Diff Eq w/ Initial Value problem

• Sep 11th 2013, 10:10 AM
shane18
Diff Eq w/ Initial Value problem
Here is the problem: {(dy/dx)+(x(y+1)^2)/(y(x+1)^2)=0, y(0)=0}

I have separated and integrated using partial fractions, which gives:
ln|y+1| + 1/(y+1) = ln|x+1| + 1/(x+1) + Constant

Maybe I've been at this for too long, but I can't figure out how to solve for y to determine my constant...
• Sep 11th 2013, 11:11 AM
topsquark
Re: Diff Eq w/ Initial Value problem
Quote:

Originally Posted by shane18
Here is the problem: {(dy/dx)+(x(y+1)^2)/(y(x+1)^2)=0, y(0)=0}

I have separated and integrated using partial fractions, which gives:
ln|y+1| + 1/(y+1) = ln|x+1| + 1/(x+1) + Constant

Maybe I've been at this for too long, but I can't figure out how to solve for y to determine my constant...

First up, you have a sign error on the RHS...the whole RHS is negative. Second, you can't solve for y explicitly. Finally, to get the constant simply plug x = 0 and y = 0 into your expression.

-Dan
• Sep 11th 2013, 11:27 AM
shane18
Re: Diff Eq w/ Initial Value problem
Thanks! Sorry for the simple error. Correcting the signs gives:

ln|y+1| + 1/(y+1) = - ln|x+1| - 1/(x+1) + C

ln|y+1| + 1/(y+1) + ln|x+1| + 1/(x+1) = C

for y(0)=0, ln|1| + 1 + ln|1| + 1 = C

1 + 1 = C = 2

so, ln|y+1| + 1/(y+1) + ln|x+1| + 1/(x+1) = 2

Thanks again, Dan.

-Shane
• Sep 12th 2013, 12:32 AM
JJacquelin
Re: Diff Eq w/ Initial Value problem
The simplest expression of the resut is on the form of the equation already given.
The closed form of y(x) requires a special function : the Lambert W function :
y =-1 -1/W(X) where X = -|x+1|*exp(1/(x+1))/eČ