A function $y(x)$ satisfies the differential equation $y^{\prime}=4\sqrt{y-x^2}$ It is known that $y(1)=2$. Find $y(3)$.

Question

A function $y(x)$ satisfies the differential equation $$y^{\prime}=4\sqrt{y-x^2}$$

It is known that $y(1)=2$. Find $y(3)$.

My attempt: Clearly $y^{\prime}=4$ at $x=1$. That's all(LOL). Any hint to proceed will be much appreciated.

Answer 1

For amusement, let us solve the ODE/IVP for the case $y(1) > 2$ the hard way and then discover the solution for the case $y(1) = 2$ as some sort of a limit.

Let $z = \sqrt{y-x^2}$, we have $y = x^2 + z^2$ and

$$y' = 4\sqrt{y-x^2} \iff (x^2+z^2)' = 4z \iff zz' + x = 2z \iff z' = 2 - \frac{x}{z}$$ Let $u = \frac{x}{z} \iff z = \frac{z}{u}$. We will temporarily assume $y(1) > 2$ so that $z > 1$ and $u < 1$. Under such an assumption, the last expression is equivalent to

$$\frac{1}{u} - \frac{xu'}{u^2} = 2 - u \iff xu' = u^2( u - 2 + \frac{1}{u}) = u(u-1)^2$$ This leads to $$\begin{align}\frac{dx}{x} &= \frac{du}{u(u-1)^2} = \left(\frac{1}{u-1}-\frac{1}{u}\right)\frac{du}{u-1} = \left(\frac{1}{(u-1)^2} - \frac{1}{u-1} + \frac{1}{u}\right)du\\ &= d\left[\log\left(\frac{u}{1-u}\right) + \frac{u}{1-u}\right] \end{align}$$ This implies $$\frac{u}{1-u}\exp\left(\frac{u}{1-u}\right) = Kx \iff \frac{u}{1-u} = W(Kx) \iff u = \frac{W(Kx)}{1+W(Kx)}$$ for some suitable chosen constant $K$ and $W(x)$ is the Lambert W function. As a result, if $y(1) > 2$, we have

$$z = \frac{x}{u} = x \left(\frac{1}{W(Kx)} + 1\right) \implies y = x^2 + z^2 = x^2\left[1 + \left(\frac{1}{W(Kx)} + 1\right)^2\right]$$ Substituting the boundary condition at $x = 1$, we have

$$K = W^{-1}\left(\frac{1}{\sqrt{y(1)-1}-1}\right) = \frac{1}{\sqrt{y(1)-1}-1} \exp\left(\frac{1}{\sqrt{y(1)-1}-1}\right)$$

When we reduce $y(1)$ to $2^{+}$, $K \to +\infty$, for any $x > 0$, $\frac{1}{W(Kx)} \to 0$. This suggests in the limit $y(1) = 2$, we have

$$y \stackrel{?}{=} x^2 \left(1 + (0 + 1)^2\right) = 2x^2$$

Substitute this back into the original ODE $y' = 4\sqrt{y - x^2}$, $y = 2x^2$ is indeed a solution of it.
So $y(3) = 2\times 3^2 = 18$.

Answer 2

Obviously, the solution of $y^{\prime}=4\sqrt{y-x^2}$ with condition $y(1)=2$ is : $$y(x)=2x^2$$ I let find $y(3)$.

What is more interresting (but outside the scope of the question) is how to solve the ODE for the general solution:

Note : the case $C=0$ corresponds to the particular solution $y=2x^2$ . The parametric form is not valid in this case because $F=2$ , meanwhile the implicit equation which is reduced to $\sqrt{y-x^2}-x=0$ continues to be valid.