How Does Laplace Transform Solve a Differential Equation with Step Functions?

[Muffin]

Homework Statement

Determine the solution for
[tex]y^{''}+81y=81U(t-\frac{\pi }{2})[/tex]
when [tex]\left\{y(0)=12,y'(0)=18\right\}[/tex]U(t) is the unit step function

The Attempt at a Solution

Laplacetransforming :
[tex]s^{2}Y(s)-sy(0)-y'(0)+81Y(s)=[/tex][tex]\frac{81e^{\frac{-\pi }{2}}}{s}[/tex]

With given data the equation becomes

[tex]s^{2}Y(s)-12s-18+81Y(s)=[/tex][tex]\frac{81e^{\frac{-\pi }{2}}}{s}[/tex]

Solving Y(s)

[tex]Y(s)=81e^{-\frac{\pi }{2}s}(\frac{1}{s(s^{2}+9^{2})})+12(\frac{s}{{s^{2}+9^{2}}})+18(\frac{1}{s^{2}+9^{2}})[/tex]

Transform again:
[tex]y(t)=81U(t-\frac{\pi }{2})sin(9t)+12cos9t+18sin9t[/tex]Problem: This is wrong but I don't know what am I doing wrong..Can someone tell me what I am doing wrong? Wolframalpha says that the solution should be: [tex]U(\frac{\pi }{2}-t)(sin(9t)-1)+sin(9t)+12cos(9t)+1[/tex]

But I don't understand how to get that answer.

Thanks

 

[mighty2000]

Hi there,

Your approach is correct, but there are a few errors in your calculations. Let's go through them step by step:

1. When taking the Laplace transform, the initial conditions should be multiplied by e^{-st}. So the correct equation would be:

s^{2}Y(s)-12se^{-s\cdot0}-18e^{-s\cdot0}+81Y(s)=\frac{81e^{\frac{-\pi }{2}}}{s}

Which simplifies to:

s^{2}Y(s)+81Y(s)=\frac{81e^{\frac{-\pi }{2}}}{s}+30

2. When solving for Y(s), you should factor out the Y(s) term, so the equation becomes:

Y(s)(s^{2}+81)=\frac{81e^{\frac{-\pi }{2}}}{s}+30

3. When solving for Y(s), you should also divide both sides by (s^{2}+81), so the equation becomes:

Y(s)=\frac{\frac{81e^{\frac{-\pi }{2}}}{s}+30}{s^{2}+81}

4. When performing the inverse Laplace transform, you should use the property that L^{-1}\left(\frac{F(s)}{s}\right)=\int_{t}^{\infty}f(\tau)d\tau, which means that the inverse Laplace transform of \frac{F(s)}{s} is the integral of f(t) from t to infinity. In this case, the integral would be:

y(t)=\int_{t}^{\infty}\frac{81e^{\frac{-\pi }{2}}}{s}\cdot\frac{1}{s^{2}+81}dt+\int_{t}^{\infty}30\cdot\frac{1}{s^{2}+81}dt

5. Now, let's solve each integral separately. For the first integral, we can use the property L^{-1}\left(\frac{F(s)}{s^{2}+a^{2}}\right)=cos(at), which means the inverse Laplace transform of \frac{F(s)}{s^{2}+a^{2}} is cos(at). In this case, the first integral would be:

y(t