2nd order with variable coefficients

**heiko** · January 13th 2011, 08:28 AM

Hello,

I want to solve the following differential equation:

y''(x) + (a+b*exp(-c*x)) y'(x) + (d+f*exp(-g*x)) y(x) = s(x)

with y(0)=0, y'(0)=0
a,b,c,d,f,g are constant and s(x) is known. Any suggestions on how I could solve it? I am even not able to solve the homogeneous equation (s(x)=0).

To find a solution to the following approximation would also help me:
y''(x) + (a-c*x) y'(x) + (d-g*x) y(x) = s(x)

Looking forward to your answers.

Regards,
Heiko

**chisigma** · January 13th 2011, 12:56 PM

You can try to use a specific property of the Laplace Tranform that extablishes that...

$\displaystyle \varphi(s)= \mathcal{L} \{y(t)\} \implies \mathcal{L} \{e^{\alpha\ t}\ y(t)\} = \varphi(s-\alpha) \implies$

$\displaystyle \implies \mathcal{L} \{e^{\alpha\ t}\ y^{'}(t)\} = (s-\alpha)\ \varphi(s-\alpha) - y(0)$ (1)

... and, considering that is $y(0)= y^{'} (0)=0$ the DE in terms of Laplace Transform becomes...

$\displaystyle (s^{2} + a\ s + d)\ \varphi(s) + b\ (s+c)\ \varphi(s+c) + f\ (s+g)\ \varphi(s+g) = \sigma(s)$ (2)

... where $\sigma(s)= \mathcal{L} \{s(t)\}$ . The solution of (2) [a 'functional equation'...] is however not trivial

...

Kind regards

$\chi$ $\sigma$

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