 # Finding general solution for a nonhomogeneous system of equations { <mtable columnal tripes3h 2022-07-04 Answered
Finding general solution for a nonhomogeneous system of equations
$\left\{\begin{array}{l}{x}_{1}^{\prime }={x}_{2}+2{e}^{t}\\ {x}_{2}^{\prime }={x}_{1}+{t}^{2}\end{array}$
I want to find the general solution for it. I started by finding the general solution for the homogeneous equations:
$\left(\begin{array}{c}{x}_{1}\\ {x}_{2}\end{array}\right)={C}_{1}\left(\begin{array}{c}{e}^{t}+{e}^{-t}\\ {e}^{t}-{e}^{-t}\end{array}\right)+{C}_{2}\left(\begin{array}{c}{e}^{t}-{e}^{-t}\\ {e}^{t}+{e}^{-t}\end{array}\right)$
Now I need to find a "specific" solution for the nonhomogeneous equations but I have problems applying the method in which I make constants ${C}_{1}$ and ${C}_{2}$ a variable.
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System is equivalent with
${x}_{2}={x}_{1}^{\prime }-2{e}^{t},{x}_{2}^{\prime }={x}_{1}+{t}^{2}$
or
${x}_{2}={x}_{1}^{\prime }-2{e}^{t},{x}_{1}^{″}={x}_{1}+{t}^{2}+2{e}^{t}.$
Firstly, by using a method of variations of constants, we will solve the second equation:
${x}_{1}^{″}-{x}_{1}={t}^{2}+2{e}^{t},$
(after that, it will be easy to determine ${x}_{2}$ from the first equation). Solution of homogenuous equation is ${x}_{1}={C}_{1}{e}^{t}+{C}_{2}{e}^{-t}$, so we have to solve the following system:
${C}_{1}^{\prime }{e}^{t}+{C}_{2}^{\prime }{e}^{-t}=0,$
${C}_{1}^{\prime }{e}^{t}-{C}_{2}^{\prime }{e}^{-t}={t}^{2}+2{e}^{t},$
where ${C}_{1}$, ${C}_{2}$ are functions of $t$. It is easy to get that
${C}_{1}^{\prime }=1+\frac{{t}^{2}}{2}{e}^{-t},$
${C}_{2}\left(t\right)={C}_{2}-\frac{{e}^{2t}}{2}-\frac{{e}^{t}}{2}\left({t}^{2}-2t+2\right).$
Finally,
${x}_{1}=\left({C}_{1}-\frac{1}{2}\right){e}^{t}+{C}_{2}{e}^{-t}+t{e}^{t}-{t}^{2}-2,$
${x}_{2}=\left({C}_{1}-\frac{1}{2}\right){e}^{t}-{C}_{2}{e}^{-t}+\left(t-1\right){e}^{t}-2t.$

We have step-by-step solutions for your answer! woowheedr
If ${x}_{1},{x}_{2}$ are your solutions to the homogeneous equation (it doesn't actually matter what they are), look for solutions to the inhomogeneous equation in the form ${y}_{i}={v}_{i}{x}_{i}.$. If you try that, you will have simple equations for ${v}_{1},{v}_{2}.$

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