2.1.7.1 Solved using first_order_ode_linear
Entering first order ode linear solver
\[\begin {aligned} v^{\prime }+u^{2} v&=\sin \left (u \right ) \end {aligned}\]
In canonical form a linear first order is
\begin{align*} v^{\prime } + q(u)v &= p(u) \end{align*}
Comparing the above to the given ode shows that
\begin{align*} q(u) &=u^{2}\\ p(u) &=\sin \left (u \right ) \end{align*}
The integrating factor \(\mu \) is
\begin{align*} \mu &= e^{\int {q\,du}}\\ &= {\mathrm e}^{\int u^{2}d u}\\ &= {\mathrm e}^{\frac {u^{3}}{3}} \end{align*}
The ode becomes
\begin{gather*} \begin {aligned} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}u}}\left ( \mu v\right ) &= \mu p \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}u}}\left ( \mu v\right ) &= \left (\mu \right ) \left (\sin \left (u \right )\right )\\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}u}} \left (v \,{\mathrm e}^{\frac {u^{3}}{3}}\right ) &= \left ({\mathrm e}^{\frac {u^{3}}{3}}\right ) \left (\sin \left (u \right )\right )\\ \mathrm {d} \left (v \,{\mathrm e}^{\frac {u^{3}}{3}}\right ) &= \left (\sin \left (u \right ) {\mathrm e}^{\frac {u^{3}}{3}}\right )\, \mathrm {d} u \end {aligned} \end{gather*}
Integrating gives
\begin{align*} v \,{\mathrm e}^{\frac {u^{3}}{3}}&= \int {\sin \left (u \right ) {\mathrm e}^{\frac {u^{3}}{3}} \,du} \\ &=\int \sin \left (u \right ) {\mathrm e}^{\frac {u^{3}}{3}}d u + c_1 \end{align*}
Dividing throughout by the integrating factor \({\mathrm e}^{\frac {u^{3}}{3}}\) gives the final solution
\[ v = {\mathrm e}^{-\frac {u^{3}}{3}} \left (\int \sin \left (u \right ) {\mathrm e}^{\frac {u^{3}}{3}}d u +c_1 \right ) \]
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| Direction field with Isoclines | Direction field |
Summary of solutions found
\[
v = {\mathrm e}^{-\frac {u^{3}}{3}} \left (\int \sin \left (u \right ) {\mathrm e}^{\frac {u^{3}}{3}}d u +c_1 \right )
\]