Example 8

\begin{align*} \frac {dy}{dt} & =\frac {-y^{2}-3ty}{t^{2}+yt}\\ y\left ( 2\right ) & =1 \end{align*}

Let \(y=ut\) or \(u=\frac {y}{t}\), hence \(\frac {dy}{dt}=t\frac {du}{dt}+u\) and the above ode becomes

\begin{align*} t\frac {du}{dt}+u & =\frac {-u^{2}t^{2}-3t^{2}u}{t^{2}+ut^{2}}\\ t\frac {du}{dt}+u & =\frac {-u^{2}-3u}{1+u}\\ t\frac {du}{dt} & =\frac {-u^{2}-3u}{1+u}-u\\ & =\frac {-u^{2}-3u-u\left ( 1+u\right ) }{1+u}\\ & =\frac {-u^{2}-3u-u-u^{2}}{1+u}\\ & =\frac {-2u^{2}-4u}{1+u}\end{align*}

Which is separable.

\begin{align*} \left ( \frac {1+u}{2u^{2}+4u}\right ) du & =-\frac {1}{t}dt\\ \frac {1}{2}\int \left ( \frac {1+u}{u^{2}+2u}\right ) du & =-\int \frac {1}{t}dt\\ \frac {1}{2}\ln \left ( 2u+u^{2}\right ) & =-2\ln t+c_{1}\\ \ln \left ( 2u+u^{2}\right ) & =-4\ln t+c_{2}\end{align*}

Or

\[ 2u+u^{2}=c_{3}\frac {1}{t^{4}}\]

But \(u=\frac {y}{t}\). Hence the above becomes

\begin{equation} 2\frac {y}{t}+\left ( \frac {y}{t}\right ) ^{2}=c_{3}\frac {1}{t^{4}} \tag {1}\end{equation}

Applying IC \(y\left ( 2\right ) =1\) the above becomes

\begin{align*} 2\frac {1}{2}+\left ( \frac {1}{2}\right ) ^{2} & =c_{3}\frac {1}{2^{4}}\\ 1+\frac {1}{4} & =\frac {c_{3}}{16}\\ c_{3} & =\frac {5}{4}\left ( 16\right ) \\ & =20 \end{align*}

Hence (1) becomes

\[ 2\frac {y}{t}+\left ( \frac {y}{t}\right ) ^{2}=\frac {20}{t^{4}}\]

Or

\begin{align*} y_{1} & =\frac {-t^{2}+\sqrt {t^{4}+20}}{t}\\ y_{2} & =\frac {-t^{2}-\sqrt {t^{4}+20}}{t}\end{align*}

Whenever we get more than one solution, we should verify each solution satisfies the ode and IC as some can be extraneous When we do this, we will find both solutions satisfy the ode itself, but \(y_{2}\) does not satisfy the IC. Hence it is now removed. The final solution is therefore

\[ y_{1}=\frac {-t^{2}+\sqrt {t^{4}+20}}{t}\]