2.6.12.1 Solved using first_order_ode_linear
Entering first order ode linear solver
\[\begin {aligned} v^{\prime }+2 v u&=2 u \end {aligned}\]
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) &=2 u\\ p(u) &=2 u \end{align*}
The integrating factor \(\mu \) is
\begin{align*} \mu &= e^{\int {q\,du}}\\ &= {\mathrm e}^{\int 2 u d u}\\ &= {\mathrm e}^{u^{2}} \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 (2 u\right )\\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}u}} \left (v \,{\mathrm e}^{u^{2}}\right ) &= \left ({\mathrm e}^{u^{2}}\right ) \left (2 u\right )\\ \mathrm {d} \left (v \,{\mathrm e}^{u^{2}}\right ) &= \left (2 u \,{\mathrm e}^{u^{2}}\right )\, \mathrm {d} u \end {aligned} \end{gather*}
Integrating gives
\begin{align*} v \,{\mathrm e}^{u^{2}}&= \int {2 u \,{\mathrm e}^{u^{2}} \,du} \\ &={\mathrm e}^{u^{2}} + c_1 \end{align*}
Dividing throughout by the integrating factor \({\mathrm e}^{u^{2}}\) gives the final solution
\[ v = {\mathrm e}^{-u^{2}} c_1 +1 \]
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Summary of solutions found
\[
v = {\mathrm e}^{-u^{2}} c_1 +1
\]