Question #91478
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.
1
Expert's answer
2019-07-10T13:31:30-0400

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

d) Two solutions.

Find the general solution of


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

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

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


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

We can separate the variables:

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

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

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

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

2y=x+C1-2\sqrt{-y}=x+C_1 \\

y=12(x+C1)\sqrt{-y}=-\frac{1}{2}(x+C_1)


y=14(x+C1)2,x+C10-y=\frac{1}{4}(x+C_1)^2, x+C_1\leq 0

y=14(x+C1)2,x+C10 (1)y=-\frac{1}{4}(x+C_1)^2, x+C_1\leq 0 \text{ } (1)


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



dydx=ydydx=y\frac{dy}{dx}=\sqrt{|y|} \Rightarrow \frac{dy}{dx}=\sqrt{y}



We can separate the variables:

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

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

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


2y=x+C12\sqrt{y}=x+C_1 \\

y=12(x+C1)\sqrt{y}=\frac{1}{2}(x+C_1)

y=14(x+C1)2,x+C10 (2)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=14(x+C1)2,x+C10y=14(x+C1)2,x+C10y=\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)=00=14(0+C1)2,C1=0y(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=14x2,x0y=14x2,x0y=-\frac{1}{4}x^2, x \geq0 \\ y=-\frac{1}{4}x^2, x \leq 0

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


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



1) y=14x2,x0y=14x2,x01)\text{ } y=\frac{1}{4}x^2, x \geq0 \\ y=-\frac{1}{4}x^2, x \leq 0

And

2) y=0.2) \text{ } y= 0 .


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