Let c >0. Show that the function $\mathbf{\varphi }\left(x\right)\mathbf{=}{\left(c^2-x^2\right)}^{\mathbf{-}\mathbf{1}}$is a solution to the initial value problem$\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{2}{\mathbf{xy}}^{\mathbf{2}}\mathbf{,}\mathbf{y}\left(0\right)\mathbf{=}\frac{\mathbf{1}}{{\mathbf{c}}^{\mathbf{2}}}\mathbf{,}$on the interval$\mathbf{-}\mathbf{c}\mathbf{<}\mathbf{x}\mathbf{<}\mathbf{c}$. Note that this solution becomes unbounded as x approaches $\mathbf{\pm }\mathit{c}$. Thus, the solution exists on the interval$\left(-\delta ,\delta \right)$ with $\mathit{\delta }\mathbf{=}\mathit{c}$, but not for larger$\mathit{\delta }$. This illustrates that in Theorem 1, the existence interval can be quite small (IFC is small) or quite large (if c is large). Notice also that there is no clue from the equation$\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{2}{\mathbf{xy}}^{\mathbf{2}}$ itself, or from the initial value, that the solution will “blow up” at$\mathit{x}\mathbf{=}\mathbf{\pm }\mathit{c}$.

First of all, take the given function as,$\varphi \left(\mathrm{x}\right)=\mathrm{y}$

Differentiating $\varphi \left(\mathrm{x}\right)={\left({\mathrm{c}}^2-{\mathrm{x}}^2\right)}^{-1}$, concerning x,

$\frac{\mathrm{dy}}{\mathrm{dx}}=-1{\left({\mathrm{c}}^2-{\mathrm{x}}^2\right)}^{-2}\times \left(-2\mathrm{x}\right)$

In step 2, we get$\frac{\mathrm{dy}}{\mathrm{dx}}=-1{\left({\mathrm{c}}^2-{\mathrm{x}}^2\right)}^{-2}\times \left(-2\mathrm{x}\right)$

$\begin{array}{l}\frac{\mathrm{dy}}{\mathrm{dx}}=2\mathrm{x}{\left({\mathrm{c}}^2-{\mathrm{x}}^2\right)}^{-2}\\ \frac{\mathrm{dy}}{\mathrm{dx}}=2\mathrm{x}{\left({\left({\mathrm{c}}^2-{\mathrm{x}}^2\right)}^{-1}\right)}^2\\ \frac{\mathrm{dy}}{\mathrm{dx}}=2{\mathrm{xy}}^2\end{array}$

Which is identical to the given differential equation.

Hence, $\varphi \left(\mathbf{x}\right)\mathbf{=}{\left({\mathbf{c}}^{\mathbf{2}}\mathbf{-}{\mathbf{x}}^{\mathbf{2}}\right)}^{\mathbf{-}\mathbf{1}}$is a solution to $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{2}{\mathbf{xy}}^{\mathbf{2}}$forrole="math" localid="1663948016242" $\mathbf{c}\mathbf{>}\mathbf{0}$.

Here, $\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)=2{\mathrm{xy}}^2$and $\frac{\partial \mathrm{f}}{\partial \mathrm{y}}=4\mathrm{xy}$ one finds thatboth of the functions $\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)$and $\frac{\partial \mathrm{f}}{\partial \mathrm{y}}$are continuous in any rectangle $\frac{\mathrm{dy}}{\mathrm{dx}}=2{\mathrm{xy}}^2$containing the point $\left(0,\frac{1}{{\mathrm{c}}^2}\right)$, so the hypotheses of Theorem 1 are satisfied. It then follows from the theorem that the given initial value problem has a unique solution in an interval about $\mathrm{x}=0$of the form $\left(0-\delta ,0+\delta \right)$, where $\delta$ is some positive number.

Hence,$\mathrm{\varphi }\left(\mathbf{x}\right)\mathbf{=}{\left({\mathbf{c}}^{\mathbf{2}}\mathbf{-}{\mathbf{x}}^{\mathbf{2}}\right)}^{\mathbf{-}\mathbf{1}}$ is a solution to the initial value problem$\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{2}{\mathbf{xy}}^{\mathbf{2}}\mathbf{,}\mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\frac{\mathbf{1}}{{\mathbf{c}}^{\mathbf{2}}}\mathbf{,}$ on the interval $\mathbf{-}\mathbf{c}\mathbf{<}\mathbf{x}\mathbf{<}\mathbf{c}$.