# find all solution to the equation $y'=Ay+b(x)$ for given $A$ and $b(x)$

I am asked to find all solution to the equation: $$y'= \left( \begin{array}{cc} 13 & 12 \\ 12 & 13 \end{array} \right) y+ \left(\begin{array}{c} x\\ 0 \end{array} \right)$$ No initial condition is specified.

My working so far: If we write the DE as $y'=Ay+b(x)$ then a solution to the homogeneous DE $y'=Ay$ is given by: $$y_h=e^{A(x-x_0)}y_0$$ where $y_0$ is a vector containing the arbitrary initial conditions. Because no initial conditions are specified could I just assume that $x_0=0$? I then probably need to use variation of parameters to find a general solution but I am unsure how to do this for a system of linear differential equations. Also, how do I calculate $e^{A(x-x_0)}$, is $A$ diagonalizable? If so I could calculate it. My linear algebra is not too great so I got stuck here too. Any help would be appreciated. Thanks!

Do you know how to find the Eigenvalues and Eigenvectors of a matrix?

To find the eigenvalues, you solve for the roots of the characteristic polynomial using $(\mathbf{A} - \lambda \mathbf{I}) = 0$?

$$\mathbf{A} = \begin{bmatrix} 13 & 12\\ 12 & 13 \end{bmatrix}$$

We form $(\mathbf{A} - \lambda \mathbf{I}) = 0$, so

$$(\mathbf{A} - \lambda \mathbf{I}) = \begin{bmatrix} 13- \lambda & 12\\ 12 & 13 - \lambda \end{bmatrix} = 0$$

$$(13 - \lambda)^{2} - 144 = 0, \text{so}, \lambda_{1,2} = 1, 25$$

To find the corresponding eigenvectors, you substitute each distinct (if they are not distinct, other approaches are needed) eigenvalue into and by solving $(\mathbf{A} - \lambda_i \mathbf{I})\mathbf{x} = 0$.

For this example, you would get:

$$\lambda_1 = 1, v1 = (1, 1)$$

$$\lambda_2 = 25, v2 = (-1, 1)$$

You could also approach this using the Jordan Normal Form and many other ways too.

Can you take it from here?

Regards

• Thanks, yeah I know about the eigenvalues and eigenvectors. So my homogeneous solution would be $y_h=c_1(1,1)^T+c_2(-1,1)^T$ right? How do I go from this to the general solution to the non-homogeneous solution? thx – Slugger Jan 25 '13 at 16:30
• See Section 3. You are forgetting the exponential terms. Regards – Amzoti Jan 25 '13 at 16:33
• How'd this go with no upvotes for so long...I made one +1 dent in that! ;-) – Namaste May 6 '13 at 1:47
• @amWhy: Not even after a followup from the OP! :-) Thanks my friend. – Amzoti May 6 '13 at 1:50

Take $x_0=0$ and $y_0=(c_1,c_2)^t\in\mathbb{R}^2$. As $A$ is real and symmetric, is diagonalizable in $\mathbb{R}$.