- #1
skyturnred
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Homework Statement
Here is the original thing:
(x[itex]^{2}[/itex]+1)y'+4x(y-1)=0, y(0)=4
Homework Equations
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..