1. Feb 14, 2012

skyturnred

1. The problem statement, all variables and given/known data

Here is the original thing:

(x$^{2}$+1)y'+4x(y-1)=0, y(0)=4

2. Relevant equations

3. The attempt at a solution

I thought I knew the procedure.. but I got it wrong. Can someone let me know where I went wrong?

First I rearrange the equation to get the following in the form y'+p(x)y=g(x)

y'+($\frac{4x}{x^{2}+1}$)y=$\frac{4x}{x^{2}+1}$

So I get the integrating factor to be

I(x)=e$^{\int\frac{4x}{x^{2}+1}}$

That integral comes out to be 2ln(x$^{2}$+1), which when raised to e, the integrating factor simply becomes

I(x)=(x$^{2}$+1)$^{2}$

So the new differential equation is:

d/dx (x$^{2}$+1)$^{2}$y=$\frac{4x}{x^{2}+1}$

integrating both sides gets me

(x$^{2}$+1)$^{2}$y=2ln(x$^{2}$+1)+C

solving for y:

y=$\frac{2ln(x^{2}+1)}{(x^{2}+1)^{2}}$+$\frac{C}{(x^{2}+1)^{2}}$

Taking the intial conditions into account I get C=4. When I input this equation into my assignment online, it says I'm wrong.

2. Feb 14, 2012

vela

Staff Emeritus
You forgot to multiply the righthand side by the integrating factor.

3. Feb 14, 2012

HallsofIvy

So you are going to multiply both sides of the equation by that?

Did you not multiply the right side by $(x^2+ 1)^2$?