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.

Answer 3

Apply the Laplace transform, and counting that $ \mathcal{L}\{y'\} = sY(s)-y(0)$ and $\mathcal{L}\{y''\} = s^2Y(s)-sy(0)-y'(0)$, where $Y(s) = \mathcal{L}\{y(t)\}(s).$

\begin{gather*} \mathcal{L}\{ty''\} - \mathcal{L}\{y'\} = \mathcal{L}\{t^2\} \\ -\frac{d}{ds}(s^2Y -sy(0) - y'(0)) - sY +y(0) = \frac{2!}{s^3}\\ -(2sY + s^2Y)-sY = \frac{2}{s^3} \\ Y(s^2 - 3s) = \frac{2}{s^3} \\ Y= \frac{2}{s^4(s-3)} \end{gather*}

Now, apply the Inverse Laplace Transform and looking the convolution product

\begin{align*} y &= 2 \mathcal{L}^{-1}\{ \frac{1}{s^4}\frac{1}{s-3} \} \\ &= 2( \frac{t^3}{3!} * e^{3t}) \\ &= \frac{1}{3} [ \int_0^t \Gamma^3e^{3(t-\Gamma)}d\Gamma] \\ &= \frac{1}{81}[2e^{3t} -9t^3 -9t^2 - 6t -2] \end{align*}

I think is correct the solution.