Question

Use the elimination method to find a general solution for the given linear system, where differentiation is with respect to t. 2x' y'-4x-4y=e^-t x' y' 2x y=e^4t

Answer 1

It looks like the system of ODEs is

$\begin{cases} 2x' + y' - 4x - 4y = e^{-t} \\ x' + y' + 2x + y = e^{4t} \end{cases}$

Differentiate both sides of both equations with respect to $t$.

$\begin{cases} 2x'' + y'' - 4x' - 4y' = -e^{-t} \\ x'' + y'' + 2x' + y' = 4e^{4t} \end{cases}$

Eliminating the exponential terms, we have

$(2x' + y' - 4x - 4y) + (2x'' + y'' - 4x' - 4y') = e^{-t} + (-e^{-t}) \\\\ \implies (2x'' - 2x' - 4x) + (y'' - 3y' - 4y) = 0$

$(x'' + y'' + 2x' + y') - 4 (x' + y' + 2x + y) = 4e^{4t} - 4\cdot e^{4t} \\\\ \implies (x'' - 2x' - 8x) + (y'' - 3y' - 4y) = 0$

Now we can eliminate $y$ and it derivatives.

$\bigg((2x'' - 2x' - 4x) + (y'' - 3y' - 4y)\bigg) - \bigg((x'' - 2x' - 8x) + (y'' - 3y' - 4y)\bigg) = 0 - 0 \\\\ \implies x'' + 4x = 0$

Solve for $x$. The characteristic equation is $r^2 + 4 = 0$ with roots at $r=\pm2i$, hence the characteristic solution is

$\boxed{x(t) = C_1 \cos(2t) + C_2 \sin(2t)}$

Solve for $y$. Substituting $x$ into the second ODE gives

$x' + y' + 2x + y = e^{4t} \\\\ \implies y' + y = e^{4t} + C_1 \cos(2t) + C_2 \sin(2t)$

The characteristic equation this time is $r + 1 = 0$ with a root at $r=-1$, hence the characteristic solution is

$y(t) = C_3 e^{-t}$

Assume a particular solution with unknown coefficients $a,b,c$ of the form

$y_p = ae^{4t} + b \cos(2t) + c \sin(2t) \\\\ \implies {y_p}' = 4ae^{4t} - 2b\sin(2t) + 2c\cos(2t)$

Substituting into the ODE gives

$5ae^{4t} + (b+2c) \cos(2t) + (-2b+c) \sin(2t) = e^{4t} + C_1 \cos(2t) + C_2 \sin(2t) \\\\ \implies \begin{cases}5a = 1 \\ b+2c = C_1 \\ -2b+c = C_2\end{cases} \\\\ \implies a=\dfrac15, b=\dfrac{C_1-2C_2}5, c=\dfrac{2C_1+C_2}5$

so that the general solution is

$\boxed{y(t) = \dfrac15 e^{4t} + \dfrac{C_1-2C_2}5 \cos(2t) + \dfrac{2C_1+C_2}5 \sin(2t) + C_3 e^{-t}}$