Differential Equation

**JacksonR** · December 11th 2010, 01:14 AM

Hello,

I have the DE....

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

I know that

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

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

The solution in the text gives

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

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

**FernandoRevilla** · December 11th 2010, 01:36 AM

Originally Posted by JacksonR

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

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

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

(Of course with $C_1=0$ )

Fernando Revilla

**chisigma** · December 11th 2010, 04:55 AM

Originally Posted by JacksonR

Hello,

I have the DE....

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

I know that

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

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

The solution in the text gives

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

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

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

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

Merry Christmas from Italy

$\chi$ $\sigma$

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