Question

solve the following initial value problem using laplace transforms 2d^2y/dt^3+3dy/dt-2y=e^-2t y=0,y*=-2

Answer 1

First apply the Laplace transform to both sides of the equation, we will note the Laplace transform of f(t) as F(s):

$2s^2F(s)-2sf(0)-2f'(0)+3sF(s)-3f(0)-2F(s)=\frac{1}{s+2}$

Now using the initial conditions we get:

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

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

Now to find the inverse transform, we will use the linearity and we will decompose the fractions into the simple elements:

$2s^2+3s-2=(2s-1)(s+2)$

$\frac{-4}{2s^2+3s-2}=\frac{A}{2s-1}+\frac{B}{s+2}$

$\begin{cases} A+2B=0 \\ 2A-B=-4 \end{cases}$ gives us $\begin{cases} A=-8/5 \\ B=4/5 \end{cases}$

$(s+2)(2s^2+3s-2)=(s+2)^2(2s-1)$

$\frac{1}{(s+2)^2(2s-1)}=\frac{Cs+D}{(s+2)^2}+\frac{E}{2s-1}$

$\begin{cases} 2C+E=0 \\ -C+4E+2D=0 \\ -D+4E=1 \end{cases}$ gives us $\begin{cases} C = -2/25 \\ D=-9/25 \\ E= 4/25\end{cases}$ , which we can write as

$\frac{-2}{25}\frac{1}{s+2}-\frac{1}{5}\frac{1}{(s+2)^2}+\frac{2}{25}\frac{1}{s-1/2}$

Now by applying the inverse transformation term by term we find:

$f(t)=-\frac{4}{5}e^{t/2}+\frac{4}{5}e^{-2t}-\frac{2}{25}e^{-2t}-\frac{1}{5}e^{-2t}t+\frac{2}{25}e^{t/2}$

$f(t)=-\frac{18}{25}e^{t/2}+(\frac{18}{25}-\frac{1}{5}t)e^{-2t}$