Use variation of parameters to find the general solution of the differential equation x2 y′′− x y′ =x3 ex if two solutions to the associated homogeneous problem are known to be 1 and x2

Sure, let's solve the given differential equation using the method of variation of parameters.

The given differential equation is x²y'' - xy' = x³eˣ.

The associated homogeneous differential equation is x²y'' - xy' = 0.

We are given that the solutions to the homogeneous differential equation are y₁ = 1 and y₂ = x².

Step 1: Write down the Wronskian of y₁ and y₂.

W = det[y₁ y₂; y₁' y₂'] = det[1 x²; 0 2x] = 2x.

Step 2: Find u₁ and u₂ using the formulas:

u₁' = -y₂f/W = -x²(x³eˣ)/(2x) = -x²eˣ/2,

u₂' = y₁f/W = (x³eˣ)/(2x) = x²eˣ/2.

Step 3: Integrate u₁' and u₂' to find u₁ and u₂:

u₁ = ∫u₁'dx = ∫-x²eˣ/2 dx = -1/2 ∫x²eˣ dx,

u₂ = ∫u₂'dx = ∫x²eˣ/2 dx = 1/2 ∫x²eˣ dx.

Step 4: Use the formula for the solution of the non-homogeneous differential equation:

y = y₁u₁ + y₂u₂ = u₁ + x²u₂.

This is the general solution of the given differential equation.

Note: The integrals ∫x²eˣ dx in the expressions for u₁ and u₂ can be computed using the method of integration by parts.