2.1.14.1 Solved using first_order_ode_linear

Entering first order ode linear solver

\[\begin {aligned} 1+2 x+\left (-t^{2}+4\right ) x^{\prime }&=0 \end {aligned}\]

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) &=-\frac {2}{t^{2}-4}\\ p(t) &=\frac {1}{t^{2}-4} \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,dt}}\\ &= {\mathrm e}^{\int -\frac {2}{t^{2}-4}d t}\\ &= \frac {\sqrt {t +2}}{\sqrt {-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 (\frac {1}{t^{2}-4}\right )\\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}t}} \left (\frac {x \sqrt {t +2}}{\sqrt {-2+t}}\right ) &= \left (\frac {\sqrt {t +2}}{\sqrt {-2+t}}\right ) \left (\frac {1}{t^{2}-4}\right )\\ \mathrm {d} \left (\frac {x \sqrt {t +2}}{\sqrt {-2+t}}\right ) &= \left (\frac {\sqrt {t +2}}{\left (t^{2}-4\right ) \sqrt {-2+t}}\right )\, \mathrm {d} t \end {aligned} \end{gather*}

Integrating gives

\begin{align*} \frac {x \sqrt {t +2}}{\sqrt {-2+t}}&= \int {\frac {\sqrt {t +2}}{\left (t^{2}-4\right ) \sqrt {-2+t}} \,dt} \\ &=-\frac {\sqrt {t +2}}{2 \sqrt {-2+t}} + c_1 \end{align*}

Dividing throughout by the integrating factor \(\frac {\sqrt {t +2}}{\sqrt {-2+t}}\) gives the final solution

\[ x = \frac {2 c_1 \sqrt {-2+t}-\sqrt {t +2}}{2 \sqrt {t +2}} \]

Direction field with Isoclines	Direction field

Summary of solutions found

\[ x = \frac {2 c_1 \sqrt {-2+t}-\sqrt {t +2}}{2 \sqrt {t +2}} \]