Problem solving separable diff EQ w/ I.Condition. , they say its wrong :x

[mr_coffee]

Hello everyone i did this one surley thing it would work out and yet another failure.
Solve the separable differential equation
11 x - 8 y*sqrt{x^2 + 1}*{dy}/{dx} = 0.
Subject to the initial condition: y(0) = 6.
y = ?I'm pretty sure where I messed up is when i tried to solve for y, i htink i screwed up there but I'm not sure, anyone know? Thanks! its #11. Ignore that top stuff please!

Here is my work:
http://img57.imageshack.us/img57/5705/lastscan8cm.jpg

 

[benorin]

Particular solution

You problem lies in where you placed your constant of integration, namely:

[tex]11\int \frac{x}{\sqrt{x^2+1}}dx = \frac{11}{2}\int \frac{1}{\sqrt{u}}du = \frac{11}{2}\int u^{-\frac{1}{2}}du = 11u^{\frac{1}{2}}+C=11\sqrt{x^2+1}+C[/tex]

C is outside the square root, from there

[tex]4y^2=11\sqrt{x^2+1}+C[/tex]

[tex]y^2=\frac{11}{4}\sqrt{x^2+1}+\frac{C}{4}[/tex]

or,

[tex]y=\pm\sqrt{\frac{11}{4}\sqrt{x^2+1}+\frac{C}{4}}[/tex]

to solve for C, use the simplest form of the solved DE, that is use

[tex]4y^2=11\sqrt{x^2+1}+C[/tex]

for x=0 and y(0)=6, this gives

[tex]4(6)^2=11\sqrt{0^2+1}+C[/tex]

which simplifies to

[tex]144=11+C[/tex]

and hence C=133, plug this into

[tex]y=\pm\sqrt{\frac{11}{4}\sqrt{x^2+1}+\frac{C}{4}}[/tex]

to get

[tex]y(x)=\pm\sqrt{\frac{11}{4}\sqrt{x^2+1}+\frac{133}{4}}=\pm\frac{1}{2}\sqrt{11\sqrt{x^2+1}+133}[/tex]

as your particular solution.

 

[leright]

yup, that constant shouldn't be under the radical!

 

[mr_coffee]

Ahhh, that worked perfectly Benorin, thanks a ton (again)! :) The step by step explanation is great!