The Laplace Transform

Solving Differential Equations with the Laplace Transform

We will use the derivative rule in terms of our dependent variabe "y" to create a function in terms of $y^{L}$ which we can then isolate, and then apply the Laplace Transform to. Then we will apply the Inverse Laplace transform to obtain our answer. The process is just as simple as I just layed out:
1. Use Derivative Rule
2. Isolate $y^{L}$
3. Apply Laplace Transform
4. Apply Inverse Laplace Transform
5. Make sure to multiply by the Heaviside function
The entire process can be done in MuPAD using all the commands that we know. It is important to note that our answer will be unique - we will be solving IVP/Cauchy problems. When we use the derivative rule, we end up with a place for the initial conditions, which works perfectly for us! The best way to learn this is by example - simply working through multiple problems.

Example Problems

Homogeneous, Characteristic Equation has Real Roots

$y''+5y'+6y = 0$ such that $y(0)=0, y'(0)=1$
First we want to apply the derivative rule to our function:
For the y'' term we obtain:
$\lambda^2*y^{L}-y'(0)-\lambda y(0)$
For the 5y' term we obtain:
$5(\lambda y^{L}-y(0))$
And finally for the 6y term we obtain:
$6y^{L}$

Now we can write this into MuPAD and begin to solve: It is important to notice here that we always end up with the initial conditions in the numerator of the fraction, and the characteristic equation in the denominator of the fraction. Also, we had to index our answer "ourtransform" with a [1] next to it because the answer was in curly brackets.

Nonhomogeneous Continuous, Characteristic Equation has Real Roots

$y''+5y'+6y = e^2t$ such that $y(0)=0, y'(0)=1$
Again, we want to apply the derivative rule to our function:
For the y'' term we obtain:
$\lambda^2*y^{L}-y'(0)-\lambda y(0)$
For the 5y' term we obtain:
$5(\lambda y^{L}-y(0))$
And finally for the 6y term we obtain:
$6y^{L}$

Now we can type it into MuPAD for a similar result: Hopefully now you get the gist of how this will work. The next few examples will use MuPAD in a similar way to illustrate the other kinds of results you may encounter:

Nonhomogeneous Continuous, Characteristic Equation has a Double Root

$y''-y = \sin(6t)$ such that $y(0)=6, y'(0)=2$

Nonhomogeneous Continuous, Characteristic Equation has Complex Roots

$y''+4y = \sin(6t)$ such that $y(0)=6, y'(0)=2$

Nonhomogeneous Piecewise Continuous, Characteristic Equation has Complex Roots

$y''+2y'+5y = = \begin{cases} t-2 &\mbox{if } 0 \leq t \leq 2 \\0 &\mbox{if } 2 \leq t \\ \end{cases}$ such that $y(0)=0, y'(0)=0$