Differential equation solved by Laplace transform

Posted March 21st, 2008 by Structure

Let begin with an example.
We want to solve initial value problem for the differential equation

$y"(x)+6y'(x)+25y(x)=f(x)u(x)$

with y(+0)=a and y'(+0)=b
Let

$Y(s)=\mathcal{L}(y(x))(s)=\int_{0}^{+\infty}y(x)e^{-sx}dx$

We have

$F(s)=\mathcal{L}(f(x)u(x))(s)$

$\mathcal{L}(y'(x))(s)=sY(s)-a;\:\mathcal{L}(y"(x))(s)=s^2Y(s)-as-b$

We have then

$s^2Y(s)-as-b+6(sY(s)-a)+25Y(s)=F(s)$

$(s^2+6s+25)Y(s)=as+6a+b+F(s);\:Y(s)=\frac{as+6a+b}{s^2+6s+25}+\frac{1}{s^2+6s+25}F(s)$

Now we can write

$Y(s)=\frac{a(s+3)}{(s+3)^2+16}+\frac{3a+b}{(s+3)^2+16}+\frac{1}{(s+3)^2+16}F(s)$

$y(x)=ae^{-3x}\cos4x\: u(x)+(3a+b)\frac{1}{4}e^{-3x}\sin4x\: u(x)+\frac{1}{4}\int_{0}^{x}e^{-3t}\sin4tf(x-t)\:u(x-t)dt$

Comments

On December 31st, 2009 alvinamvinci says:

Thank for sharing ..