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?

2 Answers
Jun 24, 2016

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

Jun 25, 2016

=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) )