# Solving a differential equation with intial conditions, please help

#### 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..

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#### vela

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

#### HallsofIvy

Science Advisor
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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