Question

Solve the following initial value problem by using Laplace transform 2d^2y/dt^2 +3 dy/dt - 2y=te^-2t, y(0)=0, y’(0)=-2

Answer 1

$2y^{\prime \prime}+3y^\prime -2y=te^{-2t},$ $y(0)=0,$ $y^\prime (0)=-2$

Using Laplace transform, we have

$2\mathcal{L}\{y^{\prime \prime}\}+3 \mathcal{L}\{ y^\prime \}-2 \mathcal{L}\{ y\}= \mathcal{L}\{te^{-2t}\}$

We will use the following formulas:

$\mathcal{L}\{y^{\prime \prime}\}= s^2Y(s)-sy(0)-y^\prime (0)$

$\mathcal{L}\{y^{\prime }\}= sY(s)-y(0)$

$\mathcal{L}\{y\}= Y(s)$

$\mathcal{L}\{t^ne^{at}\}= \frac{n!}{(s-a)^{n+1}}$

So, we have:

$2\left(s^2Y(s)-sy(0)-y^\prime (0) \right)+3\left(sY(s)-y(0) \right)-2Y(s)=\frac{1}{(s+2)^2}$

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

$Y(s)=\frac{\frac{1}{(s+2)^2}-4}{2s^2+3s-2}=\frac{-4s^2-16s-15}{(2s^2+3s-2)(s+2)^2}= \frac{-4s^2-16s-15}{(2s-1)(s+2)^3}=-\frac{192}{125(2s-1)}+\frac{96}{125(s+2)}-\frac{2}{25(s+2)^2}-\frac{1}{5(s+2)^3}$

Now we take the inverse transform:

$y(t)=-\frac{192}{125}e^{t/2}+\frac{96}{125}e^{-2t}-\frac{2}{25}te^{-2t}-\frac{1}{10}t^2e^{-2t}$

Answer: $y(t)=-\frac{192}{125}e^{t/2}+\frac{96}{125}e^{-2t}-\frac{2}{25}te^{-2t}-\frac{1}{10}t^2e^{-2t}$ .