Answer to Question #91478 in Differential Equations for Sajid

Question

Q.Choose the correct answer.
Q. The initial value value problem dy/dx=|y|1/2 ,y(0)=0 has:
a) unique solution
b) No solution
c) Infinitely many solution
d) Two solutions.

Answer 1

 The initial value value problem dy/dx=|y|1/2 ,y(0)=0 has

d) Two solutions.

Find the general solution of

$$y'=\sqrt{|y|}$$

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

1) Let $y(x) \leq 0$ :

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

We can separate the variables:

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

$$-\frac{d(-y)}{\sqrt{-y}}=dx$$

Now, integrate the left-hand side dy  and the right-hand side dx

$$-\int\frac{d(-y)}{\sqrt{-y}}=\int dx \\$$

$$-2\sqrt{-y}=x+C_1 \\$$

$$\sqrt{-y}=-\frac{1}{2}(x+C_1)$$

$$-y=\frac{1}{4}(x+C_1)^2, x+C_1\leq 0$$

$$y=-\frac{1}{4}(x+C_1)^2, x+C_1\leq 0 \text{ } (1)$$

2) Let $y( x)\geq 0$ :

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

We can separate the variables:

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

Now, integrate the left-hand side dy  and the right-hand side dx

$$\int\frac{d(y)}{\sqrt{y}}=\int dx \\$$

$$2\sqrt{y}=x+C_1 \\$$

$$\sqrt{y}=\frac{1}{2}(x+C_1)$$

$$y=\frac{1}{4}(x+C_1)^2, x+C_1\geq 0 \text{ } (2)$$

So there is our general solution (from (1), (2)): 

$$y=\frac{1}{4}(x+C_1)^2, x+C_1\geq 0\\ y=-\frac{1}{4}(x+C_1)^2, x+C_1\leq 0$$

To find the particular solution, substitute the initial condition values to obtain

$$y(0)=0 \Rightarrow 0=\frac{1}{4}(0+C_1)^2, \Rightarrow C_1=0$$

So, the particular solution that satisfies the initial condition is

$$y=-\frac{1}{4}x^2, x \geq0 \\ y=-\frac{1}{4}x^2, x \leq 0$$

But the solution to equality is $y = 0$ , which also satisfies the condition $y(0)=0$ .

The initial value value problem dy/dx=|y|1/2 ,y(0)=0 has Two solutions:

$$1)\text{ } y=\frac{1}{4}x^2, x \geq0 \\ y=-\frac{1}{4}x^2, x \leq 0$$

And

$$2) \text{ } y= 0 .$$