2.4.6.1 Solved using first_order_ode_linear

Entering first order ode linear solver

\[\begin {aligned} y^{\prime }+y \cos \left (x \right )&=\frac {\sin \left (2 x \right )}{2} \end {aligned}\]

In canonical form a linear first order is

\begin{align*} y^{\prime } + q(x)y &= p(x) \end{align*}

Comparing the above to the given ode shows that

\begin{align*} q(x) &=\cos \left (x \right )\\ p(x) &=\frac {\sin \left (2 x \right )}{2} \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,dx}}\\ &= {\mathrm e}^{\int \cos \left (x \right )d x}\\ &= {\mathrm e}^{\sin \left (x \right )} \end{align*}

The ode becomes

\begin{gather*} \begin {aligned} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu y\right ) &= \mu p \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu y\right ) &= \left (\mu \right ) \left (\frac {\sin \left (2 x \right )}{2}\right )\\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (y \,{\mathrm e}^{\sin \left (x \right )}\right ) &= \left ({\mathrm e}^{\sin \left (x \right )}\right ) \left (\frac {\sin \left (2 x \right )}{2}\right )\\ \mathrm {d} \left (y \,{\mathrm e}^{\sin \left (x \right )}\right ) &= \left (\frac {\sin \left (2 x \right ) {\mathrm e}^{\sin \left (x \right )}}{2}\right )\, \mathrm {d} x \end {aligned} \end{gather*}

Integrating gives

\begin{align*} y \,{\mathrm e}^{\sin \left (x \right )}&= \int {\frac {\sin \left (2 x \right ) {\mathrm e}^{\sin \left (x \right )}}{2} \,dx} \\ &={\mathrm e}^{\sin \left (x \right )} \left (\sin \left (x \right )-1\right ) + c_1 \end{align*}

Dividing throughout by the integrating factor \({\mathrm e}^{\sin \left (x \right )}\) gives the final solution

\[ y = {\mathrm e}^{-\sin \left (x \right )} c_1 +\sin \left (x \right )-1 \]

Direction field with Isoclines	Direction field

Summary of solutions found

\[ y = {\mathrm e}^{-\sin \left (x \right )} c_1 +\sin \left (x \right )-1 \]