solving inhomogenous $2$ x $2$ ODE?

Question

Let $y'=\begin{pmatrix}1 & 2 \\ 3 & 6\end{pmatrix}y+\begin{pmatrix}x \\ sin(x)\end{pmatrix}$ with $y_0=\begin{pmatrix}0 \\ 0\end{pmatrix}$

Now I want to solve it, but don't know how to continue:

$1)$ solve the homogeous ODE $y'=Ay$

$\Rightarrow y_{hom}(x)=e^{0x} \cdot c_1\begin{pmatrix}-2 \\ 1\end{pmatrix}+c_2 \cdot e^{7x}\begin{pmatrix}1 \\ 3\end{pmatrix}=c_1\begin{pmatrix}-2 \\ 1\end{pmatrix}+c_2 \cdot e^{7x}\begin{pmatrix}1 \\ 3\end{pmatrix}$

$2)$ solve particular ODE $y'=Ay+b$ with variation of parameters:

$y_{par}=Y_{hom}\cdot c(x)=\frac{1}{7}\begin{pmatrix}e^{7x}+6 & 3(e^{7x}-1)\\2(e^{7x}-1) & 6e^{7x}+1\end{pmatrix}\begin{pmatrix}\dot c_1(x) \\ \dot c_2(x)\end{pmatrix}=\begin{pmatrix}\ x \\ \ sin(x)\end{pmatrix}$

Now I solved for $c_1$ and $c_2$, but I guess it's wrong because I couldn't solve it without calculator and than I checked it , it was wrong :/ $c_1=\dfrac{\mathrm{e}^{-7x}\left(-1029\sin\left(x\right)+\mathrm{e}^{7x}\left(7350\cos\left(x\right)+1225x^2\right)-147\cos\left(x\right)+\left(2100x-300\right)\mathrm{e}^{14x}\right)}{17150}+C$

$c_2=\dfrac{\mathrm{e}^{-7x}\left(1029\sin\left(x\right)+\mathrm{e}^{7x}\left(1225\cos\left(x\right)-1225x^2\right)+147\cos\left(x\right)+\left(350x-50\right)\mathrm{e}^{14x}\right)}{8575}+C$

Is this way correct or is there another one ? Thanks in advance

Answer 1

We are given

$$y'= Ay + g = \begin{pmatrix}1 & 2 \\ 3 & 6\end{pmatrix}y+\begin{pmatrix}x \\ \sin x\end{pmatrix}, y_0=\begin{pmatrix}0 \\ 0\end{pmatrix}$$

You correctly found the homogeneous solution

$$y_h(x) = c_1\begin{pmatrix}-2 \\ 1\end{pmatrix}+c_2 \cdot e^{7x}\begin{pmatrix}1 \\ 3\end{pmatrix}$$

Since you want to use Variation of Parameters, we will follow Example 2 (I assume you understand the theory, but it is also on the site).

$$Y = \begin{pmatrix}-2 & e^{7 x}\\ 1 & 3 e^{7x}\end{pmatrix} \implies Y^{-1} = \begin{pmatrix} -\dfrac{3}{7} & \dfrac{1}{7} \\ \dfrac{e^{-7 x}}{7} & \dfrac{2 e^{-7 x}}{7} \\ \end{pmatrix}$$

We now form

$$Y^{-1} g = \begin{pmatrix} -\dfrac{3}{7} & \dfrac{1}{7} \\ \dfrac{e^{-7 x}}{7} & \dfrac{2 e^{-7 x}}{7} \\ \end{pmatrix}. \begin{pmatrix} x \\ \sin x \\ \end{pmatrix} = \begin{pmatrix} \dfrac{\sin x}{7}-\dfrac{3 x}{7} \\ \dfrac{1}{7}x e^{-7 x} +\dfrac{2}{7} e^{-7 x} \sin x \\ \end{pmatrix}$$

Next we integrate

$$\displaystyle \int_0^x Y^{-1}g ~dx = \begin{pmatrix} \dfrac{1}{14} \left(-3 x^2 -2 \cos x +2\right) \\ \dfrac{74-e^{-7 x} (175 x + 49 \cos x + 343 \sin x + 25)}{8575} \\ \end{pmatrix}$$

We can now write the particular solution $y_p(x) = Y \displaystyle \int_0^x Y^{-1}g~ dx$

$$y_p(x) = \begin{pmatrix} \dfrac{3675 x^2-175 x+74 e^{7 x}-343 \sin x+2401 \cos x-2475}{8575} \\ \dfrac{-3675 x^2-1050 x+444 e^{7 x}-2058 \sin x-2744 \cos x+2300}{17150} \end{pmatrix}$$