# nonhomogeneous linear system

• Feb 26th 2013, 12:17 PM
ejena
nonhomogeneous linear system
Can you help me with this system of linear equations, to find the general solution??

dx1/dt=2x1-5x2

dx2/dt=x1-2x2+cos(t)
• Feb 26th 2013, 05:03 PM
Prove It
Re: nonhomogeneous linear system
\displaystyle \displaystyle \begin{align*} \frac{dx_1}{dt} &= 2x_1 - 5x_2 \\ \frac{dx_2}{dt} &= x_1 - 2x_2 + \cos{(t)} \end{align*}

Differentiating the second equation, we get

\displaystyle \displaystyle \begin{align*} \frac{d^2 x_2}{dt^2} &= \frac{dx_1}{dt} - 2\frac{dx_2}{dt} - \sin{(t)} \\ \frac{d^2 x_2}{dt^2} &= 2x_1 - 5x_2 - 2 \left( x_1 - 2x_2 \right) - \sin{(t)} \\ \frac{d^2 x_2}{dt^2} &= - x_2 - \sin{(t)} \\ \frac{d^2 x_2}{dt^2} + x_2 &= \sin{(t)} \end{align*}

This is a second order linear constant coefficient nonhomogeneous ODE. Solve as you normally would for \displaystyle \displaystyle \begin{align*} x_2 \end{align*}.

Differentiating the first function, we have

\displaystyle \displaystyle \begin{align*} \frac{d^2 x_1}{dt^2} &= 2\frac{dx_1}{dt} - 5\frac{dx_2}{dt} \\ \frac{d^2 x_1}{dt^2} &= 2 \left( 2x_1 - 5x_2 \right) - 5 \left[ x_1 - 2x_2 + \cos{(t)} \right] \\ \frac{d^2 x_1}{dt^2} &= -x_1 - 5\cos{(t)} \\ \frac{d^2 x_1}{dt^2} + x_1 &= -5\cos{(t)} \end{align*}

This is another second order linear constant coefficient nonhomogeneous ODE. Solve as you normally would for \displaystyle \displaystyle \begin{align*} x_1 \end{align*}.
• Feb 27th 2013, 05:41 AM
ejena
Re: nonhomogeneous linear system
thanks a lot for the answer , but in fact I should solve it by using the eigenvalues and eigenvectors ... I have found the eigenvalues λ1 = i λ2=-i and the eigenvectors from
(A-λ1)α=0 ( where α is the eigenvector , and A is the matrix ( {2 , -5 } { 1 , -2} ) , but I have difficulties finding a particular solution , so i can find the general solution in the end ... :)