# Solve the integral of differential equation

1. Feb 6, 2013

### izen

1. The problem statement, all variables and given/known data

Solve the integral of differential equation

∫$^{t}_{0}$ y($\tau$) d$\tau$ -$\acute{y}$ (t) = t

for t≥0 with y(0)=4

3. The attempt at a solution

take laplace both sides

$\frac{Y}{s}$ - sY + 4 = $\frac{1}{s^{2}}$

Y $\frac{1 - s^{2}}{s}$ =$\frac{1}{s^{2}}$ - 4

Y $\frac{s^{2}-1}{s}$ =- $\frac{1}{s^{2}}$ + 4

Y = - $\frac{1}{s(s^{2}-1)}$ + $\frac{4s}{s^{2}-1}$

Partial fraction - $\frac{1}{s(s^{2}-1)}$

$\frac{1}{s}$ -$\frac{s}{s^{2}-1}$ +$\frac{4s}{s^{2}-1}$

inverse laplace transform

1-cos(t)+4cos(t) = > 1+3cos(t)

but the answer is 1+$\frac{3}{2}$ (e$^{t}$ + e$^{-t}$)

Thank you

2. Feb 6, 2013

### haruspex

I think you mean cosh, not cos. It would be cos if the denominator were s2+1.

3. Feb 6, 2013

### HallsofIvy

Staff Emeritus
I have never like "Laplace Transform"- too mechanical. In particular, for this problem, using the Laplace Transform is using a cannon to kill a fly.

Here is how I would do this problem: differentiate both sides of the equation, with respect to t, to get rid of the integral: y(t)- y''(t)= 1 or y''- y= -1. We have initial conditions y(0)= 0 and y'(0)= 0 (from setting t=0 in the orginal equation). That's a very simple linear equation with constant coefficients. The characteristic equation has real roots so the solution will involve exponentials (and so cosh as haruspex says) not trig functions.

4. Feb 6, 2013

### izen

ohh thanks 1+3cosh(t) and then i can use the complex exponential formula to change cosh(t) to 1/2 (e^t + e^-t) thanks haruspex :)

HallsofIvy << I like your saying "using the Laplace Transform is using a cannon to kill a fly" could not agree more thanks for the the comment :)

5. Feb 6, 2013

### izen

My teacher convinces the class that the Laplace transform is so powerful but like your saying maybe it is too powerful on such this simple problem. :)