The solution of the initial value problem 2y 3y2 y0 1 y0

The solution of the initial value problem 2y\" = 3y^2, y(0) = 1, y\'(0) = 1 is Select the correct answer. y = 2/(x + 2)^2 y = 4/(x + 2)^3 y = 1/(x + 1)^3 y = 1/(x + 1)^2 y = 4/(x + 2)^2

Solution

THis is a non linear differential equation

SOlving this requires use of elliptic functions

So instead we check by substitution to see which solutions satisfy the given initial conditoin

a. y(0)=1/2

b. y(0)=1/2

c. y(0)=1/8

d. y(0)=1/4

e. y(0)=1

So only e satisfies the initial condition

We check if it satisfies the other intial condition

y\'=-8/(x+2)^3

http://www.wolframalpha.com/input/?i=(4%2F(x%2B2)%5E2)%27

y\'\'=24/(x+2)^4

2y\'\'=48/(x+2)^4

3y^2=3*16/(x+2)^2=48/(x+2)^2

So it satisfies the given equation

But y\'(0)=-1

So it does not satisfy the given intial condition (Must be a typo I think)

The intial condition must ahve been: y\'(0)=-1