Question

(a) Without solving, explain why the initial-value problem

$\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\sqrt{\mathbf{y}}\mathbf{,}\mathbf{y}\mathbf{\left(}{\mathbf{x}}_{\mathbf{0}}\mathbf{\right)}\mathbf{=}{\mathbf{y}}_{\mathbf{0}}$

has no solution for ${\mathit{y}}_{\mathbf{0}}\mathbf{<}\mathbf{0}$.

(b) Solve the initial-value problem in part (a)${\mathit{y}}_{\mathbf{0}}\mathbf{>}\mathbf{0}$ for and find the largest interval on which the solution is defined.

Short answer

(a) The initial-value problem$\frac{dy}{dx}=\sqrt{y}$ has no solution for$y_0<0$ because$\sqrt{y}$ is not a real number.

(b) The interval on which the solution is defined is $(x_0-2\sqrt{y_0},\infty )$.

Step-by-step solution

Note the given data

(a)

Given the initial value problem$\frac{\mathrm{dy}}{\mathrm{dx}}=\sqrt{\mathrm{y}}$and $y_0<0$.

Since given $y_0<0$,$\sqrt{y}$is not a real number.

Calculation part and finding the required interval

(b)

We solve given IVP by using separable method as:

$\begin{array}{c}\frac{dy}{\sqrt{y}}=dx\\ \int y^{-\frac{1}{2}}dy=\int dx\\ 2\sqrt{y}=x+c\end{array}$

where c is the constant of integration

Now applying the given condition$y\left(x_0\right)=y_0$as follows:

$\begin{array}{c}2\sqrt{y_0}=x_0+c\\ c=2\sqrt{y_0}-x_0\end{array}$

Substitute $c=2\sqrt{y_0}-x_0$into $2\sqrt{y}=x+c$as:

$\begin{array}{c}2\sqrt{y}=x+2\sqrt{y_0}-x_0\\ \sqrt{y}=\frac{1}{2}(x+2\sqrt{y_0}-x_0)\\ y=\frac{1}{4}{(x+2\sqrt{y_0}-x_0)}^2\end{array}$

Differentiating $y=\frac{1}{4}{(x+2\sqrt{y_0}-x_0)}^2$on both sides with respect to x as:

$\frac{dy}{dx}=\frac{1}{2}(x+2\sqrt{y_0}-x_0)$

Since,$\sqrt{y}>0$,$\frac{dy}{dx}>0$ .

Therefore, the interval on which the solution is defined is$(x_0-2\sqrt{y_0},\infty )$ .