2.6.7.1 Solved using first_order_ode_linear

Entering first order ode linear solver

\[\begin {aligned} v^{\prime }+\frac {2 v}{u}&=3 v \end {aligned}\]

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) &=-\frac {3 u -2}{u}\\ p(u) &=0 \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,du}}\\ &= {\mathrm e}^{\int -\frac {3 u -2}{u}d u}\\ &= u^{2} {\mathrm e}^{-3 u} \end{align*}

The ode becomes

\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}u}} \mu v &= 0 \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}u}} \left (v \,u^{2} {\mathrm e}^{-3 u}\right ) &= 0 \end{align*}

Integrating gives

\begin{align*} v \,u^{2} {\mathrm e}^{-3 u}&= \int {0 \,du} + c_1 \\ &=c_1 \end{align*}

Dividing throughout by the integrating factor \(u^{2} {\mathrm e}^{-3 u}\) gives the final solution

\[ v = \frac {{\mathrm e}^{3 u} c_1}{u^{2}} \]

Direction field with Isoclines	Direction field

Summary of solutions found

\[ v = \frac {{\mathrm e}^{3 u} c_1}{u^{2}} \]