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Given the following 2x2 system of differential equations,

$\begin{cases} \dot{x} = 4x-y \\ \dot{y} = 2x+y \end{cases}$

A solution is easy to fins by computing eigenvalues and eigenvectors of the associated matrix A

$A = \begin{bmatrix} 4 & -1 \\ 2 & 1 \end{bmatrix}$.

Therefore, knowing that eigenvalues are 2 and 3, and the related eigenvectors are $\begin{bmatrix} 1 \\ 2 \end{bmatrix}$ and $\begin{bmatrix} 1 \\ 1 \end{bmatrix}$ respectively, the solution is of the form

$Ae^{2t}\begin{bmatrix} 1 \\ 2 \end{bmatrix}+Be^{3t}\begin{bmatrix} 1 \\ 1 \end{bmatrix}$.

What about if I make the initial system of equations non-autonomous? Resulting in something as follows

$\begin{cases} \dot{x} = 4x-y \\ \dot{y} = 2x+y +f(t) \end{cases}$

where f(t) can be either a polynomial on t, an exponential ($e^{at}$) or a trigonometric function ($sin(at)$ or $cos(at)$).

What is now the procedure to follow this system of differential equations? For example, let $f(t)=t^2$.

How do I solve the following system of differential equations?

$\begin{cases} \dot{x} = 4x-y \\ \dot{y} = 2x+y+t^2 \end{cases}$

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  • $\begingroup$ Hint: Use Laplace transform. $\endgroup$
    – xpaul
    Dec 5 '20 at 18:12
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$$\begin{cases} x' = 4x-y \\ {y'} = 2x+y +f(t) \end{cases}$$ Note that if you substract both DE you have: $$x'-y' =2(x-y) -f(t) $$ $$u' -2u= -f(t) $$ $$(u(t)e^{-2t})'= -e^{-2t}f(t) $$ $$u(t)e^{-2t}= -\int e^{-2t}f(t) \,dt $$ Where $u(t)=x-y$. $$x(t)=y(t)-e^{2t}\int e^{-2t}f(t) \,dt $$ Plug this in the second equation and solve.


Continuing from your result of integration in the comment:

$$u(t)=\frac{1}{4}(2t^2+2t+1)+Ce^{2t}$$ $$x-y=\frac{1}{4}(2t^2+2t+1)+Ce^{2t}$$ We have the first differential equation: $$x'=4x-y$$ $$x'-4x=-x+\frac{1}{4}(2t^2+2t+1)+Ce^{2t}$$ $$x'-3x=\frac{1}{4}(2t^2+2t+1)+Ce^{2t}$$ $$(xe^{-3t})'=e^{-3t}\frac{1}{4}(2t^2+2t+1)+Ce^{-t}$$ Integrate. Don't forget the second constant of integration.

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    $\begingroup$ @AnindyaPrithvi Integration factor $\endgroup$
    – imranfat
    Dec 5 '20 at 20:10
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    $\begingroup$ then i reach to $u(t)=\frac{1}{4}(2t^2+2t-1)$, and this can't help me to find $x(t)$ and $y(t)$, and these have to be dependent on constants determined by some initial conditions, and $u(t)$ is not dependent on constants. where am i going wrong? $\endgroup$ Dec 6 '20 at 16:33
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    $\begingroup$ Add a constant for $u$ @JenaRayner You always need a constant for integration when you integrate. $\endgroup$ Dec 6 '20 at 16:36
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    $\begingroup$ I added some lines and corrected also the result of the integration it's $u(t)=\frac{1}{4}(2t^2+2t+1)+Ce^{2t}$ @JenaRayner $\endgroup$ Dec 6 '20 at 16:47
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    $\begingroup$ Yes when you have intial conditions with the system of DE'syou can use that information to find the constants of integration @JenaRayner $\endgroup$ Dec 6 '20 at 17:09
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You can express your non$-$autonomous differential equation in the form $$\frac{d\vec{x}}{dt}-A\vec{x}=\vec{f}(t)$$ where $\vec{x}=\big(\begin{smallmatrix} x \\ y \end{smallmatrix}\big)$, $A=\big(\begin{smallmatrix} 4 & -1\\ 2 & 1 \end{smallmatrix}\big)$, and $f(t)=\big(\begin{smallmatrix} 0 \\ t^2 \end{smallmatrix}\big)$. It turns out that $$\vec{x}=e^{At}\int e^{-At}f(t)dt$$ just as you might expect!

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  • $\begingroup$ how am i supposed to integrate the exponential of a matrix? $\endgroup$ Dec 6 '20 at 16:14
  • $\begingroup$ Do you know how to compute $e^{At}$? $\endgroup$
    – Matthew H.
    Dec 6 '20 at 16:58
  • $\begingroup$ i dont, i've never seen the exponential of a matrix $\endgroup$ Dec 6 '20 at 17:03
  • $\begingroup$ Note $A=\big(\begin{smallmatrix} 1 & 1\\ 2 & 1\end{smallmatrix}\big)\big(\begin{smallmatrix} 2 & 0\\ 0 & 3\end{smallmatrix}\big)\big(\begin{smallmatrix} 1 & 1\\ 2 & 1 \end{smallmatrix}\big)^{-1}$ and so $$e^{At}=\big(\begin{smallmatrix} 1 & 1\\ 2 & 1\end{smallmatrix}\big)\big(\begin{smallmatrix} e^{2t} & 0\\ 0 & e^{3t}\end{smallmatrix}\big)\big(\begin{smallmatrix} 1 & 1\\ 2 & 1 \end{smallmatrix}\big)^{-1}$$ $\endgroup$
    – Matthew H.
    Dec 6 '20 at 17:26
  • $\begingroup$ Also $$e^{-At}=\big(\begin{smallmatrix} 1 & 1\\ 2 & 1\end{smallmatrix}\big)\big(\begin{smallmatrix} e^{-2t} & 0\\ 0 & e^{-3t}\end{smallmatrix}\big)\big(\begin{smallmatrix} 1 & 1\\ 2 & 1 \end{smallmatrix}\big)^{-1}$$ $\endgroup$
    – Matthew H.
    Dec 6 '20 at 17:31

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