Find the General Second Order Differential Equation Using Substitution (A2 Further Maths)

Find the General Second Order Differential Equation 4x²d²y/dx² - 8xdy/dx +(8+4x²)y = x⁴, Transform Using y = vx. 1) Differentiate both sides of y = vx (apply Product Rule to right hand side) dy/dx = xdv/dx + v ... Equation 12) Differentiate dy/dx = xdv/dx + v (apply Product Rule to xdv/dx) d²y/dx² = xd²v/dx² + dv/dx + dv/dx d²y/dx² = xd²v/dx² + 2dv/dx ... Equation 23) Substitute Equation 1, Equation 2 & y = vx into the original equation and simplify 4x²(xd²v/dx² + 2dv/dx) - 8x(xdv/dx + v) + vx(8 + 4x²) = x⁴ 4x³d²v/dx² + 8x²dv/dx - 8x²dv/dx - 8vx + 8vx + 4x³v = x⁴ 4x³d²v/dx² + 4x³v = x⁴ 4d²v/dx² + 4v = x4) Find the Particular integral (right side of original equation is x) Particular Intergal: v = ax + b dv/dx = a d^2vdx² = 05) Substitute Particular Integrals into 4d²v/dx² + 4v = x 4(0) + 4(ax + b) = x 4ax +4b = x equating coefficients, a = 1/4 & b = 0, therefore particular integral is 1/4x6) Form Auxiliary Quadratic Equation 4m² + 4 = 0 4m² = -4 m² = -1 therefore m = i therefore Auxiliary Quadratic Equation is Acos(x) + Bsin(x) therefore General Equation : v = Acos(x) + Bsin(x) +1/4x 7) Revert back to original form (in terms of y) v = y/x therefore, y/x = Acos(x) + Bsin(x) +1/4x y = x[Acos(x) + Bsin(x) +1/4x]