1.3.5.1 \(y^{\prime \prime }+y^{\prime }=0\) with IC \(y\left ( 0\right ) =-2,y^{\prime }\left ( 0\right ) =8\)

The system to solve is

\begin{align*} y^{\prime \prime }+y^{\prime } & =0\\ y\left ( 0\right ) & =-2\\ y^{\prime }\left ( 0\right ) & =8 \end{align*}

Hence in the form of \(y\left ( x_{0}\right ) =y_{0}\) and \(y^{\prime }\left ( x_{1}\right ) =y_{1}^{\prime }\) then we see that \(x_{0}=0,x_{1}=0,y_{0}=-2,y_{1}^{\prime }=8\). The general solution is

\[ y=c_{1}+c_{2}e^{-x}\]

We start by generating one equation for each \(c_{i}\) by solving for these from general solution. This gives

\begin{align} c_{1} & =y-c_{2}e^{-x}\tag {1}\\ c_{2} & =\frac {y-c_{1}}{e^{-x}} \tag {2}\end{align}

Applying the limit

\begin{align} c_{1} & =\lim _{y\rightarrow y_{0}}\lim _{x\rightarrow x_{0}}y-c_{2}e^{-x}=-2-c_{2}e^{0}=-2-c_{2}\tag {1A}\\ c_{2} & =\lim _{y\rightarrow y_{0}}\lim _{x\rightarrow x_{0}}\frac {y-c_{1}x}{x^{2}}=\frac {-2-c_{1}0}{0} \tag {2A}\end{align}

The second equation fails. We end up with only one equation which is

\begin{equation} c_{1}=-2-c_{2} \tag {1A}\end{equation}

Now we apply the second IC. For this we have to first differentiate the solution which gives

\[ y^{\prime }=-c_{2}e^{-x}\]

Solving for the remaining constant in the above gives

\begin{equation} c_{2}=\frac {-y^{\prime }}{e^{-x}} \tag {3}\end{equation}

And now we apply the second IC to (3,4) giving

\begin{equation} c_{2}=\lim _{y^{\prime }\rightarrow y_{1}^{\prime }}\lim _{x\rightarrow x_{1}}\frac {-y^{\prime }}{e^{-x}}=\frac {-8}{e^{0}}=-8 \tag {3A}\end{equation}

Hence we end up with two equations

\begin{align*} c_{1} & =-2-c_{2}\\ c_{2} & =-8 \end{align*}

Hence the solution is \(c_{1}=6,c_{2}=-8\) and therefore the solution now becomes

\[ y=6-8e^{-x}\]

This secondary algorithm was not needed in this case, since main algorithm will also work here. To see this lets solve this again for the \(c_{1},c_{2}\) using the main algorithm. In main algorithm the solution itself is plugged into each IC. Using the first IC here \(y\left ( 0\right ) =-2\) and since the solution is \(y=c_{1}+c_{2}e^{-x}\), then the first IC becomes

\begin{align} -2 & =c_{1}+c_{2}e^{0}\nonumber \\ & =c_{1}+c_{2} \tag {1B}\end{align}

To use the second IC, we have to first differentiate the solution which gives

\[ y^{\prime }=-c_{2}e^{-x}\]

Applying the second IC given by \(y^{\prime }\left ( 0\right ) =8\) then the second IC gives

\begin{align} 8 & =-c_{2}e^{0}\nonumber \\ & =-c_{2} \tag {2B}\end{align}

From (1B,2B) we see we have \(c_{c}=-8,c_{1}=6\) which is the same result found above. Hence the solution is

\[ y=6-8e^{-x}\]

The only reason to try the secondary algorithm, is when the main algorithm fail for some cases.

The secondary algorithm works only when IC are given in the form \(y\left ( x_{0}\right ) =y_{0}\). It does not work for mixed IC for example \(y\left ( x_{0}\right ) +y^{\prime }\left ( x_{0}\right ) =A\).

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