3.6.1.1 Examples

3.6.1.1.1 Example 1

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

Hence

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

Taking derivative w.r.t. \(t~\) gives

\begin{equation} y^{\prime }=-\frac {x^{\prime }}{2}+\frac {1}{2}-\frac {1}{2}e^{t} \tag {4}\end{equation}

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

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

This is linear ode. Its solution is

\begin{equation} x=c_{1}e^{t}-3t-2 \tag {6}\end{equation}

Substituting this in (3) gives

\begin{align*} y & =-\frac {1}{2}\left ( c_{1}e^{t}-3t-2\right ) +\frac {1}{2}t-\frac {1}{2}e^{t}\\ & =2t-\frac {1}{2}e^{t}-\frac {1}{2}c_{1}e^{t}+1 \end{align*}

3.6.1.1.2 Example 2

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

Let \(x^{\prime }=A,y^{\prime }=B\) then

\begin{align} A+B & =x+y+t\tag {1}\\ 2A+B & =2x+3y+e^{t} \tag {2}\end{align}

From (1), \(B=x+y+t-A\). Substituting in (2) gives

\begin{align} 2A+\left ( x+y+t-A\right ) & =2x+3y+e^{t}\nonumber \\ A & =x-t+2y+e^{t} \tag {3}\end{align}

Now we plugin the above in (1) which gives

\begin{align} \left ( x-t+2y+e^{t}\right ) +B & =x+y+t\nonumber \\ B & =2t-y-e^{t} \tag {4}\end{align}

Hence we have the following two linear ode’s of standard form now. These are (3,4)

\begin{align*} x^{\prime } & =x-t+2y+e^{t}\\ y^{\prime } & =2t-y-e^{t}\end{align*}

And now these can be solved using standard methods.

3.6.1.1.3 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*}