Solving a differential equation with intial conditions,

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SUMMARY

The discussion centers on solving the differential equation (x²+1)y' + 4x(y-1) = 0 with the initial condition y(0) = 4. The user correctly identifies the integrating factor as I(x) = (x² + 1)² but fails to apply it correctly to the right-hand side of the equation. The correct approach requires multiplying both sides of the rearranged equation by the integrating factor before integrating, which was overlooked. This mistake leads to an incorrect solution for the constant C when applying the initial condition.

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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..
 
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You forgot to multiply the righthand side by the integrating factor.
 
skyturnred said:

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

So the new differential equation is:

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

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