Question

By separation of variables, find a solution of the equation $\mathit{y}\mathbf{\text{'}}\mathbf{=}\sqrt{\mathbf{y}}$ containing one arbitrary constant. Find a particular solution satisfying y=0 when x=0. Show that y=0 is a solution of the differential equation which cannot be obtained by specializing the arbitrary constant in your solution above. Computer plot a slope field and some of the solution curves. Show that there are an infinite number of solution curves passing through any point on the xaxis, but just one through any point for which y>0 . Hint:See Example 3. Problems 17 and 18 are physical problems leading to this differential equation.

Short answer

The general solution is $y={\left(\frac{x+C}{2}\right)}^2$. Using the boundary condition C=0 and the particular solution is $y=\frac{x^2}{4}$ . The singular solution is y=0.

Step-by-step solution

Given Information.

The givenequation is$y\text{'}=\sqrt{y}$ .Particular solution and the singular solution is to be found out.

Definition of Differential equation 

A differential equation is an equation that contains at least one derivative of an unknown function, either an ordinary derivative or a partial derivative.

Find General solution

First find the general solution by separation of variables

$$\begin{gathered} \frac{dy}{dx}=y^{\frac{1}{2}} \\ \Rightarrow y^{-\frac{1}{2}}dy=dx \\ \int y^{-\frac{1}{2}}dy=\int dx \\ 2y^{\frac{1}{2}}=x+C \end{gathered}$$

Therefore, the general solution is$y={\left(\frac{x+C}{2}\right)}^2$

Find particular solution

Particular solution is to be found out by putting x=0 in the general solution

$$\begin{gathered} y={\left(\frac{x+C}{2}\right)}^2 \\ x=0 \\ \Rightarrow C=0 \end{gathered}$$

Therefore, the particular solution is

$y=\frac{x^2}{4}$

Find singular solution

By separating the variables in the differential equation

$\frac{dy}{\sqrt{y}}=dx$

The general solution is

$y={\left(\frac{x+C}{2}\right)}^2$

So, there is a solution which cannot be obtained from the general solution by putting any value of C.

Therefore the singular solution is y=0

Plot slope field and solution curves

In the slope field below, the particular solution satisfying y=0 when x=0 is shown by the blue curve, the singular solution shown by y=0 is shown by the green line and the particular solution satisfying y=1 when x=0 is shown by the green curve

The differential equation$\frac{dy}{dx}=y^{\frac{1}{2}}$, has the solution y=0 for $x=-\infty$to$+\infty$ as shown by green curve in the figure above.

The general solution $y={\left(\frac{x+C}{2}\right)}^2$, depends on the value of constant C(which could exist at any point for which $y⩾0$

Thus it can be concluded that there are an infinite number of solutions through any point of the x-axis (that is y=0 ), and only one solution through any point for which y>0

Therefore, the general solution is $y={\left(\frac{x+C}{2}\right)}^2$. Using the boundary condition C=0 and the particular solution is $y=\frac{x^2}{4}$. The singular solution is y=0.