How should I solve this simultaneous differential equations$$\begin{cases} \frac{dx}{dt}-\frac{dy}{dt} = -x+y+4t^2+10t+3 \\ \frac{dx}{dt} = -x-y+8e^{2t} \end{cases}$$
where $x=x(t),y=y(t)$.
I would like to solve for $x(t)$ and $y(t)$. Can someone help me to solve ?
Thank you in advance.
Here the system is
\begin{cases} \frac{dx}{dt}-\frac{dy}{dt} = -x+y+4t^2+10t+3 \\ \frac{dx}{dt} = -x-y+8e^{2t} \end{cases}
Putting the value of $\frac{dx}{dt}$ in the first equation we have,
$\frac{dy}{dt}+2y=8e^{2t}-4t^2-10t-3$
This is a first order differential equation, so multiplying both sides by integrating factor (I.F) is $e^{2t}$ and then integrating we have
$y e^{2t}= \int (8e^{4t}-4t^2 e^{2t}-10t e^{2t}-3 e^{2t})dt$
$\implies y e^{2t}= 2e^{4t}-4[\frac{t^2}{2}e^{2t}-\int te^{2t}dt]-\int 10t e^{2t}dt - \frac{3}{2}e^{2t}$
$\implies y e^{2t}= 2e^{4t}-2{t^2}e^{2t}-\int 6t e^{2t}dt - \frac{3}{2}e^{2t}$
$\implies y e^{2t}= 2e^{4t}-2{t^2}e^{2t}- 6[\frac{t}{2}e^{2t}-\frac{e^{2t}}{4} ] - \frac{3}{2}e^{2t} + c$, where $c$ is integrating constant.
$\implies y e^{2t}= 2e^{4t}-2{t^2}e^{2t}- 3{t}e^{2t}+c$
$\implies y = 2e^{2t}-2{t^2}- 3{t}+c e^{-2t}$
Now $\frac{dx}{dt} = -x-y+8e^{2t}\implies \frac{dx}{dt} + x = 2{t^2}+ 3{t}-c e^{-2t}+6e^{2t}$
This is again a first order differential equation, so multiplying both sides by integrating factor (I.F) is $e^{t}$ and then integrating we have
$xe^t=\int [2{t^2}e^t+ 3{t}e^t-c e^{-t}+6e^{3t}]dt$
$\implies xe^t= 2[t^2 e^{t}-\int 2t e^{t} dt] + \int [3{t}e^t-c e^{-t}+6e^{3t}]dt$
$\implies x e^t = 2 t^2 e^t-\int t e^{t} dt + c e^{-t}+2 e^{3t}$
$\implies x e^t = 2 t^2 e^t-[t e^t-e^t] + c e^{-t}+2 e^{3t}+d$, $d$ is integrating constant.
$\implies x=2 e^{2t}+2t^2-t+1+c e^{-2t}+de^{-t}$