Question

Solve the following system of equations

dx/dt-dy/dt-2x-4y=t²

dx/dt+dy/dt-x-y=1

Answer 1

$$(x+y)'-(x+y)=1$$

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

$$e^{-t}(x+y)'-e^{-t}(x+y)=e^{-t}$$

$$(e^{-t}(x+y))'=e^{-t}$$

Integrate

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

$$e^{-t}(x+y)=-e^{-t}+C_1$$

$$x+y=-1+C_1e^{t}$$

$$y=-x-1+C_1e^{t}$$

$$y'=-x'+C_1e^{t}$$

$$x'+x'-C_1e^{t}-2x+4x+4-4C_1e^{t}=t^2$$

$$2x'+2x=5C_1e^{t}+t^2-4$$

$$x'+x=\dfrac{5}{2}C_1e^{t}+\dfrac{1}{2}t^2-2$$

$$\mu(t)=e^t$$

$$e^{t}x'+e^{t}x=\dfrac{5}{2}C_1e^{2t}+\dfrac{1}{2}e^{t}t^2-2e^{t}$$

$$d(e^{t}x)=\dfrac{5}{2}C_1e^{2t}+\dfrac{1}{2}e^{t}t^2-2e^{t}$$

Integrate

$$\int d(e^{t}x)=\int (\dfrac{5}{2}C_1e^{2t}+\dfrac{1}{2}e^{t}t^2-2e^{t})dt$$

$$\int e^{t}t^2dt=t^2e^t-2te^t+2e^{t}+C_3$$

$$e^{t}x=\dfrac{5}{4}C_1e^{2t}+\dfrac{1}{2}e^{t}t^2-e^{t}t+e^{t}-2e^{t}+C_2$$

$$x(t)=\dfrac{5}{4}C_1e^{t}+\dfrac{1}{2}t^2-t-1+C_2e^{-t}$$

$$y(t)=-\dfrac{5}{4}C_1e^{t}-\dfrac{1}{2}t^2+t+1+C_2e^{-t}-1+C_1e^{t}$$

$$y(t)=-\dfrac{1}{4}C_1e^{t}-\dfrac{1}{2}t^2+t+C_2e^{-t}$$