Difficulty with an initial value problem.

Snowdog2112

New member
Joined
Apr 30, 2006
Messages
7
I am trying to solve the initial value problem:

(x+1)dy/dx-2(x^2+x)y=(e^x^2)/(x+1) x>-1 y(0)=5


I begin going through the motions and I get down to:

(e^x^2)+y=int((e^2x^2)/(x+1)dx)

I can't figure out how to integrate this, or if I missed cancelling something in the beginning or totally loused something up to this point. Any help would be appreciated.
 

galactus

Super Moderator
Staff member
Joined
Sep 28, 2005
Messages
7,216
Divide through by x+1 to make the coefficient of y' unity. You get:

\(\displaystyle \L\\\frac{dy}{dx}-2xy=\frac{e^{x^{2}}}{(x+1)^{2}}\)

Find the integrating factor:

\(\displaystyle \L\\e^{-2\int{x}dx}=e^{-x^{2}}\)

Multiply by integrating factor:

\(\displaystyle \L\\\frac{d}{dx}[e^{-x^{2}}y]=\frac{1}{(x+1)^{2}}\)

Integrate:

\(\displaystyle \L\\e^{-x^{2}}y=\frac{-1}{x+1}+C\)

Solve for y:

\(\displaystyle \L\\y=(\frac{-1}{x+1}+C)e^{x^{2}}\)

Now, using the initial condition, set x=0 and y=5, solve for C.

Then you have:

\(\displaystyle \H\\y=(6-\frac{1}{x+1})e^{x^{2}}\)
 

Snowdog2112

New member
Joined
Apr 30, 2006
Messages
7
Thank you for pointing out my missing negative sign!
 
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