# Differential Equation

Printable View

• Dec 11th 2010, 01:14 AM
JacksonR
Differential Equation
Hello,

I have the DE....

$\displaystyle y''(t) + \lambda^2 y(t) - x(t) = 0$

I know that

$\displaystyle y''(t) + \lambda^2 y(t) = 0$

gives $\displaystyle y(t) = A \cos \lambda t + B \sin \lambda t$

The solution in the text gives

$\displaystyle y(t) = A \cos \lambda t + B \sin \lambda t + \frac{x(t)}{\lambda^2}$

but Im not sure why the solution has the $\displaystyle \frac{x(t)}{\lambda^2}$ part.
• Dec 11th 2010, 01:36 AM
FernandoRevilla
Quote:

Originally Posted by JacksonR
but Im not sure why the solution has the $\displaystyle \frac{x(t)}{\lambda^2}$ part.

Necessarily, $\displaystyle y(t)=x(t)/\lambda^2$ should be a solution for the complete equation. Equivalently:

$\displaystyle \dfrac{x''(t)}{\lambda^2}+ \lambda^2\dfrac{x(t)}{\lambda^2}-x(t)=0 \Leftrightarrow x''(t)=0 \Leftrightarrow x(t)=C_1t^2+C_2t+C_3$

(Of course with $\displaystyle C_1=0$ ) :)

Fernando Revilla
• Dec 11th 2010, 04:55 AM
chisigma
Quote:

Originally Posted by JacksonR
Hello,

I have the DE....

$\displaystyle y''(t) + \lambda^2 y(t) - x(t) = 0$

I know that

$\displaystyle y''(t) + \lambda^2 y(t) = 0$

gives $\displaystyle y(t) = A \cos \lambda t + B \sin \lambda t$

The solution in the text gives

$\displaystyle y(t) = A \cos \lambda t + B \sin \lambda t + \frac{x(t)}{\lambda^2}$

but Im not sure why the solution has the $\displaystyle \frac{x(t)}{\lambda^2}$ part.

The function $\displaystyle \displaystyle y^{*} (t)= \frac{x(t)}{\lambda^{2}}$ is a particular solution of the DE $\displaystyle \displaystyle y^{''}(t) + \lambda^{2}\ y(t)= x(t)$ only if is $\displaystyle x^{''}(t)=0$, i.e. $\displaystyle x(t)= c_{0} + c_{1}\ t$. For all other $\displaystyle x(t)$ this particular solution is...

$\displaystyle \displaystyle y^{*} (t) = x(t) * \frac{\sin \lambda t}{\lambda} = \int_{0}^{t} \frac{\sin \lambda \tau}{\lambda}\ x(t-\tau)\ d\tau$ (1)

http://digilander.libero.it/luposaba...ato&#91;1].jpg

Merry Christmas from Italy

$\displaystyle \chi$ $\displaystyle \sigma$