Need help with solving this question i have no idea how to even start so any help would be appreciated thanks
Ok, then here's how I'd do (A):
$\displaystyle xy'+2y=3x,\quad y(0)=y_0$ and dividing by $\displaystyle x$:
$\displaystyle y'+2/x y=3$ for which the integrating factor is $\displaystyle x^2$ so we have:
$\displaystyle x^2(y'+2/x y)=3x^2$ which the left side is the differential of $\displaystyle x^2y$ so:
$\displaystyle \int d(x^2y)=\int 3x^2$ or:
$\displaystyle y(x)=x+\frac{c}{x^2}$ and if $\displaystyle y_0=0$ then I can let $\displaystyle c=0$ and obtain the solution $\displaystyle y=x$. However, if $\displaystyle y_0\neq 0$ say $\displaystyle 3$ for example, then I'd obtain $\displaystyle 3=0+\frac{c}{0}$ which has no solution.
Ok, so I'm a little uneasy about dividing through by $\displaystyle x$ in the first step since this is valid if $\displaystyle x\ne 0$ then obtain the solution and solve it for $\displaystyle x=0$.