Question

What is a solution to the differential equation `(x+1)y'-2(x^2+x)y=e^(x^2)/(x+1)` where x>-1 and y(0)=5?

Answer 1

`y(x) = (6-1/(x+1))e^{x^2}`

Explanation:

The differential equation is first order linear nonhomogeneus. In this case the solution is composed from the homogeneus solution `y_h(x)` and a particular solution `y_p(x)`.

`(x + 1) y'_h(x) - 2 (x^2 + x) y(x)= 0`
`(x + 1) y'_p(x) - 2 (x^2 + x) y_p(x)=e^{x^2}/(x + 1)`

Finally

`y(x) = y_h(x)+y_p(x)`

1) Obtaining the homogeneus solution `y_h(x)`
Simplifying we obtain

`y'_h(x) - 2 x xx y_h(x)= 0`

grouping variables

`(y'_h(x))/(y_h(x)) = 2x`

The solution is

`log_e(y_h(x))=x^2+C_0->y_h(x) = C_1e^{x^2}`

2) Obtaining the particular solution `y_p(x)`
For this purpose we will suppose that

`y_p(x) = C_1(x)e^{x^2}`

This method was due to Euler and Lagrange and can be seen in
https://en.wikipedia.org/wiki/Variation_of_parameters

Introducing `y_p(x)` into the complete equation we obtain after simplifications

`e^{x^2} (1 + x) C'_1(x) = e^{x^2}/(1 + x)`

Solving for `C_1(x)` we obtain

`C_1(x)=-1/(x+1)+C_2`

Now, putting all together

`y(x) = (C_2-1/(x+1))e^{x^2}`

With the initial conditions we find the `C_2` value as

`y(0) = (C_2-1)=5->C_1=6`

Answer 2

`=e^{x^2} ( (6x + 5)/(x+1) )`

Explanation:

this is linear so we can start by just moving stuff around.

`(x+1)y'-2(x^2+x)y=e^(x^2)/(x+1)`

`y'- 2((x(x+1))/(x+1))y= e^(x^2)/(x+1)^2`

`y'- 2x \ y= e^(x^2)/(x+1)^2`

we solve with inregrating factor `I = exp( -2 int \ x \ dx) = e^{-x^2}`

`e^{-x^2} * y'- (e^{-x^2} )*2x y= e^{-x^2} *e^(x^2)/(x+1)^2`

`(e^{-x^2} * y)'= 1/(x+1)^2`

`e^{-x^2} * y = int 1/(x+1)^2 \ dx`

`e^{-x^2} * y = - 1/(x+1) + alpha`

`y =e^{x^2} * ( - 1/(x+1) + alpha )`

From `y(0) = 5`

`5 =1* ( - 1/(1) + alpha ) \implies alpha = 6`

`y =e^{x^2} ( 6 - 1/(x+1) )`

`=e^{x^2} ( (6x + 5)/(x+1) )`