2.3.7.2 ✓ Maple. Time used: 0.008 (sec). Leaf size: 79
ode:=(T(t)+1/(t^2-T(t)^2)^(1/2))*diff(T(t),t) = T(t)/t/(t^2-T(t)^2)^(1/2)-t;
dsolve(ode,T(t), singsol=all);
\begin{align*} \frac {\left (\frac {t^{2}}{2}+\frac {T^{2}}{2}+c_1 \right ) \sqrt {-T^{2}}+T \left (\ln \left (\frac {-T^{2}+\sqrt {-T^{2}}\, \sqrt {t^{2}-T^{2}}}{t}\right )+\ln \left (2\right )-\ln \left (T\right )\right )}{\sqrt {-T^{2}}} = 0 \end{align*}
Maple trace
Methods for first order ODEs:
--- Trying classification methods ---
trying homogeneous types:
differential order: 1; looking for linear symmetries
trying exact
<- exact successful
Maple step by step
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (T \left (t \right )+\frac {1}{\sqrt {t^{2}-T \left (t \right )^{2}}}\right ) \left (\frac {d}{d t}T \left (t \right )\right )=\frac {T \left (t \right )}{t \sqrt {t^{2}-T \left (t \right )^{2}}}-t \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d t}T \left (t \right ) \\ \square & {} & \textrm {Check if ODE is exact}\hspace {3pt} \\ {} & \circ & \textrm {ODE is exact if the lhs is the total derivative of a}\hspace {3pt} C^{2}\hspace {3pt}\textrm {function}\hspace {3pt} \\ {} & {} & \frac {d}{d t}G \left (t , T \left (t \right )\right )=0 \\ {} & \circ & \textrm {Compute derivative of lhs}\hspace {3pt} \\ {} & {} & \frac {\partial }{\partial t}G \left (t , T\right )+\left (\frac {\partial }{\partial T}G \left (t , T\right )\right ) \left (\frac {d}{d t}T \left (t \right )\right )=0 \\ {} & \circ & \textrm {Evaluate derivatives}\hspace {3pt} \\ {} & {} & -\frac {1}{t \sqrt {-T^{2}+t^{2}}}-\frac {T^{2}}{t \left (-T^{2}+t^{2}\right )^{{3}/{2}}}=-\frac {t}{\left (-T^{2}+t^{2}\right )^{{3}/{2}}} \\ {} & \circ & \textrm {Simplify}\hspace {3pt} \\ {} & {} & -\frac {t}{\left (-T^{2}+t^{2}\right )^{{3}/{2}}}=-\frac {t}{\left (-T^{2}+t^{2}\right )^{{3}/{2}}} \\ {} & \circ & \textrm {Condition met, ODE is exact}\hspace {3pt} \\ \bullet & {} & \textrm {Exact ODE implies solution will be of this form}\hspace {3pt} \\ {} & {} & \left [G \left (t , T\right )=\mathit {C1} , M \left (t , T\right )=\frac {\partial }{\partial t}G \left (t , T\right ), N \left (t , T\right )=\frac {\partial }{\partial T}G \left (t , T\right )\right ] \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} G \left (t , T\right )\hspace {3pt}\textrm {by integrating}\hspace {3pt} M \left (t , T\right )\hspace {3pt}\textrm {with respect to}\hspace {3pt} t \\ {} & {} & G \left (t , T\right )=\int \left (t -\frac {T}{t \sqrt {-T^{2}+t^{2}}}\right )d t +\textit {\_F1} \left (T \right ) \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & G \left (t , T\right )=\frac {t^{2}}{2}+\frac {T \ln \left (\frac {-2 T^{2}+2 \sqrt {-T^{2}}\, \sqrt {-T^{2}+t^{2}}}{t}\right )}{\sqrt {-T^{2}}}+\textit {\_F1} \left (T \right ) \\ \bullet & {} & \textrm {Take derivative of}\hspace {3pt} G \left (t , T\right )\hspace {3pt}\textrm {with respect to}\hspace {3pt} T \\ {} & {} & N \left (t , T\right )=\frac {\partial }{\partial T}G \left (t , T\right ) \\ \bullet & {} & \textrm {Compute derivative}\hspace {3pt} \\ {} & {} & T +\frac {1}{\sqrt {-T^{2}+t^{2}}}=\frac {\ln \left (\frac {-2 T^{2}+2 \sqrt {-T^{2}}\, \sqrt {-T^{2}+t^{2}}}{t}\right )}{\sqrt {-T^{2}}}+\frac {T^{2} \ln \left (\frac {-2 T^{2}+2 \sqrt {-T^{2}}\, \sqrt {-T^{2}+t^{2}}}{t}\right )}{\left (-T^{2}\right )^{{3}/{2}}}+\frac {T \left (-4 T -\frac {2 \sqrt {-T^{2}+t^{2}}\, T}{\sqrt {-T^{2}}}-\frac {2 \sqrt {-T^{2}}\, T}{\sqrt {-T^{2}+t^{2}}}\right )}{\sqrt {-T^{2}}\, \left (-2 T^{2}+2 \sqrt {-T^{2}}\, \sqrt {-T^{2}+t^{2}}\right )}+\frac {d}{d T}\textit {\_F1} \left (T \right ) \\ \bullet & {} & \textrm {Isolate for}\hspace {3pt} \frac {d}{d T}\textit {\_F1} \left (T \right ) \\ {} & {} & \frac {d}{d T}\textit {\_F1} \left (T \right )=T +\frac {1}{\sqrt {-T^{2}+t^{2}}}-\frac {\ln \left (\frac {-2 T^{2}+2 \sqrt {-T^{2}}\, \sqrt {-T^{2}+t^{2}}}{t}\right )}{\sqrt {-T^{2}}}-\frac {T^{2} \ln \left (\frac {-2 T^{2}+2 \sqrt {-T^{2}}\, \sqrt {-T^{2}+t^{2}}}{t}\right )}{\left (-T^{2}\right )^{{3}/{2}}}-\frac {T \left (-4 T -\frac {2 \sqrt {-T^{2}+t^{2}}\, T}{\sqrt {-T^{2}}}-\frac {2 \sqrt {-T^{2}}\, T}{\sqrt {-T^{2}+t^{2}}}\right )}{\sqrt {-T^{2}}\, \left (-2 T^{2}+2 \sqrt {-T^{2}}\, \sqrt {-T^{2}+t^{2}}\right )} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} \textit {\_F1} \left (T \right ) \\ {} & {} & \textit {\_F1} \left (T \right )=\frac {T^{2}}{2}-\frac {\ln \left (T \right ) T}{\sqrt {-T^{2}}} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} \textit {\_F1} \left (T \right )\hspace {3pt}\textrm {into equation for}\hspace {3pt} G \left (t , T\right ) \\ {} & {} & G \left (t , T\right )=\frac {t^{2}}{2}+\frac {T \ln \left (\frac {-2 T^{2}+2 \sqrt {-T^{2}}\, \sqrt {-T^{2}+t^{2}}}{t}\right )}{\sqrt {-T^{2}}}+\frac {T^{2}}{2}-\frac {\ln \left (T \right ) T}{\sqrt {-T^{2}}} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} G \left (t , T\right )\hspace {3pt}\textrm {into the solution of the ODE}\hspace {3pt} \\ {} & {} & \frac {t^{2}}{2}+\frac {T \ln \left (\frac {-2 T^{2}+2 \sqrt {-T^{2}}\, \sqrt {-T^{2}+t^{2}}}{t}\right )}{\sqrt {-T^{2}}}+\frac {T^{2}}{2}-\frac {\ln \left (T \right ) T}{\sqrt {-T^{2}}}=\mathit {C1} \end {array} \]