Initial Value Problem using Laplace Transforms

Question

Solve using Laplace Transform: $$y''(t)+2y'(t)+5y(t)=xf(t), \\ y(0)=1,y'(0)=1$$ where $x$ is a constant.
Once the solution is found, evaluate the limit as $t \to\infty$.

Progress: If I have done it right, I have ended up with $Y(s)=(3+sF(s))/(s^2+s+5)$. I am having trouble with how to take the inverse since $F(s)$ is involved.

Answer 1

Partial solution: Let $Y(s)=\mathcal{L}\left \{ y(t) \right \}$. Taking the Laplace transform of the constant coefficient ODE results in the following $$\left [s^2Y(s)-sy(0)-y'(0) \right ]+2\left [ sY(s)-y(0)) \right ]+5\left [Y(s) \right ]=x F(s)$$ Collecting like terms gives $$(s^2+2s+5)Y(s)-(s+3)=xF(s)$$ or after rearrangement $$Y(s)=\frac{s+3+xF(s)}{s^2+2s+5}$$

APPROACH #1:

Completing the square of the denominator gives $$Y(s)=\frac{s+3+xF(s)}{(s+1)^2+4}=\frac{s+3}{(s+1)^2+4}+\frac{xF(s)}{(s+1)^2+4}$$ Algebraic manipulation of each term gives an equivalent equation in a more appropriate form (reasons behind this step will become apparent shortly) $$Y(s)=\frac{s+3+1-1}{(s+1)^2+4}+\frac{x}{2}F(s)\frac{2}{(s+1)^2+4}=\frac{s+1+2}{(s+1)^2+4}+\frac{x}{2}F(s)\frac{2}{(s+1)^2+4}$$ or $$Y(s)=\frac{s+1}{(s+1)^2+4}+\frac{2}{(s+1)^2+4}+\frac{x}{2}F(s)\frac{2}{(s+1)^2+4}$$

From the transform tables, we use the pairs:

• $e^{at}cos(bt)\leftrightarrow \frac{s-a}{(s-a)^2+b^2}$
• $e^{at}sin(bt)\leftrightarrow \frac{b}{(s-a)^2+b^2}$

to invert the first two terms.

Moreover, the property of multiplication in the $s$-domain

• $\int_{0}^{t}f(t-\tau)g(\tau)d\tau \leftrightarrow F(s)G(s)$

can be used to invert the third term.

Now, we identify: $a=-1$ and $b=2$. Moreover, since $G(s)=\frac{2}{(s+1)^2+4}$, we find $$ g(t) = \mathcal{L^{-1}} \left \{ G(s) \right \} =\mathcal{L^{-1}}\left \{ \frac{2}{(s+1)^2+4}\right \} = e^{-t}sin(2t)$$

Therefore

$$ y(t) = \mathcal{L^{-1}} \left \{ Y(s) \right \} = e^{-t}\left [ cos(2t) + sin(2t) \right ]+ \int_{0}^{t}f(t-\tau)e^{-\tau}sin(2\tau)d\tau$$

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APPROACH #2:

Invoking the idea of partial fraction decomposition, we write $$\frac{s+3+xF(s)}{s^2+2s+5}=\frac{A}{s+1-2i}+\frac{B}{s+1+2i}$$ or after multiplying by $s^2+2s+5$ $$(s+1+2i)A+(s+1-2i)B=s+3+xF(s)$$

• If $s=-1-2i$: $$0+(-1-2i+1-2i)B=-1-2i+3+xF(s)\Rightarrow B=\frac{i}{4}(2-2i+xF(s))$$

• If $s=-1+2i$: $$(-1+2i+1+2i)A+0=-1+2i+3+xF(s)\Rightarrow A=\frac{-i}{4}(2+2i+xF(s))$$

Therefore $$Y(s)=\frac{-i}{4}(2+2i+xF(s))\frac{1}{s+1-2i}+\frac{i}{4}(2-2i+xF(s))\frac{1}{s+1+2i}$$ or $$Y(s)=\frac{i}{4}\left [ \frac{2-2i}{s+1+2i} - \frac{2+2i}{s+1-2i} + \frac{xF(s)}{s+1+2i} - \frac{xF(s)}{s+1-2i}\right ]$$

From the transform tables, we use the pair:

• $\alpha e^{-(a+jb)t}\leftrightarrow \frac{\alpha}{s+a+jb}$

to invert the first two terms.

Moreover, the property of multiplication in the $s$-domain

• $\int_{0}^{t}f(t-\tau)g(\tau)d\tau \leftrightarrow F(s)G(s)$

can be used to invert the last two term.

the rest is similar to approach 1.

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Hint: instead of evaluating the limit as $t \to \infty$, use the Final Value Theorem $$\lim_{t\rightarrow \infty }y(t)=\lim_{s\rightarrow 0}sY(s)$$ however, you might need to consider the various possibilities as $F(s)$ is arbitrary.