Problem:
I came across the following differential equation set:
$$ \begin{cases} \dot{x}&=3x-2y+t\\ \dot{y}&=4x-y+t^2\\ \end{cases} $$
My solution
I followed normal practices and found the general solution to the homogeneous equation $$ \left[ \begin{array}{c} x\\ y\\ \end{array} \right] '=\left[ \begin{matrix} 3& -2\\ 4& -1\\ \end{matrix} \right] \cdot \left[ \begin{array}{c} x\\ y\\ \end{array} \right] \Rightarrow \left[ \begin{array}{c} x\\ y\\ \end{array} \right] =\left[ \begin{array}{c} e^t\left( c_1\cos 2t+c_2\sin 2t \right)\\ e^t\left( \left( c_1-c_2 \right) \cos 2t+\left( c_1+c_2 \right) \sin 2t \right)\\ \end{array} \right] $$
I selected $(c_1, c_2)=(1,0)$ and $(0,1)$ to form a fundamental matrix $X=\left[ \begin{matrix}
e^t\cos 2t& e^t\sin 2t\\
e^t\left( \cos 2t+\sin 2t \right)& e^t\left( \sin 2t-\cos 2t \right)\\
\end{matrix} \right] $, and plugged it into the variation of constants formula to get a particular solution $y(t)=X(t)\int_{0}^{t}\left(X^{-1}(\tau)\cdot \left[ \begin{array}{c}
\tau\\
\tau ^2\\
\end{array} \right]\right)\mathrm{d}\tau $, and after a one-page long calculation and the help of Wolframalpha, I got the following solution of $y$:
Question
I wonder what's wrong with my selection since the what wolfram gave me if I typed the original equation set would be like this:
So I guess there must be something in my selection of the fundamental matrix that makes the calculations really complex, but how can I improve it?
Also, what's your favourite way of solving this kind of nonhomogeneous linear differential equation sets?
We want to solve for $x,y$ such that $$ \frac{d}{dt}\left[\begin{array}{c}x \\ y\end{array}\right] = \left[\begin{array}{cc}3 & -2 \\ 4 & -1\end{array}\right]\left[\begin{array}{c}x \\ y\end{array}\right] +\left[\begin{array}{c}t \\ t^2\end{array}\right]. $$ Let $C$ be the constant coefficient matrix on the right. Then the above may be written as $$ \frac{d}{dt}\left[\begin{array}{c}x\\y\end{array}\right]-C\left[\begin{array}{c}x\\y\end{array}\right]=\left[\begin{array}{c} t \\ t^2\end{array}\right] \\ \frac{d}{dt}\left(e^{-tC}\left[\begin{array}{c} x \\ y\end{array}\right]\right) = e^{-tC}\left[\begin{array}{c} t \\ t^2\end{array}\right] $$ Integrating over $[0,t]$ and rearranging terms gives $$ e^{-tC}\left[\begin{array}{c} x \\ y\end{array}\right] = \left[\begin{array}{c}x_0\\ y_0\end{array}\right]+\int_0^t e^{-sC}\left[\begin{array}{c} s \\ s^2 \end{array}\right]ds \\ \left[\begin{array}{c} x \\ y\end{array}\right]= e^{tC}\left[\begin{array}{c}x_0 \\ y_0\end{array}\right]+\int_0^t e^{(t-s)C}\left[\begin{array}{c}s \\ s^2\end{array}\right]ds $$ Now the task is reduced to finding $e^{uC}$ for a real parameter $u$. You can use complex variables to write $$ e^{uC} = \frac{1}{2\pi i}\oint_{\gamma} \frac{e^{uz}}{zI-C}dz, $$ where the contour $\gamma$ is a positively-oriented, simple, closed, rectifiable curve enclosing all eigenvalues of $C$ in its interior. The coefficient matrix $C$ has characteristic polynomial $$ p(z)= (z-3)(z+1)+8=z^2-2z+8=(z-4)(z+2) $$ Therefore, $p(C)=0$, which gives \begin{align} p(z)I&=p(z)I-p(C)\\ &=(z^2 I-C^2)-2(z I-C) \\ &=(z I-C)(z I+C)-2(z I-C) \\ &=(z I-C)(z I+C-2I) \end{align} Hence, $z I-C$ is invertible for all $z$ for which $p(z)\ne 0$, and the inverse of $z I-C$ is given by $$ (z I-C)^{-1}=\frac{1}{p(z)}((z-2)I+C). $$ And that gives an integral which may be directly computed by residues at $z=-2,4$, assuming that the contour $C$ encloses $z=2$ and $z=4$ in its interior: $$ e^{uC} = \frac{1}{2\pi i}\oint_{\gamma}\frac{e^{uz}}{(z-4)(z+2)}((z-2)I+C)dz. $$ This is a relatively straightforward integral to evaluate by residues because the integrand has only poles of order $1$ at $z=-2,4$. Hence, $e^{uC}$ is a linear combination of matrices $I$ and $C$: \begin{align} e^{uC}&= \frac{e^{-2u}}{-6}(-4I+C)+\frac{e^{4u}}{6}(2I+C) \\ &= \left(\frac{2}{3}e^{-2u}+\frac{1}{3}e^{4u}\right)I+\left(-\frac{1}{6}e^{-2u}+\frac{1}{6}e^{4u}\right)C. \end{align} Therefore, using the above, the solution is known at this point, even though the fully expanded expression is messy: $$ \left[\begin{array}{c} x \\ y\end{array}\right]= e^{tC}\left[\begin{array}{c}x_0 \\ y_0\end{array}\right]+\int_0^t e^{(t-s)C}\left[\begin{array}{c}s \\ s^2\end{array}\right]ds $$