Answer: f(x) = a₀, where a₀ is a constant
Explanation:
To find the finite polynomial solution to the given differential equation, let's assume that the solution is in the form of a power series:
f(x) = ∑[n=0 to ∞] aₙxⁿ
where aₙ are constants to be determined.
Now, let's differentiate f(x) twice and substitute it into the differential equation:
d²f/dx² = ∑[n=0 to ∞] aₙn(n-1)xⁿ⁻²
df/dx = ∑[n=0 to ∞] aₙnxⁿ⁻¹
Substituting these derivatives into the differential equation, we get:
x(∑[n=0 to ∞] aₙn(n-1)xⁿ⁻²) + x(∑[n=0 to ∞] aₙnxⁿ⁻¹) - 4(∑[n=0 to ∞] aₙxⁿ) = 0
Next, let's combine the terms with the same power of x and simplify the equation:
∑[n=0 to ∞] aₙn(n-1)xⁿ + ∑[n=0 to ∞] aₙnxⁿ - ∑[n=0 to ∞] 4aₙxⁿ = 0
Now, we can equate the coefficients of each power of x to zero to obtain a system of equations:
a₀(0(0-1) - 4) = 0 => -4a₀ = 0
a₁(1(1-1) - 4) + a₁ = 0 => -4a₁ + a₁ = 0 => -3a₁ = 0
a₂(2(2-1) - 4) + 2a₂ - 4a₁ = 0 => -2a₂ + 2a₂ - 4a₁ = 0 => -4a₁ = 0
We observe that all the coefficients except a₀ satisfy the equation -4aₙ = 0, which means they must be zero. However, since we are looking for a non-trivial polynomial solution, a₀ cannot be zero.
Therefore, the finite polynomial solution to the given differential equation is f(x) = a₀, where a₀ is a constant.
I hope this helps :)