skyturnred

1. Homework Statement

Here is the original thing:

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

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

Related Calculus and Beyond Homework Help News on Phys.org

vela

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

HallsofIvy

Homework Helper
1. Homework Statement

Here is the original thing:

(x$^{2}$+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'+($\frac{4x}{x^{2}+1}$)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 you are going to multiply both sides of the equation by that?

So the new differential equation is:

d/dx (x$^{2}$+1)$^{2}$y=$\frac{4x}{x^{2}+1}$
Did you not multiply the right side by $(x^2+ 1)^2$?

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.

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

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