2.6.1.1 Example 1 \(y^{\prime }=y^{3}\sin x\)

Solve

\begin{align*} y^{\prime } & =y^{3}\sin x\\ y\left ( 0\right ) & =0 \end{align*}

From uniqueness and existence theory we see that solution to \(y^{\prime }=y^{3}\sin x\) exist and is unique. This is because \(f=y^{3}\sin x\) is continuous everywhere (hence solution exist) and \(f_{y}=3y^{2}\sin x\) is also continuous everywhere (hence uniqueness is guaranteed).

This is little more tricky than it looks. Notice that \(y=0\) at \(x=0\). This is special IC, because this means if we start by dividing both sides by \(y^{3}\) to separate them as we normally do, this gives

\[ \frac {dy}{y^{3}}=\sin xdx \]

But when we get to later on (after integration and adding constant of integration) to solve for \(c\) we will have problems. The reason is,  we should not divide by \(y\,\) in first place, since \(y=0\) at initial conditions. In this special IC case, then at \(x=0\) the ode is

\[ y^{\prime }=0 \]

Hence \(y=C_{1}\). But since the solution is guaranteed to be unique, then \(C_{1}\) must be zero to give \(y=0\) as only one value of \(y\left ( x\right ) \) can exist. Hence this is the solution. This way we do not even have to integrate or solve for constant of integration. If we were not given IC, then we do as normal and now can divide by \(y\). Assuming \(y\neq 0\) then the ode becomes

\[ \frac {dy}{y^{3}}=\sin xdx\qquad y\neq 0 \]

Integrating gives

\begin{align} -\frac {1}{2y^{2}} & =-\cos x+c\nonumber \\ \frac {1}{y^{2}} & =2\cos x-2c\nonumber \\ \frac {1}{y^{2}} & =2\cos x+c_{1} \tag {1}\end{align}

Hence

\[ y^{2}=\frac {1}{2\cos x+c_{1}}\]

Therefore

\begin{equation} y=\pm \frac {1}{\sqrt {2\cos x+c_{1}}} \tag {2}\end{equation}

So we should always start, when IC are given, by checking uniqueness and existence and never divide by \(y\) if \(y=0\) at initial conditions. In all other cases, we can divide to separate. Lets do more examples on this to practice.