# Difficulty with an initial value problem.

#### Snowdog2112

##### New member
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
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
Thank you for pointing out my missing negative sign!