undetermined coefficients

[walt89]

Hello everyone,

could anyone share some insight on how should i go about solving this equation using the undetermined coefficients methode ?

Thanks in advance

 

[Prof]

x and y would be related to e^2t. Let x = Ae^2t and y = Be^2t. Substitute them in. Figure out dx/dt and dy/dt and substitute those in also. Then you will get two algebraic equations and you can solve them to find A and B.

 

[topsquark]

Have you taken Linear Algebra? The usual steps to take are this:
[math]\dfrac{d}{dt} \left ( \begin{matrix} x \\ y \end{matrix} \right ) - \left ( \begin{matrix} \text{-2} & 1 \\ \text{-3} & 2 \end{matrix} \right ) \left ( \begin{matrix} x \\ y \end{matrix} \right ) = \left ( \begin{matrix} \text{-}e^{2t} \\ 6 e^{2t} \end{matrix} \right )[/math]
Call the matrix [math]M = \left ( \begin{matrix} 2 & -1 \\ 3 & -2 \end{matrix} \right )[/math]. The (homogeneous) solution to the equation
[math]\dfrac{d}{dt} \left ( \begin{matrix} x \\ y \end{matrix} \right ) + M \left ( \begin{matrix} x \\ y \end{matrix} \right ) = 0[/math] is [math]\left ( \begin{matrix} x \\ y \end{matrix} \right ) = A e^{ \lambda _1 t} \vec{u} + B e^{ \lambda _2 t} \vec{v}[/math]where [math]\lambda _{1,2}[/math] are the eigenvalues of M, [math]\vec{u},~ \vec{v}[/math] are the associated eigenvectors, and A and B are arbitrary constants.

Using undetermined coefficients, then, your trial particular solution will be [math]C \left ( \begin{matrix} \text{-} e^{2t} \\ 6 e^{2t} \end{matrix} \right )[/math], where C is to be dtermined.

Can you finish from here?

-Dan