Solution to 2nd O.D.E. using power seris

[Jayzar_1]

Can anyone show me a complete solution to the problem below using power series?

y"+xy=x^2 y(0)=1 , y'(0)=0

 

[Galileo]

Start with assuming a solution in the form of a power series:
$$y=\sum_{n=0}^{\infty}a_nx^n$$

Plug it into your differential equation and simplify. I think you should get a recursive relation between your coefficients whose initial values $a_0$ and $a_1$ can be determined from your initial conditions.

 

[M98Ranger]

Yes, I can provide a complete solution to this problem using power series. First, we will rewrite the equation as a series expansion:

y" + xy = x^2
= y" + x(y-1+1) = x^2
= y" + xy - x + x = x^2
= y" + x(y-1) + (x-1) = x^2
= y" + x(y-1) + (x-1)^2 = x^2

Next, we will assume that y can be written as a power series in x:

y(x) = a0 + a1x + a2x^2 + a3x^3 + ...

Substituting this into the equation, we get:

y" + x(y-1) + (x-1)^2 = x^2
= (a2 + 2a3x + 3a4x^2 + ...) + x(a0 + a1x + a2x^2 + ...) - a0 - a1x + (x^2 - 2x + 1) = x^2

Equating coefficients of each power of x, we get the following equations:

a2 + a0 - a0 = 0
2a3 + a1 - a1 - 2 = 0
3a4 + a2 - a2 - 2 = 0
...

Solving for each coefficient, we get:

a0 = 1
a1 = 0
a2 = 2
a3 = 1
a4 = 2/3
a5 = 1/6
...

Therefore, the solution to the given equation is:

y(x) = 1 + 2x + x^2 + (2/3)x^3 + (1/6)x^4 + ...

To satisfy the initial conditions, we can use the formula for the Maclaurin series to find the derivatives of y at x = 0:

y'(x) = a1 + 2a2x + 3a3x^2 + ...
= 0 + 2(2) + 3(1)x + ...
= 4 + 3x + ...

y"(x) = 2a2 + 2(3)a3x + ...
= 2(2) + 2