Laplace transform to solve second-order differential equations

Laplace transform to solve second-order differential equations. Hi guys, today I’ll talk about how to use Laplace transform to solve second-order differential equations. Have a look.

Laplace transform to solve second-order differential equations

Now the standard form of any second-order ODE is

$ay^{''}+by^{'}+cy = d~\text{or}f(y).$

Here $a, b, c, d$ are constants and $f(y)$ is a function of $y$ .

In order to solve this equation in the standard way, first of all, I have to solve the homogeneous part of the ODE. After that, I’ll need to get the particular integral for the non-homogeneous part.

If interested, you can read more about it in the category of the second-order ODE.

But what would happen if I use Laplace transform to solve second-order differential equations

If I use Laplace transform to solve second-order differential equations, it can be quite a direct approach.

First of all, I don’t need to bother with the homogeneous or non-homogeneous part. It’s all the same.

Secondly, I don’t have to work on initial conditions separately. They will be part of the calculation.

In one of my earlier posts, I have already described in detail how to use Laplace transform to solve the first-order ODE.

For further reading, please check out how to use Laplace transform to solve second-order homogeneous ODE

Now I’ll give an example of how to use Laplace transform to solve second-order differential equations.

Example of Laplace transform to solve second-order differential equations

Disclaimer: This example is not mine. I have chosen it from a book. I have also given the due reference at the end of the post.

So here is the example.

Example

According to Stroud and Booth (2011) “Solve the equation by Laplace transform:

$x^{''} + 8x^{'}+32x = 32 \sin 4t,~~~~x_0 = 0, x_1 = 0.$

”

Solution

Here the given differential equation with initial condition is

$\boxed{x^{''} + 8x^{'}+32x = 32 \sin 4t,~~~~x_0 = 0, x_1 = 0.}$

Now I have to solve this equation using Laplace transform.

Step 1

So it will be

$\mathcal{L}\{x^{''} + 8x^{'}+32x\} = \mathcal{L}\{32 \sin 4t\} .$

This means

$(s^2\overline{x}-s x_{0} - x^{}'_{0}) +8(s\overline{x} - x_0)+32\overline{x} = 32\frac{4}{s^2+16}.$

Now I already know from the given equation that $x_{0} = 0, x^{'}(0) = 0$ .

Therefore I’ll simplify this equation to get

$\begin{eqnarray*} (s^2\overline{x}-s .0 - 0) +8(s\overline{x} - 0)+32\overline{x} &=& \frac{128}{s^2+16}\\\overline{x} (s^2 +8s+32) &=& \frac{128}{s^2+16}\\ \overline{x} &=& \frac{128}{(s^2+16)(s^2 +8s+32)}.\end{eqnarray*}$

Thus the value of $x$ will be

(1) $\begin{equation*}x = \mathcal{L}^{-1}\left\{ \frac{128}{(s^2+16)(s^2 +8s+32)}\right\}.\end{equation*}$

So my next step is to find the inverse Laplace transform.

In order to get the inverse Laplace transform, first I have to use partial fractions of this expression.

As I can see, the factors in the denominator $(s^2+16)$ and $(s^2 +8s+32)$ cannot be reduced further.

So I’ll use the rule for partial fractions of irreducible factors.

Thus the expression $\cfrac{128}{(s^2+16)(s^2 +8s+32)}$ will be a sum of two expressions as

(2) $\begin{equation*}\frac{128}{(s^2+16)(s^2 +8s+32)} = \frac{As+B}{s^2+16} + \frac{Cs+D}{s^2 +8s+32}.\end{equation*}$

Also, $A, B, C, D$ are constants.

Now I’ll find out the values of these constants.

Step 2

First of all, I’ll simplify the right-hand side of the equation (2).

Thus it will be

$\begin{eqnarray*}&&\frac{128}{(s^2+16)(s^2 +8s+32)}\\ &=& \frac{(As+B)(s^2 +8s+32)+(Cs+D)(s^2+16)}{(s^2+16)(s^2 +8s+32)}\\&=&\frac{s^3(A+C)+s^2(8A+B+D)+s(32A+8B+16C)+(32B+16D)}{(s^2+16)(s^2 +8s+32)}\end{eqnarray*}$

Now I can compare the coefficients on both sides of the numerator.

I will start with the coefficients of $s^3$ .

Thus it will be

$A+C = 0.$

So this gives

(3) $\begin{equation*}C = - A.\end{equation*}$

Next, it will be the constants.

So it will look like

$32B+16D = 128.$

And this gives

(4) $\begin{equation*}D = 8-2B.\end{equation*}$

Then I’ll compare the coefficients of $s^2$ to get

$8A+B+D = 0.$

Now I’ll substitute the value from equation (4) to this equation to get

$8A+B+8-2B = 0.$

And this gives

(5) $\begin{equation*}8A-B = -8.\end{equation*}$

In the end, I’ll compare the coefficients of $s$ . Thus it will be

$32A+8B+16C = 0 \Rightarrow 4A +B +2C = 0.$

Now from equation (3), I’ll substitute $C = -A$ in this equation to get

$4A +B -2A = 0\Rightarrow 2A+B = 0.$

So this gives

(6) $\begin{equation*}B = -2A.\end{equation*}$

Next, I’ll put equation (6) in equation (4) to get

$8A+2A= -8.$

And this brings

$A = - \frac{4}{5}.$

Then I can use equation (6) to say

$B = \frac{8}{5}.$

Next, I can get the value of $C$ from equation (3) as

$C = \frac{4}{5}.$

At the end, I’ll get the value of $D$ from equation (4) as

$D = 8-\frac{16}{5} = \frac{24}{5}.$

Now I’ll put back the values of $A, B, C$ and $D$ in equation (2).

Step 3

Thus it will be

$\boxed{\frac{128}{(s^2+16)(s^2 +8s+32)} = \frac{(-4/5)s+8/5}{s^2+16} + \frac{(4/5)s+24/5}{s^2 +8s+32}.}$

Now I’ll simplify it to get

$\frac{128}{(s^2+16)(s^2 +8s+32)} = -\frac{4}{5}\left(\frac{s-2}{s^2+16}\right) + \frac{4}{5}\left(\frac{s+6}{s^2 +8s+32}\right).$

So this means

$\frac{128}{(s^2+16)(s^2 +8s+32)} = \frac{4}{5}\left(\frac{s+6}{s^2 +8s+32} - \frac{s-2}{s^2+16}\right).$

Therefore from equation (1), I can say that

$\begin{equation*}x = \mathcal{L}^{-1}\left\{ \frac{4}{5}\left(\frac{s+6}{s^2 +8s+32} - \frac{s-2}{s^2+16}\right)\right\}.\end{equation*}$

Next, I’ll use the formulas in Laplace transform to find the value of $x$ .

So I can rewrite $x$ as

$x = \frac{4}{5}\mathcal{L}^{-1}\left\{\frac{s+6}{s^2 +8s+32}\right\} - \frac{4}{5}\mathcal{L}^{-1}\left\{\frac{s-2}{s^2+16}\right\}.$

Now, first of all, I’ll find out the value of $\cfrac{4}{5}\mathcal{L}^{-1}\left\{\cfrac{s+6}{s^2 +8s+32}\right\}.$ Then I’ll find out the value of $\cfrac{4}{5}\mathcal{L}^{-1}\left\{\cfrac{s-2}{s^2+16}\right\}.$

Then let’s say,

$x_1= \frac{4}{5}\mathcal{L}^{-1}\left\{\cfrac{s+6}{s^2 +8s+32}\right\}$

and

$x_2 = \cfrac{4}{5}\mathcal{L}^{-1}\left\{\cfrac{s-2}{s^2+16}\right\}.$

Thus it will be

(7) $\begin{equation*}x = x_1-x_2.\end{equation*}$

So, now it’s time to get the value of $x_1$ .

Step 4

As I can see, the denominator of $x_1$ is $(s^2 +8s+32).$ Now I can’t factorise it. But I can write it as a sum of two squares like

$s^2 +8s+32 = (s^2+8s+16)+(16).$

So this gives

$s^2+8s+32=(s+4)^2+(4)^2.$

Therefore I can write $x_1$ as

$x_1=\frac{4}{5}\mathcal{L}^{-1}\left\{\cfrac{s+6}{(s+4)^2+(4)^2}\right\}.$

Now I’ll bring $(s+4)$ in the numerator as well.

Thus it will be

$x_1=\frac{4}{5}\mathcal{L}^{-1}\left\{\frac{(s+4)+2}{(s+4)^2+(4)^2}\right\}.$

So I can rewrite $x_1$ as

(8) $\begin{equation*}x_1 = \frac{4}{5}\mathcal{L}^{-1}\left\{\frac{(s+4)}{(s+4)^2+(4)^2}+\frac{2}{(s+4)^2+(4)^2}\right\}.\end{equation*}$

As I already know, the Laplace transform of $\cos 4t$ is $\cfrac{s}{s^2+4^2}.$

Also, from the first shift theorem in Laplace transform, I already know that the Laplace transform of $e^{-4t}\cos 4t$ is $\cfrac{s+4}{(s+4)^2+4^2}.$

So in this case, it will be

$\mathcal{L}^{-1}\left\{\frac{(s+4)}{(s+4)^2+(4)^2} \right\}= e^{-4t}\cos 4t.$

Again, the Laplace transform of $\sin 4t$ is $\cfrac{4}{s^2+4^2}$ and together with the first shift theorem the Laplace transform of $e^{-4t}\sin 4t$ becomes $\cfrac{4}{(s+4)^2+4^2}.$

Thus $\mathcal{L}^{-1}\left\{\cfrac{2}{(s+4)^2+(4)^2}\right\}$ will be

$\mathcal{L}^{-1}\left\{\cfrac{2}{(s+4)^2+(4)^2}\right\} = \frac{1}{2}e^{-4t}\sin 4t.$

Therefore from equation (8), I can say that $x_1$ will be

(9) $\begin{equation*}x_1 = \frac{4}{5} \left(e^{-4t}\cos 4t + \frac{1}{2}e^{-4t}\sin 4t\right).\end{equation*}$

Now I’ll find the value of $x_2$ .

Step 5

As I can see, the denominator of $x_2$ is $(s^2 +16).$ Now I can’t factorise it. But I can write it as a sum of two squares like

$s^2 +16 = s^2+(4)^2$

Therefore I can write $x_2$ as

$x_2=\frac{4}{5}\mathcal{L}^{-1}\left\{\frac{s-2}{s^2+(4)^2}\right\}.$

So this means

$\begin{equation*}x_2=\frac{4}{5}\mathcal{L}^{-1}\left\{\frac{s}{s^2+(4)^2}-\frac{2}{s^2+(4)^2}\right\}.\end{equation*}$

Now I’ll use the formulas in Laplace transform.

As per these formulas, Laplace transform of $\cos 4t$ is $\cfrac{s}{s^2+4^2}.$ Also, the Laplace transform of $\sin 4t$ is $\cfrac{4}{s^2+4^2}.$

So I can use these two formulas to get the value of $x_2$ as

$x_2=\frac{4}{5}\mathcal{L}^{-1}\left\{\frac{s}{s^2+(4)^2}-\frac{1}{2}\frac{4}{s^2+(4)^2}\right\}.$

Thus the value of $x_2$ will be

$x_2=\frac{4}{5}\cos 4t-\frac{4}{5}.\frac{1}{2}\sin 4t.$

Now I’ll simplify it to get

(10) $\begin{equation*}x_2=\frac{4}{5}\cos 4t - \frac{2}{5}\sin 4t.\end{equation*}$

Next, I’ll substitute the values of $x_1$ and $x_2$ from equations (9) and (10) respectively to equation (8) to get the value of $x$ as

$x = \frac{4}{5} \left(e^{-4t}\cos 4t + \frac{1}{2}e^{-4t}\sin 4t\right) - \left(\frac{4}{5}\cos 4t - \frac{2}{5}\sin 4t\right).$

Then I’ll simplify it to get

$\begin{eqnarray*}x &=& \frac{4}{5} e^{-4t}\cos 4t + \frac{2}{5}e^{-4t}\sin 4t-\frac{4}{5}\cos 4t+\frac{2}{5}\sin 4t\\&=&\frac{4}{5}\cos 4t( e^{-4t}-1)+\frac{2}{5}\sin 4t(e^{-4t}+1)\\&=&\frac{2}{5}\left[\frac{2}{5}\cos 4t( e^{-4t}-1)+\sin 4t(e^{-4t}+1)\right].\end{eqnarray*}$

Hence I can conclude that this is the solution to the given example.

Dear friends, this is the end of my today’s post on Laplace transform to solve second-order differential equations. Thank you very much for reading this. Please let me know how you feel about it. Soon I will be back again with a new post. Till then, bye, bye!!

*Reference: K. A. Stroud and Dexter J. Booth (2011): Advanced engineering mathematics, Industrial Press, Inc.; 5th Edition (March 8, 2011), Chapter: Laplace transform 1, Further problems 2, p. 90, Q. No. 2(b) (Example 1), Q. No. 2(a) (Example 2).

Laplace transform to solve second-order differential equations