Solve the differential equation 6y dy/dx = 6x*(1+x^2)^(1/2)*(1+2y^2)^(1/2) subject to the initial condition...

subject to the initial condition that x = 0 when y = 0 .

The answer is 3(2y^2+1)^(3/2) = 2(x^2+1)^3/2.

i need help with getting there. THANX !!! <3

Step #1: Bring the x/dx's to one side of the equation and the dy/y's to the other side of the equation
6y*dy/(1+2y^2)^(1/2) = 6x*(1 + x^2)*dy

Step #2: Use a "u" substitution to simplify if you can't see the integration (in this case you will most likely need a "w" substitution as well)

Set u = 1+2y^2
Take the derivative
du = 4y*dy
Divide both sides by 4 and you get
1/4*du = y*dy

Set w = 1 +x^2
Take the derivative
dw = 2x*dx
Divide both sides by 2
1/2*dw = x*dx

Now substitue and you get
3/2*du/(u)^(1/2) = 3*dw*(w)^(1/2)

Step #3: Integrate both sides (by using a "u" substitution you have an integration that can be solved simply by using the power rule)
int[3/2*du/(u)^(1/2)] = int[3*dw*(w)^(1/2)] and you get
3*(u)^(1/2) + C = 2*(u)^(3/2) + C, where C is a constant

Step #4: Substitue "u" an "w" with the expressions (expressed in terms of y and x respectively) defined earlier in Step #2 and you get
3*(1+2y^2)^(1/2) + C = 2*(1+x^2)^(3/2) + C

Step #5: Simplify
i) Remeber that C is just any constant. When you add (or subtract) two constants you will ALWAYS get another constant (for example 6 - 8 = -2 (another constant)) so you can rewrite the equation as:
3*(1+2y^2)^(1/2) = 2*(1+x^2)^(3/2) + C

Step #6: Plug in the intial point (0,0) and solve for C and you get
3 = 2 + C
C = 1

Step #7: Substitute the numerical value of C into the equation found in Step #7 and circle ypour answer.
3*(1+2y^2)^(1/2) = 2*(1+x^2)^(3/2) + 1

Unfortunately I got a slightly different answer from yours. Even though the answer is different i hope you understood the logic behind the steps. I will say though that I am quite confident in my answer and I believe that it is correct