4.1.1.0.3 Example 3

\begin{align} x^{\prime }\left ( t\right ) +y^{\prime }\left ( t\right ) & =x+2y+2e^{t}\tag {1}\\ x^{\prime }\left ( t\right ) +y^{\prime }\left ( t\right ) & =3x+4y+e^{2t}\tag {2}\end{align}

Hence

\begin{align} x+2y+2e^{t} & =3x+4y+e^{2t}\nonumber \\ y & =-x-\frac {1}{2}e^{2t}+e^{t}\tag {3}\end{align}

Taking derivative w.r.t. \(t~\) gives

\begin{equation} y^{\prime }=-x^{\prime }-e^{2t}+e^{t}\tag {4}\end{equation}

Substituting (3,4) in (1) to eliminate \(y,y^{\prime }\) gives

\begin{align} x^{\prime }+\left ( -x^{\prime }-e^{2t}+e^{t}\right ) & =x+2\left ( -x-\frac {1}{2}e^{2t}+e^{t}\right ) +2e^{t}\nonumber \\ x^{\prime }-x^{\prime }-e^{2t}+e^{t} & =x-2x-e^{2t}+2e^{t}+2e^{t}\nonumber \\ 0 & =-x+3e^{t}\nonumber \\ x & =3e^{t}\tag {5}\end{align}

Substituting this in (3) gives

\begin{align*} y & =-3e^{t}-\frac {1}{2}e^{2t}+e^{t}\\ & =-2e^{t}-\frac {1}{2}e^{2t}\end{align*}

Hence the solution is

\begin{align*} x & =3e^{t}\\ y & =-2e^{t}-\frac {1}{2}e^{2t}\end{align*}

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