Differential Equations

IVP's with Laplace Transforms<!-- --> | Differential Equations

IVP's with Laplace Transforms

Sharan Sajiv Menon - November 15th, 2021

You can use Laplace Transforms to solve Initial Value Problems. Let's start off with some examples.

Read the previous article on Laplace Transforms if you havent already read that.

Example 1: Solve the following IVP using Laplace Transforms.

y′′−10y′+9y=5t,y(0)=−1,y′(0)=2y''-10y'+9y=5t, y(0)=-1, y'(0)=2

Solution: The first step is to take the Laplace Transform of both sides.

L{y′′−10y′+9y}=L{5t}→L{y′′}−L{10y′}+L{9y}=L{5t}L\{y''-10y'+9y\}=L\{5t\} \\ \to \\ L\{y'' \} - L\{10y'\} + L\{9y\}=L\{ 5t\}

Use the table of Laplace transforms

s2Y(s)−sy(0)−y′(0)−10sY(s)−y(0)+9Y(s)=5s2s^2Y(s)-sy(0)-y'(0)-10sY(s)-y(0)\\+9Y(s)=\frac{5}{s^2}

Plug in the initial contidions

s2Y(s)+s−2−10sY(s)+1+9Y(s)=5s2s^2Y(s)+s-2-10sY(s)+1+9Y(s)=\frac{5}{s^2}

Collect all the Y(s)Y(s) terms.

Y(s)(s2−10s+9)+s−1=5s2→Y(s)(s−9)(s−1)+s−1=5s2Y(s)(s^2-10s+9)+s-1=\frac{5}{s^2} \\ \to \\ Y(s)(s-9)(s-1)+s-1=\frac{5}{s^2}

Solve for Y(s)Y(s)

Y(s)=5+12s2−s3s2(s−9)(s−1)Y(s)=\frac{5+12s^2-s^3}{s^2(s-9)(s-1)}

Now, we have to take the inverse Laplace transform of Y(s)Y(s) to find y(t)y(t), but it is easier first to use decomposition of partial fractions to decompose Y(s)Y(s) into partial fractions. This makes it a lot easier to take the inverse transform

You should know how to take the partial fraction decomposition of a function

Our partial fraction will look like the following.

Y(s)=As+Bs2+C(s−9)+D(s−1)Y(s)=\frac{A}{s} + \frac{B}{s^2} + \frac{C}{(s-9)} + \frac{D}{(s-1)}

Set the numerators equal

5+12s2−s3=As(s−9)(s−1)+B(s−9)(s−1)+Cs2(s−1)+Ds2(s−9)5+12s^2-s^3=As(s-9)(s-1) + B(s-9)(s-1)\\ + Cs^2(s-1) + Ds^2(s-9)

Plug in the different values of S to get the coefficents

  • s=0→9B=5→B=59s=0 \to 9B=5 \to B= \frac{5}{9}
  • s=1→−8D=16→D=−2s=1 \to -8D=16 \to D=-2
  • s=9→648C=248→C=3181s=9 \to 648C=248 \to C=\frac{31}{81}

Setting s=0,1,9s=0, 1, 9 all remove the AA term, so we use a different number, like s=2s=2 now that we have our coefficents BB, CC, DD.

s=2→45=−14A+434581→A=5081s=2 \to 45=-14A+\frac{4345}{81} \to A=\frac{50}{81}

Plug in the 4 constants to get the following

Y(s)=5081s+59s2+3181(s−9)+−2(s−1)Y(s)=\frac{\frac{50}{81}}{s} + \frac{\frac{5}{9}}{s^2} + \frac{\frac{31}{81}}{(s-9)} + \frac{-2}{(s-1)}

Taking the Inverse Laplace Transform of this is a lot easier.

y(t)=L{5081s}+L{59s2}+L{3181(s−9)}+L{−2(s−1)}y(t)=L\{\frac{\frac{50}{81}}{s}\} + L\{\frac{\frac{5}{9}}{s^2}\} + L\{\frac{\frac{31}{81}}{(s-9)}\} + L\{\frac{-2}{(s-1)}\}

Simplify and you get the solution to the IVP.

y(t)=5081+59t+3181e9t−2ety(t)= \frac{50}{81} + \frac{5}{9}t + \frac{31}{81}e^{9t} -2e^{t}

Created by Sharan Sajiv Menon, © 2022