Solving DE eqn: x + 2ysqrt[x^2 + 1](dy/dx) = 0, y(0) = 1

jman2807

New member
Joined
Sep 4, 2006
Messages
29
I am having trouble with the following Differential equation:

mathai0.gif


I tried seperating to x/sqrt(x^2+1) = -2y but when I integrate that I dont know what to do next. Any help appreciated.

Edit: Sorry about posting this in the calculus forums and not the DE one, This is a calc II problem though so i just put it here.
 
Separate your variables:

\(\displaystyle \L\\2ydy+\frac{x}{\sqrt{x^{2}+1}dx\)

\(\displaystyle \L\\2ydy=\frac{-x}{\sqrt{x^{2}+1}dx\)

Integrate:

\(\displaystyle \L\\\int{2y}dy=\int\frac{-x}{\sqrt{x^{2}+1}}dx\)

For the right side, use a u-substitution:

\(\displaystyle u=x^{2}+1\)
\(\displaystyle \frac{du}{2}=xdx\)
 
Yeah I've gotten that far. But I dont know what to do when you have y^2 = -sqrt(x^2+1)
 
Don't forget your C.

You have:

\(\displaystyle \L\\y^{2}={-}\sqrt{x^{2}+1}+C\)

Use your initial condition to find C:

\(\displaystyle 1^{2}={-}\sqrt{(0)^{2}+1}+C\)

C=2

You have:

\(\displaystyle y^{2}={-}\sqrt{x^{2}+1}+2\)

That's an acceptable answer.

You can go ahead and

\(\displaystyle \L\\y=\sqrt{-\sqrt{x^{2}+1}+2}\)
 
Ok thank you, the only thing I didnt understand was the square root of a negative square root, but that was the right answer. Thanks.
 
Top