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**1. Homework Statement**

Here is the original thing:

(x[itex]^{2}[/itex]+1)y'+4x(y-1)=0, y(0)=4

**2. Homework 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'+([itex]\frac{4x}{x^{2}+1}[/itex])y=[itex]\frac{4x}{x^{2}+1}[/itex]

So I get the integrating factor to be

I(x)=e[itex]^{\int\frac{4x}{x^{2}+1}}[/itex]

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

I(x)=(x[itex]^{2}[/itex]+1)[itex]^{2}[/itex]

So the new differential equation is:

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

integrating both sides gets me

(x[itex]^{2}[/itex]+1)[itex]^{2}[/itex]y=2ln(x[itex]^{2}[/itex]+1)+C

solving for y:

y=[itex]\frac{2ln(x^{2}+1)}{(x^{2}+1)^{2}}[/itex]+[itex]\frac{C}{(x^{2}+1)^{2}}[/itex]

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

Please help, I thought I knew this process pretty well..