left\{ y\cdot \frac{1}{x} \right\}'&=2x.
\end{aligned}$
In the third step I found the "integrating factor." Do you know what the I.F. is?
When you put your equation in standard form
$\displaystyle y'+a(x)y=f(x)$
$\displaystyle y'=\frac{y}{x} +2x^{2}$
you need to move the $\displaystyle \frac{y}{x}$ term to the left side of the equation
$\displaystyle y'-\frac{y}{x}=2x^2$
Good luck