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Find the first four nonzero terms in a power series expansion about x0 for a general solution to the given differential equation. 0

Sagot :

Answer:

Start with equation,

         y''(x) = f(x) = 6y/(10x-x²)               about point x₀ = 5

Then,

       y(x) = a₀ + a₁(x-5) + a₂(x-5)²+ a₃(x-5)³ + a₄(x-5)⁴...….

by inspection,

           y(5) = a₀      and       y'(5) = a₁

then,

y''(5) = 6a₀/(50-25) = 6a₀/25              

y'''(x) = 6y'/(10x-x2) - 6y(10-2x)/(10x-x2)2

then,                      

                              y'''(5) = 6a₁/25 - 0 = 6a₁/25                

y''''(x) = 6y''/(10x-x2) -12y'(10-2x)/(10x-x2)² + 12y/(10x-x2)² + 12y(10-2x)2/(10x-x2)³

then,

    y''''(5) = (6/25)(6a₀/25) + 0 + 12a₀/625 + 0

              = 48a₀/625

So,

y(x) = f(x) = f(5) + f'(5)(x-5) + f''(5)(x-5)2/2! + f'''(5)(x-5)3/3! + f''''(5)(x-5)4/4! …

                                                       or

y(x) = f(x) = a₀ + a₁(x-5) + (6ao/25)(x-5)2/2! + (6a1/25)(x-5)3/3! + (48a₀/625)(x-5)4/4!...

                                                         or

y(x) = f(x) = a₀ + a₁(x-5) + (3a₀/25)(x-5)²+ (a₁/25)(x-5)³+ (2a₀/625)(x-5)⁴.....

Learn more about equation here-

https://brainly.com/question/3391581

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