1.2.9 Example 10 \(y^{\prime }=-y-\sin \left ( x\right ) \)

\[ p=-y-\sin \left ( x\right ) \]

This example was added to show that parametric method works also for standard first order ode’s such as linear, separable and so on. But it should not be used for these, as the generated ode can be more complicated than the original ode and there is no point of using this method here. This method should be used only for ode’s which are non-linear first order as in all the examples above.

Isolating \(y\) gives

\begin{align} y & =-\sin x-p\tag {1}\\ & =f\left ( x,p\right ) \nonumber \end{align}

Using (2A)

\begin{align*} \frac {dx}{dp} & =\frac {\frac {df}{dp}}{p-\frac {\partial f}{\partial x}}\\ & =\frac {-1}{p-\cos x}\end{align*}

We see that this ode is much more complicated to solve that the original ode (which is linear ode and can be easily solved using integrating factor). Solving this ode gives

\begin{equation} p-\frac {\cos x}{2}-\frac {\sin x}{2}-e^{-x}c=0 \tag {2}\end{equation}

Eliminating \(p\) from (1,2) gives

\begin{align*} y & =-\frac {e^{x}\left ( 3\sin x+\cos x\right ) +2c}{2e^{x}}\\ & =-\frac {1}{2}\left ( 3\sin x+\cos x\right ) +ce^{-x}\end{align*}

Which is the same solution we could found by solving the original ode using standard integrating factor method much more easily.