Given that x(d²y/dx²) + (2x+2)(dy/dx) + (2-3x)y = 10e^(2x) and that v = xy, show that d²v/dx² + 2(dv/dx) – 3v = 10e^(2x). Find the general solution for y in terms of x.

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2014). The mark or mark type is shown in brackets.

1. [B1B1] Given v = xy, then y = v/x. Differentiating once: dv/dx = y + x(dy/dx). Differentiating again: d²v/dx² = dy/dx + (dy/dx + x(d²y/dx²)) = 2(dy/dx) + x(d²y/dx²).
2. [M1A1] From these relations: dy/dx = (dv/dx - y)/x = (dv/dx - v/x)/x = (1/x)dv/dx - v/x². And x(d²y/dx²) = d²v/dx² - 2(dy/dx). Substitute these into the given differential equation: (d²v/dx² - 2(dy/dx)) + (2x+2)(dy/dx) + (2-3x)y = 10e^(2x). d²v/dx² + 2x(dy/dx) + 2(dy/dx) - 2(dy/dx) + (2-3x)y = 10e^(2x). d²v/dx² + 2x(dy/dx) + (2-3x)y = 10e^(2x). Substitute dy/dx = (dv/dx - y)/x: d²v/dx² + 2x((dv/dx - y)/x) + (2-3x)y = 10e^(2x). d²v/dx² + 2(dv/dx) - 2y + (2-3x)y = 10e^(2x). d²v/dx² + 2(dv/dx) - 2y + 2y - 3xy = 10e^(2x). d²v/dx² + 2(dv/dx) - 3(xy) = 10e^(2x). Since v = xy, this gives d²v/dx² + 2(dv/dx) - 3v = 10e^(2x). (AG)
3. [M1] The homogeneous equation is d²v/dx² + 2(dv/dx) - 3v = 0. The auxiliary equation is m² + 2m - 3 = 0.
4. [A1] Factoring, (m+3)(m-1) = 0, so m = -3, 1. The complementary function (CF) is v_c = Ae^(-3x) + Beˣ.
5. [M1] For the particular integral (PI), assume v_p = ke^(2x). Then dv_p/dx = 2ke^(2x) and d²v_p/dx² = 4ke^(2x).
6. [M1A1] Substitute into the differential equation: 4ke^(2x) + 2(2ke^(2x)) - 3(ke^(2x)) = 10e^(2x). 4ke^(2x) + 4ke^(2x) - 3ke^(2x) = 10e^(2x). 5ke^(2x) = 10e^(2x). So 5k = 10 ⇒ k = 2. The PI is v_p = 2e^(2x).
7. [M1A1] The general solution for v is v = v_c + v_p = Ae^(-3x) + Beˣ + 2e^(2x). Since v = xy, the general solution for y in terms of x is y = (1/x)(Ae^(-3x) + Beˣ + 2e^(2x)).