Solving 4x=2y(x²+1)^(1/2)(dy/dx) with y(0)=-3

[beanryu]

okay... i got this problem
sovle the separable differential equation
4x-2y(x^2+1)^(1/2)(dy/dx)=0
using the following intial condition: y(0) = -3
y^2 = ? (function of x)

I guess that means the constant is -3

so i put all the x on 1 side and all the y on one side

4x = 2y(x^2+1)^(1/2)(dy/dx)
(4x)(dx) = 2y(x^2+1)^(1/2)(dy)
(4xdx)/(x^2+1)^(1/2) = 2ydy

integral both sides I got
4(x^2+1)^(1/2) = y^2

i tried the following answers
y^2 = 4(x^2+1)^(1/2)
y^2 = 4(x^2+1)^(1/2)+9
y^2 = 4(x^2+1)^(1/2)-3

they are all wrong!

WHAT IS WRONG?! IS MY WAY OF DOING IT TATALLY WRONG?!

 

[Jameson]

This is correct. $$y^2=4\sqrt{x^2+1}+C$$

Now plug in your initial condition to solve for C.

 

[mathmike]

well you got most of it but i don't know why you are trying 9 and -3 as c.

it says y(0) = -3

y = +-4*(x^2+1) + c

so y(0) = +-(0^2+1) + c = -3

can you figure it out from here

 

[beanryu]

if
y^2 = 4(x^2+1)^(1/2)+c
y = sqrt(4(x^2+1)^(1/2)+c)
c = 5 is the correct answer.

THANX!