Laplace transform

Question

Hello how can I solve this problem $ty''-y'=t^2$, $y (0)=0 $

I though about several ways like $y''-y'/t =t$ but I did not know how to solve the integral of $( y'/t) $.

Then I thought about integration of $y''t$ which $= ty'-y+C$ but then I reach to $ty'-y+C=y+t^3/3$ but then what?

Answer 1

Try the substitution $y'=u$ so that $u'-\frac{u}{t}=t$ and dividing by $t$ $$ \frac{u'}{t}-\frac{u}{t^2}=1 $$ or $$ \left(\frac{u}{t}\right)'=1\quad \Longrightarrow\;\frac{u}{t}=t+A $$ and then $$u(t)=t^2+At$$ So from $y'=u$ you'll have $$ y(t)=\int u\mathrm dt=\frac{t^3}{3}+At^2+B $$ From $y(0)=0$, we have $B=0$
$$ y(t)=\frac{t^3}{3}+At^2 $$ To find $A$ you need another extra condition.

Answer 2

Do we need to use a laplace transform?

$t y'' -y' = t^2\\ v = y', v' = y''\\ tv' - v = t^2\\ v' - (1/t) v = t\\ e^{\int \frac{-1}{t} dt} = \frac{1}{t}\\ \frac{1}{t}v' - \frac{1}{t^2}v = 1\\ \frac{1}{t}y' = t + c $

And I will let you take it from here.