Lesson 23 - Solving IVPs using Laplace Transforms

Now that we understand how to take the Laplace transform of a function, let's apply this directly to solving differential equations.

We'll introduce a differential equation with initial values.

$y^{″}+6y^{\prime }+5y=12e^t$, $y(0)=-1$, $y^{\prime }(0)=7$

Before jumping in to solving with Laplace transform, let's recall a previous method of solving the IVP.

We recognize this as a nonhomogeneous equation with constant coefficients. We can solve this equation by finding the homogeneous solution of the left hand side of the equation and then solving for the complementary solution of the right hand side. Let's do this now.

Characteristic equation

$m^2+6m+5=0$

$(m+5)(m+1)=0$

Our homogeneous solution is:

$y_h(t)=C_1{e}^{-5t}+C_2{e}^{-t}$

Now for the complementary solution, we use the method of undetermined coefficients. Since $f(t)=12e^t$, we can set up $y_c(t)=A{e}^t$.

Our full solution, without solving for the coefficients nor taking into account the initial values, becomes

$y(t)=y_h(t)+y_c(t)=C_1{e}^{-5t}+C_2{e}^{-t}+A{e}^t$

Now recall that we would need to take the first and second derivative of $y(t)$ in order to solve for the coefficient $A$ and then use our derivative equations to solve for $C_1$ and $C_2$.

Try this on your own. For a reminder on how to solve an equation using the method of undetermined coefficients, visit Lesson 10 - Method of Undetermined Coefficients.

Your solution should end up as $y(t)=e^t-e^{-5t}-e^{-t}$.

Now, let's see if we can arrive to the same solution using Laplace transforms.

First, we must take the Laplace transform of both sides of the equation. In order to do so, I will introduce you to some new definitions that we will need to take the Laplace transform of derivatives.

We recall that the Laplace transform of any function $f(t)$ is $F(s)$. Let's set up definitions of the Laplace transform of the derivatives of this function.

$\mathcal{L}[y^{\prime }]=sY(s)-y(0)$

$\mathcal{L}[y^{″}]=s^{2}Y(s)-sy(0)-y^{\prime }(0)$

Notice, we are utilizing our initial conditions here rather than later. Let's apply these definitions to the equation below, starting with the left side.

Left side:

$y^{″}+6y^{\prime }+5y=12e^t$

$\mathcal{L}[y^{″}]=s^{2}Y(s)-sy(0)-y^{\prime }(0)$

$\mathcal{L}[6y^{\prime }]=6\mathcal{L}[y^{\prime }]=6(sY(s)-y(0))=6sY(s)-6y(0)$

$\mathcal{L}[5y]=5\mathcal{L}[y]=5Y(s)$

Right side:

$\mathcal{L}[12e^t]=12\mathcal{L}[e^t]=12\left(\frac{1}{s-1}\right)=\frac{12}{s-1}$

Together, we have

$s^{2}Y(s)-sy(0)-y^{\prime }(0)+6sY(s)-6y(0)+5Y(s)=\frac{12}{s-1}$

Note: From this point on, we are simply performing algebraic manipulation on the differential equation. If you are not comfortable with algebra, it is highly recommended to practice those skills and become comfortable with algebra in order to accurately perform these calculations.

Next, I would recommend that we plug in our initial conditions. This will eliminate some of the variables and help see things clearer.

Our initial conditions were $y(0)=-1$, and $y^{\prime }(0)=7$.

$s^{2}Y(s)-s(-1)-7+6sY(s)-6(-1)+5Y(s)=\frac{12}{s-1}$

Let's clean this up

$s^{2}Y(s)+s-7+6sY(s)+6+5Y(s)=\frac{12}{s-1}$

$s^{2}Y(s)+s+6sY(s)-1+5Y(s)=\frac{12}{s-1}$

At this point, I recommend lumping together the like terms. Notice that we have three terms that contain $Y(S).$ Let's group these together. I'm also going to put parentheses around this group to hopefully make the next step clear.

$(s^{2}Y(s)+6sY(s)+5Y(s))+s-1=\frac{12}{s-1}$

Hopefully you see the value of those parentheses. We can now factor out the $Y(S)$.

$Y(s)(s^2+6s+5)+s-1=\frac{12}{s-1}$

Now, the goal here is to get $Y(s)$ by itself.

$Y(s)(s^2+6s+5)=\frac{12}{s-1}-s+1$

Before continuing, we should combine everything into one fraction on the right side.

$\frac{12}{s-1}-s+1=\frac{12}{s-1}+\frac{(-s+1)(s-1)}{s-1}=\frac{12}{s-1}+\frac{-s^2+s+s-1}{s-1}=\frac{-s^2+2s+11}{s-1}$

We're in a good spot here. Let's see what's left to do.

$Y(s)(s^2+6s+5)=\frac{-s^2+2s+11}{s-1}$

Now, remember our nonhomogeneous solution which started with the characteristic equation of the form $m^2+6m+5$. This is a good place to check if you are going in the right direction because notice that the characteristic equation matches the one above. If you see this, then you are likely going in the right direction. If not, it is good to double check if you made an error somewhere so far. It's better to check here because going forward, there is more algebra that would be very difficult to backtrack on if you made an error, especially during a quiz or exam.

Let's move on by checking to see if we can factor the quadratic. In this case, we can.

$Y(s)(s^2+6s+5)=Y(s)(s+1)(s+5)$

Last, let's divide by the binomials.

$Y(s)=\frac{-s^2+2s+11}{(s-1)(s+1)(s+5)}$

Now, let's take a deep breath. We did a lot of work, but we're only about halfway done.

We started by taking the Laplace transform of each side of the equation, utilizing our initial conditions, and performed some algebraic manipulation to solve for Y(s). What more is there to do? Well this is technically a solution, $Y(s)$. However, we want to put $Y(s)$ into a form that allows us to take the inverse Laplace transform in order to revert it back to $y(t)$. We do this using partial fractions. If you need a review of partial fractions, please review Lesson 0.1 - Review of Partial Fraction Decomposition.

Let's do this now.

To set up $Y(s)$ in partial fractions, we should have a fully factored denominator, which is why we did the work we had done before. We can see that we can set up our partial fractions as such:

$\frac{-s^2+2s+11}{(s-1)(s+1)(s+5)}=\frac{A}{s-1}+\frac{B}{s+1}+\frac{C}{s+5}$

$-s^2+2s+11=A(s+1)(s+5)+B(s-1)(s+5)+C(s-1)(s+1)$

At this point, we can solve for $A,B$ and $C$ by setting $s$ equal to $-1,-5$ and $1$, respectively.

Since we should be comfortable with this process, I'm going to skip this step and provide the values for $A,B$ and $C$, but I encourage you to try this for yourself.

$A=1,B=-1,C=-1$

So now we can say that $Y(s)=\frac{1}{s-1}-\frac{1}{s+1}-\frac{1}{s+5}$.

We can see that this is in a form that we can take the inverse Laplace transform of. Let's do so now.

$\mathcal{L}[Y(s)]=\mathcal{L}\left[\frac{1}{s-1}\right]-1\mathcal{L}\left[\frac{1}{s+1}\right]-1\mathcal{L}[s+5]$

$\mathcal{L}\left[\frac{1}{s-1}\right]=e^t$

$-1\mathcal{L}\left[\frac{1}{s+1}\right]=-e^{-t}$

$-1\mathcal{L}\left[\frac{1}{s+5}\right]=-e^{-5t}$

$y(t)=e^t-e^{-t}-e^{-5t}$

Notice that this matches the solution that we came up with using the method of undetermined coefficients. I know that this was a longwinded example, but the Laplace transform tends to make solving differential equations much simpler.

To recall, here are the steps to solving an IVP using Laplace transforms.

1. Take Laplace transform of the DE on both sides.
2. Perform algebraic manipulation to solve for $Y(s)$.
3. If necessary, perform partial fractions to take inverse Laplace transform.
4. Take ${\mathcal{L}}^{-1}[Y(s)]$.

Try For Yourself

1. $y^{″}-4y;+5y=4e^{3t}$, $y(0)=2,$ $y^{\prime }(0)=7$
2. $y^{″}-4y=4t-8e^{-2t},$ $y(0)=0$, $y^{\prime }(0)=5$

Next Lesson: Lesson 24 - Introduction to Discontinuous Functions (Heaviside Functions)

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