Using Laplace Transform to solve Differential equations

**Paymemoney** · April 10th 2011, 01:11 AM

Hi
The following question i am having trouble solving
Use Laplace Transforms to solve each system of differential equations

$\frac{dx}{dt} = -2y+1$
$\frac{dy}{dt} = 2x-t$

given x(0) = -1, y(0) = 1

This is what i have done:

$sX+2Y = 0$
$sY-2X=1-\frac{1}{s^2}$

I used cramers rule to find X and Y
$Y = \frac{s^2-1}{s(s^2+4)}$
$Y = \frac{s^2}{s(s^2+4)}-\frac{1}{s(s^2+4)}$
$Y = cos(2t)-\frac{1}{4}+\frac{cos(2t)}{4}$

$X = \frac{-2s^2+2}{s^2(s^2+4)}$
$X = \frac{-2s^2}{s^2(s^2+4)}+\frac{2}{s^2(s^2+4)}$
$X = -sin(2t) + \frac{2t}{4}-\frac{2sin(2t)}{8}$

Answer:

$X = \frac{1}{2}(t - \frac{3sin(2t)}{2})-cos(2t)$
$Y = \frac{1}{4}(1+3cos(2t))-sin(2t)$

P.S

**Prove It** · April 10th 2011, 01:15 AM

You don't need Laplace Transforms...

Take the second derivative of the first DE to get

$\displaystyle \frac{d^2x}{dt^2} = -2\frac{dy}{dt}$ , which means

$\displaystyle \frac{d^2x}{dt^2} = -2\left(2x - t\right)$

$\displaystyle \frac{d^2x}{dt^2} = -4x + 2t$

$\displaystyle \frac{d^2x}{dt^2} + 4x = 2t$ .

This is a second order linear non homogeneous constant coefficient ODE. You should know how to solve this.

**Paymemoney** · April 10th 2011, 01:44 AM

yes i know you can use that method but the question specify the use of Laplace Transform

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