Question

Solve the following differential equation

dx/dt+dy/dt-2x-4y=e^t

dx/dt+dy/dt-y=e^4t

Answer 1

$$2x+4y+e^{t}=y+e^{4t}$$

$$x=-\dfrac{3}{2}y-\dfrac{1}{2}e^{t}+\dfrac{1}{2}e^{4t}$$

$$x'=-\dfrac{3}{2}y'-\dfrac{1}{2}e^{t}+2e^{4t}$$

$$-\dfrac{3}{2}y'-\dfrac{1}{2}e^{t}+2e^{4t}+y'+3y+e^{t}-e^{4t}-4y=e^{t}$$

$$\dfrac{1}{2}y'+y=-\dfrac{1}{2}e^{t}+e^{4t}$$

$$y'+2y=-e^{t}+2e^{4t}$$

$$\mu(t)=e^{2t}$$

$$e^{2t}y'+2e^{2t}y=-e^{3t}+2e^{6t}$$

$$(e^{2t}y)'=-e^{3t}+2e^{6t}$$

Integrate

$$\int d(e^{2t}y)=\int(-e^{3t}+2e^{6t})dt$$

$$e^{2t}y=-\dfrac{1}{3}e^{3t}+\dfrac{1}{3}e^{6t}+C_1$$

$$y(t)=-\dfrac{1}{3}e^{t}+\dfrac{1}{3}e^{4t}+C_1e^{-2t}$$

$$x(t)=\dfrac{1}{2}e^{t}-\dfrac{1}{2}e^{4t}-\dfrac{3C_1}{2}e^{-2t}-\dfrac{1}{2}e^{t}+\dfrac{1}{2}e^{4t}$$

$$x(t)=-\dfrac{3C_1}{2}e^{-2t}$$

$$y(t)=-\dfrac{1}{3}e^{t}+\dfrac{1}{3}e^{4t}+C_1e^{-2t}$$