# Particular integral, Matrix method

• Apr 29th 2010, 07:06 AM
bobred
Particular integral, Matrix method
Hi

I have found the complementary function but I am not sure how to work out the particluar integral. Here are the equations

$\displaystyle \dot{x}=3x+8y-2t$
$\displaystyle \dot{y}=-5x+-10y+1$

Which is

$\displaystyle \left[\begin{array}{c} \dot{x}\\ \dot{y}\end{array}\right]=\left[\begin{array}{cc} 3 & 8\\ -5 & -10\end{array}\right]\cdot\left[\begin{array}{c} x\\ y\end{array}\right]+\left[\begin{array}{c} -2t\\ 1\end{array}\right]$

I have solved the complementary function

$\displaystyle \left[\begin{array}{c} x\\ y\end{array}\right]=\alpha\left[\begin{array}{c} 1\\ -5/8\end{array}\right]e^{-2t}+\beta\left[\begin{array}{c} 1\\ -1\end{array}\right]e^{-5t}$

But I am not sure how to solve for the particular integral, any ideas

Thanks
• Apr 29th 2010, 01:20 PM
shawsend
Use variation of parameter by substituting:

$\displaystyle \left(\begin{array}{c} x \\ y\end{array}\right)_p=\alpha(t)\left(\begin{array} {c} 1 \\-5/8\end{array}\right)e^{-2t}+\beta(t)\left(\begin{array}{c} 1 \\ -1\end{array}\right)e^{-5t}$

into:

$\displaystyle \left(\begin{array}{c} x \\ y\end{array}\right)^{'}=\left(\begin{array}{cc} 3 & 8 \\ -5 & -10\end{array}\right)\left(\begin{array}{c} x \\ y\end{array}\right)+\left(\begin{array}{c} -2t \\ 1\end{array}\right)$

obtaining a system of two equations in the unknowns $\displaystyle \alpha'(t)$ and $\displaystyle \beta'(t)$ which you can solve for each by elementary means.
• Apr 29th 2010, 11:31 PM
bobred
Please excuse me for being dumb, but could you elaborate.

James
• Apr 30th 2010, 05:36 AM
bobred
HI

Sorted it out, I was been dim. Thanks