Laplace Transformations

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April 30th 2011, 09:55 PM
Naples

Laplace Transformations

Solve using Laplace Transformations

1. $y' + y = sin(x)$ $where$ $y(0) = 1$

$sY(s) - y(0) + 4Y(s) = 1/(s^2 + 1)$
$Y(s)(s + 4) = 1/(s^2 + 1) - 1$
$Y(s) = (s^2 + 2)/[(s + 4)(s^2 +1)]$

2. $y^(^3^) + y' = e^x$ $where$ $y(0)=y'(0)=y"(0)=0$

$s^3Y(s) - s^2y(0) - sy'(0) - y"(0) + Y(s) - y(0) = 1/(s - 1)$
$Y(s)(s^3 + s) = 1/(s - 1)$
$Y(s) = 1/[x(x + 1)(x^2 + 1)]$

I'd greatly appreciate it if someone could assist me in finishing these problems as I'm unsure of what to do.
April 30th 2011, 10:13 PM
TheEmptySet

You have a slight error in the first one

$y'+y=\sin(x), \quad y(0)=1$

Taking the Laplace transform gives

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

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

Now from here you need to use partial fractions on

$\frac{s^2+2}{(s+1)(s^2+1)}=\frac{\frac{3}{2}}{s+1} +\frac{1}{2}\cdot\frac{-s+1}{s^2+1}=\frac{3}{2}\cdot \frac{1}{s+1}-\frac{1}{2}\cdot \frac{s}{s^2+1}+\frac{1}{2}\cdot \frac{1}{s^2+1}$

Each of these should be known transforms so the inverse is

$\frac{3}{2}e^{-t}-\frac{1}{2}\cos(t)+\frac{1}{2}\sin(t)$

for the 2nd one the first two lines are correct but the third should be

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

Now use partial fractions to find the inverse transform.
May 1st 2011, 09:04 PM
Naples

$Y(s) = 1/(s(s^2 + 1)(s - 1))$
$Y(s) = (1/2)*1/(s-1) - 1/s + (1/2)*s/(s^2 + 1) - 1/2*(s^2 + 1)$
$y = (1/2)e^x - 1 + (1/2)cos(x) - (1/2)sin(x)$

Is that the right answer for the second problem?
May 1st 2011, 09:09 PM
TheEmptySet

Quote:

Originally Posted by Naples

$Y(s) = 1/(s(s^2 + 1)(s - 1))$
$Y(s) = (1/2)*1/(s-1) - 1/s + (1/2)*s/(s^2 + 1) - 1/2*(s^2 + 1)$
$y = (1/2)e^x - 1 + (1/2)cos(x) - (1/2)sin(x)$

Is that the right answer for the second problem?

Yes, Also remember you can take some derivative and plug back into the ODE to check that the answer is correct.