How Do You Solve a Second Order Differential Equation Using Laplace Transforms?

[PiRho31416]

[Solved] Laplace Transform

Homework Statement

$$\frac{d^{2}y}{dt^{2}}+4y=sin(t),\quad y(0)=0,\quad\frac{dy}{dt}(0)=0$$

Homework Equations

Laplace transform is defined as:
$$\mathcal{L}\{f(t)\} = \int_{-\infty}^{\infty}f(t)e^{st}dt$$

The Attempt at a Solution

$$\frac{d^{2}y}{dt^{2}}+4y=sin(t),\quad y(0)=0,\quad\frac{dy}{dt}(0)=0$$
$$s^{2}\mathcal{L}\{y\}-sy'(0)-y(0)-4\mathcal{L}\{y\}=\frac{1}{s^{2}+1}$$
$$\mathcal{L}\{y\}(s^{2}-4)=\frac{1}{s^{2}+1}$$
$$\Rightarrow \frac{1}{s^{2}+1}\cdot \frac{1}{s^{2}-4}=\frac{As+B}{s^2+1}\cdot\frac{Cs+D}{s^2-4}$$
$$1 = (As+B)(s^2-4)\cdot(Cs+D)(s^2+1)$$
$$1 = As^3-4As+Bs^2-4B+Cs^3+Cs+Ds^2+D$$
$$1 = s^3(A+C)+s^2(B+D)+s(C-4A)+D-4B$$
$$\begin{bmatrix} 1 & 0 & 1 & 0\\ 0 & 1 & 0 & 1\\ -4 & 0 & 1 & 0\\ 0 & -4 &0 & 1 \end{bmatrix} \begin{bmatrix} A\\ B\\ C\\ D \end{bmatrix} =\begin{bmatrix} 0\\ 0\\ 1\\ 0 \end{bmatrix} \\\ \begin{bmatrix} A\\ B\\ C\\ D \end{bmatrix} = \begin{bmatrix} -1/5 \\ 0 \\ 1/5 \\ 0 \end{bmatrix}$$


$$\mathcal{L}\{Y\}=\frac{1}{5}\frac{1}{s^2-1}-\frac{1}{5}\frac{1}{s^2+1}$$
$$\mathcal{L}\{Y\}=\frac{1}{5}\frac{1}{s^2-1}-\frac{1}{5}\frac{1}{s^2+1}$$
$$\Rightarrow\mathcal{L}\{\frac{1}{5}\frac{1}{s^2+1}\}=\frac{1}{5}sin(t)$$
$$\Rightarrow\mathcal{L}\{\frac{-1}{5}\frac{1}{s^2-1}\}=\mathcal{L}\{\frac{-1}{5}\frac{1}{(s+1)(s-1)}\}$$
So this is where I get stuck using convolution. Since we know
$$\Rightarrow \mathcal{L}^{-1}\{\frac{1}{s+1}\} = e^{-t}$$
$$\Rightarrow \mathcal{L}^{-1}\{\frac{1}{s-1}\} = e^{t}$$
$$f(t)=e^{t}$$
$$g(t)=e^{t}$$
$$f(t) \ast g(t) = \int_{0}^{t}f(\tau)g(\tau-t)\,\,d\tau$$
$$\frac{1}{5}\int_{0}^{t}e^{-\tau}e^{\tau-t}\,\,d\tau = \frac{1}{5}\int_{0}^{t}e^{-\tau+\tau-t},\,\,d\tau = \frac{1}{5}\int_{0}^{t}e^{-t}\,\,d\tau \\ =\frac{1}{5}\tau e^{-t} \bigg|_{\tau=0}^{\tau=t} = \frac{1}{5}te^{-t}$$

 

[Ray Vickson]

Homework Statement

$$\frac{d^{2}y}{dt^{2}}+4y=sin(t),\quad y(0)=0,\quad\frac{dy}{dt}(0)=0$$

Homework Equations

Laplace transform is defined as:
$$\mathcal{L}\{f(t)\} = \int_{-\infty}^{\infty}f(t)e^{st}dt$$

The Attempt at a Solution

$$\frac{d^{2}y}{dt^{2}}+4y=sin(t),\quad y(0)=0,\quad\frac{dy}{dt}(0)=0$$
$$s^{2}\mathcal{L}\{y\}-sy'(0)-y(0)-4\mathcal{L}\{y\}=\frac{1}{s^{2}+1}$$
$$\mathcal{L}\{y\}(s^{2}-4)=\frac{1}{s^{2}+1}$$
$$\Rightarrow \frac{1}{s^{2}+1}\cdot \frac{1}{s^{2}-4}=\frac{As+B}{s^2+1}\cdot\frac{Cs+D}{s^2-4}$$
$$1 = (As+B)(s^2-4)\cdot(Cs+D)(s^2+1)$$
$$1 = As^3-4As+Bs^2-4B+Cs^3+Cs+Ds^2+D$$
$$1 = s^3(A+C)+s^2(B+D)+s(C-4A)+D-4B$$
$$\begin{bmatrix} 1 & 0 & 1 & 0\\ 0 & 1 & 0 & 1\\ -4 & 0 & 1 & 0\\ 0 & -4 &0 & 1 \end{bmatrix} \begin{bmatrix} A\\ B\\ C\\ D \end{bmatrix} =\begin{bmatrix} 0\\ 0\\ 1\\ 0 \end{bmatrix} \\\ \begin{bmatrix} A\\ B\\ C\\ D \end{bmatrix} = \begin{bmatrix} -1/5 \\ 0 \\ 1/5 \\ 0 \end{bmatrix}$$


$$\mathcal{L}\{Y\}=\frac{1}{5}\frac{1}{s^2-1}-\frac{1}{5}\frac{1}{s^2+1}$$
$$\mathcal{L}\{Y\}=\frac{1}{5}\frac{1}{s^2-1}-\frac{1}{5}\frac{1}{s^2+1}$$
$$\Rightarrow\mathcal{L}\{\frac{1}{5}\frac{1}{s^2+1}\}=\frac{1}{5}sin(t)$$
$$\Rightarrow\mathcal{L}\{\frac{-1}{5}\frac{1}{s^2-1}\}=\mathcal{L}\{\frac{-1}{5}\frac{1}{(s+1)(s-1)}\}$$
So this is where I get stuck using convolution. Since we know
$$\Rightarrow \mathcal{L}^{-1}\{\frac{1}{s+1}\} = e^{-t}$$
$$\Rightarrow \mathcal{L}^{-1}\{\frac{1}{s-1}\} = e^{t}$$
$$f(\tau)=e^{-\tau}$$
$$g(\tau-t)=e^{\tau-t]$$
$$\frac{1}{5}\int_{0}^{t}e^{-\tau}e^{\tau-t}\,\,d\tau = \frac{1}{5}\int_{0}^{t}e^{-\tau+\tau-t},\,\,d\tau = \frac{1}{5}\int_{0}^{t}e^{-t}\,\,d\tau \\ =\frac{1}{5}\tau e^{-t} \bigg|_{\tau=0}^{\tau=t} = \frac{1}{5}te^{-t}$$

Why not use 1/[(s+1)(s-1)] = (1/2)[1/(s-1) - 1/(s+1)]?

RGV

 

[Wingeer]

Uhm. The Laplace transform of +4y is not -4Y.

 

[PiRho31416]

Why not use 1/[(s+1)(s-1)] = (1/2)[1/(s-1) - 1/(s+1)]?

RGV

Thanks! Duh!

 

[PiRho31416]

Uhm. The Laplace transform of +4y is not -4Y.

Pretty sure it is.

 

[HallsofIvy]

Pretty sure it's NOT!
If
$$\mathcal{L}(y)= \int_0^\infty e^{-st}y(t)dt$$
then
$$\mathcal{L}(4y)= \int_0^\infty e^{-st}4y(t)dt= 4\int_0^\infty e^{-st}y(t)dt= 4\mathcal{L}(y)$$
not $-4\mathcal{L}(y)$.

 

[PiRho31416]

Oh crap. The negative sign! Sorry it was late last night and somehow I thought he was saying $$\mathcal{L}\{y(t)\}\neq Y$$

Sorry about that. He even says it's +4y not -4y. Geez :-(

Pretty sure it's NOT!
If
$$\mathcal{L}(y)= \int_0^\infty e^{-st}y(t)dt$$
then
$$\mathcal{L}(4y)= \int_0^\infty e^{-st}4y(t)dt= 4\int_0^\infty e^{-st}y(t)dt= 4\mathcal{L}(y)$$
not $-4\mathcal{L}(y)$.

 

[Ray Vickson]

For what it's worth: the convolution of exp(t) and exp(-t) is sinh(t), not c*t*exp(-t) or c*t*exp(t). You made an error in your convolution integral.

RGV

 

[vela]

Why not use 1/[(s+1)(s-1)] = (1/2)[1/(s-1) - 1/(s+1)]

For what it's worth: the convolution of exp(t) and exp(-t) is sinh(t), not c*t*exp(-t) or c*t*exp(t). You made an error in your convolution integral.

The OP made several mistakes. There should be no 1/(s²-1) term in the problem in the first place.