My exam question reads as follows:

Given the system:

$y'_1=y_1-y_2+\frac{1}{\cos \left(x\right)}$


Solve it using the method of variation of parameters, before that describe that method for solving systems of non-h. differential equations.

I'm not sure on how to solve this system, that is only remember learning about homogeneous systems so not sure even on how to describe the general case. Either way after some googling I find out that as always the solutions is given as $Y=Y_h+Y_p$ and finding the solution to the homogeneous system is something I know how to do. But then how can I find the particular solution?

  • $\begingroup$ @Moo none, (buffer text to reach enough chars) $\endgroup$ Feb 5 '17 at 21:48

I will map it out and you can fill in the details. We are given

$$y'= Ay + g = \begin{pmatrix}1 & -1 \\ 2 & -1\end{pmatrix}y+\begin{pmatrix}\dfrac{1}{\cos x} \\ 0\end{pmatrix}$$

Using eigenvalues / eigenvectors (or other approaches), we find the homogeneous solution

$$Y_h(x) = c_1\begin{pmatrix}\sin x + \cos x \\ 2\sin x\end{pmatrix} + c_2 \begin{pmatrix}-\sin x \\ \cos x - \sin x\end{pmatrix}$$

For Variation of Parameters (there are other approaches too), we will follow Example 2

$$Y = \begin{pmatrix}\sin x + \cos x & -\sin x \\ 2\sin x & \cos x - \sin x \end{pmatrix} \implies Y^{-1} = \begin{pmatrix} \cos x-\sin x & \sin x \\-2 \sin x & \sin x + \cos x \end{pmatrix}$$

We now form

$$Y^{-1} g = \begin{pmatrix} \cos x -\sin x & \sin x \\-2 \sin x & \sin x + \cos x \end{pmatrix} \begin{pmatrix}\dfrac{1}{\cos x} \\ 0\end{pmatrix} = \begin{pmatrix} \sec x ~(\cos x - \sin x) \\ -2 \tan x \end{pmatrix} $$

Next we integrate the previous result

$$\displaystyle \int Y^{-1}g ~dx = \begin{pmatrix} x+\ln (\cos x) \\ 2 \ln (\cos x) \\ \end{pmatrix}$$

We can now write the particular solution $Y_p(x) = Y \displaystyle \int Y^{-1}g~ dx$

$$Y_p(x) = \begin{pmatrix} \cos x (x+\ln (\cos x))+\sin x (x-\ln (\cos x)) \\2 (x \sin x +\cos x \ln (\cos x))\end{pmatrix}$$

Now write

$$Y(x) = Y_h(x) + Y_p(x)$$


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