Find the first four nonzero terms in a power series expansion about x_0 for a general solution to the given differential equation with the given value for x_0.

Question

y' - 5xy = 0; x₀ = -2 (Expression must be in terms of a₀ that includes all terms up to order 3.)

y(x) =

This one's a doozy, if possible please include a step-by-step solution. Thank you!

Answer 1

Hi,

x₀ = -2----> x + 2

AND

f(x) = a₀+ a₁(x +2) +a₂(x+2)^2 +a₃(x+2)^3 +a₄(x+2)^4 +....----> f(-2) = a₀

_{Take derivative of given function and plug in x = -2}

y' - 5xy = 0 ------> f'(-2) +10y = 0 ----> f'(-2)= -10 a₀

y'' -5y -5xy' = 0 -

f"(-2) -5y +10y' = 0 ----> f"(-2) = 5y -10y'= 5 a₀ -10(-10a₀) =>y"=f"(-2)= 105 a₀

y"' - 5y' -5y' -5xy" = 0 ---> y"' - 10y' -5xy" = 0 the same method

plug in y', y" and x = -2 you will have:

y"' - 10y' -5xy" = 0 ----> y"' = -1150 a₀

y"" -10y" -5y" -5xy"' = 0 ---> y"" -15y" -5xy"' = 0 one more time plug in all above information then you will have:

y"" -15y" -5xy"' = 0 ---> y"" = 13075 a₀

_Now f(x) = a₀+ a₁(x +2) +a₂(x+2)^2 +a₃(x+2)^3 +a₄(x+2)^4 +....

find f'(x) , f"(x), f'"(x) and f""(x)

THEN plug in x = -2 and plug in f'(x) , f"(x), f'"(x) and f""(x) that you will have

f'(-2)= -10 a₀ =f'(-2) = a₁

f"(-2)= 105 a₀ =f"(-2) = 2a₂ ====> a₂ = 105/2 a₀

y"' = -1150 a₀ = f'"(-2)= 6a₃ ====> a₃= -1150/6 a₀

y"" = 13075 a₀ = f""(-2) =24 a₄====> a₄= 13075/24 a₀

Now all coefficients plug in to the function f(x)

f(x) = a₀ -10 a₀ (x +2) +105/2 a₀(x+2)^2 -1150/6 a₀(x+2)^3 +13075/24a₀(x+2)^4

I hope it is useful, please one more time check all parts I hope no mistake.

Minoo