4.4.2.1 Example 1

Solve

\[ \left ( -y\sin y+\cos y\right ) y^{\prime \prime }-\left ( y^{\prime }\right ) ^{2}\left ( 2\sin y+y\cos y\right ) =\sin x \]

Comparing the above to (1) shows that

\begin{align*} a_{2} & =-y\sin y+\cos y\\ a_{1} & =-\left ( 2\sin y+y\cos y\right ) y^{\prime }\\ a_{0} & =-\sin x \end{align*}

Checking the exactness conditions in (2) shows they are all satisfied. Since no initial conditions are given, then we will use (4). This gives

\begin{align*} \int _{0}^{x}-\sin \left ( \alpha \right ) d\alpha +\int _{0}^{y}-\left ( 2\sin \beta +\beta \cos \beta \right ) y^{\prime }d\beta +\int _{0}^{y^{\prime }}\left ( -\left ( 0\right ) \sin \left ( 0\right ) +\cos \left ( 0\right ) \right ) d\gamma & =c_{1}\\ -\int _{0}^{x}\sin \left ( \alpha \right ) d\alpha -\int _{0}^{y}\left ( 2\sin \beta +\beta \cos \beta \right ) y^{\prime }d\beta +\int _{0}^{y^{\prime }}d\gamma & =c_{1}\\ -\int _{0}^{x}\sin \left ( \alpha \right ) d\alpha -y^{\prime }\int _{0}^{y}\left ( 2\sin \beta +\beta \cos \beta \right ) d\beta +\int _{0}^{y^{\prime }}d\gamma & =c_{1}\\ \left ( -1+\cos x\right ) -y^{\prime }\left ( 1+y\sin y-\cos y\right ) +y^{\prime } & =c_{1}\\ y^{\prime }\left ( 1-\left ( 1+y\sin y-\cos y\right ) \right ) & =1-\cos x+c_{1}\\ y^{\prime }\left ( \cos y-y\sin y\right ) & =1-\cos x+c_{1}\end{align*}

Solving gives

\[ y\cos y=c_{1}x+x-\sin x+c_{2}\]

And this is the solution to original ode.