Using Laplace to solve for Initial Value Problem

Question

Use Laplace to solve IVP:

$y'' - 2y' 3y = 3e^t ; y(0)=-1, y'(0)=2$

I solved for:

$\mathscr{L}[3e^t] = \dfrac{3}{s-1}$

$\mathscr{L}[y''] = s^2Y(s) + s - 2$

$\mathscr{L}[y'] = sY(s) + 1$

and got as far as:

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

How do I go further to solve for $y(t)$?

Answer 1

Hints:

• Solve for $y(s)$.
• Combine all terms and then write out using Partial Fractions
• Find the Inverse Laplace Transform of the previous steps

We have:

• $\mathscr{L}[3e^t] = \dfrac{3}{s-1}$
• $\mathscr{L}[y''] = s^2y(s) -sy(0) - y'(0) = s^2 y(s) + s - 2$
• $\mathscr{L}[-2y'] = -2(sY(s) -y(0)) = -2s y(s) - 2$
• $\mathscr{L}[-3y] = -3y(s)$

We now find $y(s)$ as:

$$(s^2 -2s - 3)y(s) = \dfrac{3}{s-1} - s +4$$

So:

$$y(s) =- \dfrac{3}{4 (s-1)}-\dfrac{7}{8 (s+1)} + \dfrac{5}{8 (s-3)}$$

Now just find the inverse of each of those.

Final result:

$$y(t) = \dfrac{1}{8} e^{-t} (-6 e^{2 t}+9 e^{4 t}+5)$$