Laplace Differential Equation: $ty'' - y' = 2t^2$

Question

Solve $ty'' - y' = 2t^2$ using Laplace Transforms.

I took the Laplace transform of both sides. This yielded me in the end:

$$\frac3s \,Y(s) +\frac{d}{ds}\,Y(s) = -\frac4{s^5}$$

Is there anyway I can group the Y(s) by itself and then take the inverse Laplace? If so how? Also, can you do this without integrating factor?

Answer 1

HINT: Observe that $$ \mathscr L\{ty''\}=-\frac{d}{ds}L\{y''\}=-\frac{d}{ds}(s^2Y (s)-sy (0)-y'(0))=-2sY (s)-s^2Y'(s)-y (0)=-2sY (s)-s^2Y'(s) $$ and $$ \mathscr L\{y'\}=sY (s)-y (0)=sY (s) $$ so we don't need $y'(0)$, but only $y (0) =0$.

The equation becomes $$ -2sY (s)-s^2Y'(s)-sY (s)=\frac {4}{s^3} $$ that is $$ 3s^4Y (s)+s^5Y'(s)+4=0 $$ This is first order linear ODE with solution $$ Y (s)=\frac {c}{s^3}+\frac {4}{s^4} $$ where $c $ is a constant (to be determined).

Answer 2

Your problem statement (as pointed out in the comments) should also have the condition of $y(0)=0$. I will answer your bolded question, but just to be thorough, I will show the solution to where you got incase anyone else had questions about it.

Using a Laplace transform, solve: $ty'' - y' = 2t^2$ ; $y(0)=0$

$\mathscr L\{ty''\} - \mathscr L\{y'\} = 2\mathscr L\{t^2\}$

You can use a theorem that states: $\mathscr L\{t^nf(t)\}$ = $(-1)^n \frac {d^n}{ds^n}F(s)$

Remember that derivatives are still functions of their independent variables (here t is the independent variable).

Take the Laplace transforms of the above:

$(-1)\frac {d}{ds}[s^2Y(s) - sy(0) - y'(0)] - [sY(s) - y(0)] = 2(\frac {2}{s^3})$

We know from the information given that $y(0) = 0$. We also know that $y'(0) = 0$ since the derivative of $y(0)$ would also be $0$. This simplifies what we have to:

$(-1)\frac {d}{ds}[s^2Y(s)] - [sY(s)] = 2(\frac {2}{s^3})$

Now we just take the derivative:

$(-1)[2sY(s) + s^2Y'(s)] - sY(s) = \frac {4}{s^3}$

Simplify:

$-s^2Y'(s) - 3sY(s) = \frac {4}{s^3}$

Put into Standard Form:

$Y'(s) + \frac{3}{s}Y(s) = -\frac{4}{s^5}$

So, this is a first order, linear, nonhomogeneous, variable coefficient DE. It doesn't provide us with many options to solve since we are not dealing with constant coefficients. It is also not a separable equation. You can use a power series solution method to solve. You can also use the method of Variation of Parameters, although that is built using the method of the integrating factor. Using the integrating factor would be the simplest method from this point. I know your instructions were to use the Laplace Transform to solve the original DE, but you should also note that $ty'' - y' = 2t^2$ ; $y(0)=0$ is in the form of an Euler-Cauchy DE. You could solve the DE using the E-C method from the start if you didn't want to use integrating factors.

However, since we are in standard form I will use the integrating factor method because the power series solution is tedious and not necessary here.

From the Standard Form: $P(s) = \frac{3}{s} , f(s) = -\frac{4}{s^5}$

Integrating factor: $e^{\int P(s)ds}$ = $e^{3ln(s)}$ = $s^3$

$\frac{d}{ds}[s^3Y(s)]= -4(\frac{1}{s^5})(s^3)$

$\int d[s^3Y(s)] = -4\int \frac{1}{s^2}ds$

$s^3Y(s) = 4(\frac{1}{s}) + c$

$Y(s) = 4(\frac{1}{s^4}) + c(\frac{1}{s^3})$

Take the Inverse Laplace Transform:

$\mathscr{L}^{-1}\{Y(s)\} = 4\mathscr{L}^{-1}\{\frac{1}{s^4}\} + c\mathscr{L}^{-1}\{\frac{1}{s^3}\}$

$y(t) = 4(\frac{1}{3!})\mathscr{L}^{-1}\{\frac{3!}{s^4}\} + c(\frac{1}{2!})\mathscr{L}^{-1}\{\frac{2!}{s^3}\}$

$y(t) = \frac{2}{3}t^3 + \frac{c}{2}t^2$

Which simplifies to:

$y(t) = \frac{2}{3}t^3 + c_1t^2$