 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2013/paper-2/14/a/solution

y_{h} = C_{1} + C_{2} e^{3 x}

. Since the forcing is quadratic and the homogeneous solution contains a constant, use the polynomial trial

y_{p} = A x^{3} + B x^{2} + C x

. Substitution gives

y_{p}^{''} - 3 y_{p}^{'} = - 9 A x^{2} + (6 A - 6 B) x + 2 B - 3 C .

Matching coefficients gives

A = - 1/3

B = - 1/3

, and

C = - 8/9

. Therefore

y = C_{1} + C_{2} e^{3 x} - \frac{x ^{3}}{3} - \frac{x ^{2}}{3} - \frac{8 x}{9} .

 New to topics? Read the docs here!