2.1.19.1 Solved using first_order_ode_linear

Entering first order ode linear solver

\[\begin {aligned} x^{\prime }+2 x&={\mathrm e}^{t} \end {aligned}\]

In canonical form a linear first order is

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

Comparing the above to the given ode shows that

\begin{align*} q(t) &=2\\ p(t) &={\mathrm e}^{t} \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,dt}}\\ &= {\mathrm e}^{\int 2d t}\\ &= {\mathrm e}^{2 t} \end{align*}

The ode becomes

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

Integrating gives

\begin{align*} x \,{\mathrm e}^{2 t}&= \int {{\mathrm e}^{t} {\mathrm e}^{2 t} \,dt} \\ &=\frac {{\mathrm e}^{3 t}}{3} + c_1 \end{align*}

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

\[ x = \frac {{\mathrm e}^{t}}{3}+c_1 \,{\mathrm e}^{-2 t} \]

Direction field with Isoclines	Direction field

Summary of solutions found

\[ x = \frac {{\mathrm e}^{t}}{3}+c_1 \,{\mathrm e}^{-2 t} \]