Laplace Transforms with Non-repeated Irreducible quadratic factors

I'm solving the following differential equation with a Laplace Transform:

$$x″+ 9x = \cos(t) + \delta(t-\pi)$$

The initial conditions are that x(0) and x'(0) are equal to 0.

After after the transform, we obtain:

$$F(S) = \frac{S}{(S^2+9)[(S^2+1)+9]} + \frac{e^{-\pi\cdot t}}{S^2+9}$$

The problem is that there are two unique irreducible quadratic factors in the denominator of that first term: therefore I cannot seem to find a way to simplify this to find equivalent components which would make the inverse transform easier. How do I find the inverse of that first term?

$$\frac{S}{(S^2+9)[(S^2+1)+9]}$$

The answer includes simple cos(t) and cos(3t) terms with coefficients of $$\frac{1}{8}$$ and $$\frac{-1}{8}$$ respectively. What is the process of going from this complex laplace transform to something so simple?

• You can still factor into linear factors with complex roots. Alternatively, you can find a partial fraction decomposition of the form $(AS + B)/(S^2 + 9) + (CS + D)/(S^2 + 1)$ for some $A,B,C,D$ real and use facts about Laplace transforms of $\sin$ and $\cos$ May 2, 2019 at 1:18
• Thank you- I will try solving this way. May 2, 2019 at 1:24
• The problem has changed now: there is an extra 9 in the denominator which I didn't expect: How might I approach the problem now? May 2, 2019 at 1:47
• The way you wrote it, you can rewrite $(S^2+1)+9 = S^2 + 10$. If you mean $(S+1)^2 + 9$, you can look at Laplace transforms of functions of the form $e^{at}\cos(bt)$ and $e^{bt}\sin(bt)$. May 2, 2019 at 2:07
• Thank you- I'll try this. May 2, 2019 at 2:34

Alternative Method

$$F(s)=\underbrace{\frac{s}{(s^2+1)(s^2+9)}}_{\mathbf{(I)}}+\underbrace{\frac{e^{-\pi s}}{s^2+9}}_{\mathbf{(II)}}$$

Alternative (and fast) method for $$\mathbf{(I)}$$:

We can also decompose $$\mathbf{(I)}$$ using partial fractions by the following way:

$$\frac{s}{(s^2+1)(s^2+9)}\equiv\frac{Es+F}{s^2+1}+\frac{Gs+H}{s^2+9}$$ Finding the constants of partial fractions decomposition $$\frac{s}{(s^2+1)(s^2+9)}\equiv\frac{(Es+F)\cdot(s^2+9)+(Gs+H)\cdot(s^2+1)}{(s^2+1)(s^2+9)}$$

Now, working only with the numerators: $$s\equiv (E+G)\cdot s^3+(F+H)\cdot s^2+(9E+G)\cdot s+(9F+H)$$ Thus, we can find $$E$$, $$F$$, $$G$$ and $$H$$ solving the following system of equations:

\mbox{ \begin{cases} \begin{alignat}{1} E+G&=0 \\F+H&=0 \\9E+G&=1 \\9F+H&=0 \end{alignat} \end{cases} } \implies \mbox{ \begin{cases} \begin{alignat}{1} E&=1/8 \\F&=0 \\G&=-1/8 \\H&=0 \end{alignat} \end{cases} }

Thus, we get: $$\frac{s}{(s^2+1)(s^2+9)}\equiv \frac{s}{8\cdot(s^2+1)}-\frac{s}{8\cdot(s^2+9)}$$

Now we can apply the Inverse Laplace Transform:

Remember that the Laplace transform of cosine function is: $$\cos(\omega t) \iff \frac{s}{s^2+\omega^2}$$ So, we have:

\begin{alignat}{1} \mathscr{L}^{-1}\left[\frac{s}{8\cdot(s^2+1)}-\frac{s}{8\cdot(s^2+9)}\right]&=\frac 1 8 \,\mathscr{L}^{-1}\left[\frac{s}{s^2+1^2}\right]-\frac 1 8 \,\mathscr{L}^{-1}\left[\frac{s}{s^2+3^2}\right] \\&=\frac{1}{8}\left[cos(t)-cos(3t)\right] \end{alignat}

$$\bbox[5px,border:1.1px solid black] { x(t)=\frac{1}{8}\left[cos(t)-cos(3t)\right]-\frac 1 3\sin(3t)\cdot\operatorname{H}(t-\pi) }$$

Laplace Transform

We have, as pointed in the question, the following equation: $$x''(t)+ 9x(t) = \cos(t) + \delta(t-\pi)$$ With $$x(0)=0$$ and $$x'(0)=0$$. So applying the Laplace Transform: \require{cancel} \begin{alignat}{1} \mathscr{L}\left[x''(t)+9x(t)\right]&=\mathscr{L}\left[ \cos(t) + \delta(t-\pi)\right] \\ s^2F(s)-s\cancelto{0}{f(0)}-\cancelto{0}{f'(0)}+9F(s)&=\frac{s}{s^2+1^2}+e^{-\pi s}\cdot1 \\s^2F(s)+9F(s)&=\frac{s}{s^2+1}+e^{-\pi s} \\(s^2+9)\cdot F(s)&=\frac{s+(s^2+1) e^{-\pi s}}{s^2+1} \end{alignat}

So, after the transform, we obtain: $$\bbox[5px,border:1.1px solid black] { F(s)=\frac{s+(s^2+1) e^{-\pi s}}{(s^2+1)(s^2+9)}=\frac{s}{(s^2+1)(s^2+9)}+\frac{e^{-\pi s}}{s^2+9} }$$

Your transform is wrong in two places: $$F(S) = \frac{S}{(S^2+9)[(S^2+1) \bbox[yellow,4px,border:1.5px solid red]{+9}\,]} + \frac{e^{-\pi\cdot \, \bbox[yellow,4px,border:1.5px solid red]t}}{S^2+9}$$

Inverse Laplace Transform

Now, we have two parts in the right side to apply the Inverse Laplace Transform: $$F(s)=\underbrace{\frac{s}{(s^2+1)(s^2+9)}}_{\mathbf{(I)}}+\underbrace{\frac{e^{-\pi s}}{s^2+9}}_{\mathbf{(II)}}$$

For $$\mathbf{(I)}$$, we need to decompose it via partial fractions first:

$$\,\,\,\,\,\,\,\mathbf{1.}$$ Finding zeros of $$s^2+1$$ and $$s^2+9$$ \begin{alignat}{1} s^2+1=0 &\iff s=\pm\sqrt{-1}=\pm \,i \\s^2+9=0 &\iff s=\pm\sqrt{-3}=\pm \,3i \end{alignat} Now that we know that $$s^2+1=(s+i)\cdot(s-i)$$ and $$s^2+9=(s+3i)\cdot(s-3i)$$, we can proceed the decomposition.

$$\,\,\,\,\,\,\,\mathbf{2.}$$ Finding the constants of partial fractions decomposition $$\frac{s}{(s^2+1)(s^2+9)}\equiv\frac{A}{s+i}+\frac{B}{s-i}+\frac{C}{s+3i}+\frac{D}{s-3i}\equiv \\\equiv\frac{A\cdot (s-i)(s^2+9)+B\cdot(s+i)(s^2+9)+C\cdot (s^2+1)(s-3i)+D\cdot (s^2+1)(s+3i)}{(s^2+1)(s^2+9)}$$ Now, working only with the numerators: $$s\equiv A\cdot (s-i)(s^2+9)+B\cdot(s+i)(s^2+9)+C\cdot (s^2+1)(s-3i)+D\cdot (s^2+1)(s+3i)$$ Since we're dealing with an identity (i.e. valid for every $$s$$), we can arbitrarily chose values and put them into the identity to find $$A$$, $$B$$, $$C$$ and $$D$$:

\begin{cases} \begin{alignat}{1} s=-i &\implies A=\frac{1}{16} \\ s=i &\implies B=\frac{1}{16} \\ s=-3i &\implies C=-\frac{1}{16} \\ s=3i &\implies D=-\frac{1}{16} \end{alignat} \end{cases} Thus, we get: $$\frac{s}{(s^2+1)(s^2+9)}\equiv\frac{1/16}{s+i}+\frac{1/16}{s-i}-\frac{1/16}{s+3i}-\frac{1/16}{s-3i}$$

Now we can apply the Inverse Laplace Transform: $$\mathscr{L}^{-1}\left[\frac{1/16}{s+i}+\frac{1/16}{s-i}-\frac{1/16}{s+3i}-\frac{1/16}{s-3i}\right]= \\=\frac{1}{16}\,\mathscr{L}^{-1}\left[\frac{1}{s+i}\right]+\frac{1}{16}\,\mathscr{L}^{-1}\left[\frac{1}{s-i}\right]-\frac{1}{16}\,\mathscr{L}^{-1}\left[\frac{1}{s+3i}\right]-\frac{1}{16}\,\mathscr{L}^{-1}\left[\frac{1}{s-3i}\right]$$ Remembering that the Laplace transform of exponential function is: $$e^{-at} \iff \frac{1}{s+a}$$ We finally have: $$\mathscr{L}^{-1}\left[\frac{1/16}{s+i}+\frac{1/16}{s-i}-\frac{1/16}{s+3i}-\frac{1/16}{s-3i}\right]= \frac{1}{16}e^{-i\cdot t}+\frac{1}{16}e^{i\cdot t}-\frac{1}{16}e^{-3i\cdot t}-\frac{1}{16}e^{3i\cdot t}$$ Using Euler's identity ($$e^{i\cdot\theta}\equiv \cos\theta+i\sin\theta)$$, we can write the above result in terms of sines and cosines: \begin{cases} \begin{alignat}{1} e^{-i\cdot t}&=\cos(t)-i\sin(t) \\e^{i\cdot t}&=\cos(t)+i\sin(t) \\e^{-3i\cdot t}&=\cos(3t)-i\sin(3t) \\e^{3i\cdot t}&=\cos(3t)+i\sin(3t) \end{alignat} \end{cases} \implies \frac{1}{16}\left(e^{-i\cdot t}+e^{i\cdot t}-e^{-3i\cdot t}-e^{3i\cdot t}\right)=\frac{1}{8}\left[cos(t)-cos(3t)\right]

For $$\mathbf{(II)}$$, we need to use the following:

$$\,\,\,\,\,\,\,\mathbf{1.}$$ Laplace time shift property $$f(t-a) \cdot \operatorname{H}(t-a) \iff e^{-a s}\cdot F(s)$$ (where $$\operatorname{H}(t)$$ is the Heaviside step function)

$$\,\,\,\,\,\,\,\mathbf{2.}$$ Laplace transform of sine function $$\sin(\omega t) \iff \frac{\omega}{s^2+\omega^2}$$

Combining these two things, we can find the Inverse Laplace Transform of $$\mathbf{(II)}$$:

$$\mathscr{L}^{-1}\left[\frac{e^{-\pi s}}{s^2+9}\right]=\mathscr{L}^{-1}\left[\frac{1} {\color{red}3}\cdot\frac{\color{red}3\cdot e^{-\pi s}}{s^2+3^2}\right]=\frac 1 3 \cdot \mathscr{L}^{-1}\left[e^{-\pi s}\cdot\frac{3}{s^2+3^2}\right]=\frac 1 3\sin(3t-\pi)\cdot\operatorname{H}(t-\pi)$$

Remembering that $$\sin(3t-\pi)=-\sin(3t)$$, we have: $$\bbox[5px,border:1.1px solid black] { x(t)=\frac{1}{8}\left[cos(t)-cos(3t)\right]-\frac 1 3\sin(3t)\cdot\operatorname{H}(t-\pi) }$$
• The partial fraction decomposition goes easier if you first separate the quadratic factors, treating $s^2$ like a single variable, $$\frac{1}{(s^2+1)(s^2+9)}=\frac{(s^2+9)-(s^2+1)}{8(s^2+1)(s^2+9)}=\frac1{8(s^2+1)}-\frac1{8(s^2+9)}.$$ Such an intermediate decomposition is always possible if the polynomial factors do not have common roots. As one obtains terms that have a direct inverse Laplace transform, no further factorization is necessary. May 2, 2019 at 8:28