Questions on second order equations

Question 1

Write down the auxiliary equation for ay'' + by' + cy = 0, and state where it comes from.

Accepted Answer

am² + bm + c = 0. Substituting the trial solution y = e^mx puts a factor of e^mx in every term; cancelling it leaves the quadratic. B1 for the auxiliary equation, B1 for where it comes from.

Question 2

Find the general solution of y'' − 7y' + 12y = 0.

Accepted Answer

Auxiliary: m² − 7m + 12 = (m − 3)(m − 4) = 0, roots 3 and 4. General solution y = Ae^3x + Be^4x: two distinct real roots, two independent exponentials. M1 for the auxiliary equation, A1 for the roots, A1 for the general solution.

Question 3

Write down the general solution of ay'' + by' + cy = 0 in each of the three cases for the roots of the auxiliary equation: two distinct real roots m₁ and m₂; a repeated root m; a complex pair p ± qi.

Accepted Answer

Distinct real roots: y = Ae^m₁x + Be^m₂x. B1. Repeated root: y = (A + Bx)e^mx, the extra factor of x needed because a single exponential cannot carry two independent constants. B1. Complex pair: y = e^px(A cos qx + B sin qx), oscillation from the imaginary part inside an envelope from the real part. B1.

Question 4

Without solving it, classify the roots of the auxiliary equation m² + 4m + 13 = 0 as two distinct real roots, a repeated root, or a complex pair.

Accepted Answer

Discriminant = 4² − 4(1)(13) = 16 − 52 = −36 < 0, so the roots are a complex pair. B1 for the discriminant, B1 for the classification.

Question 5

Find the general solution of y'' + 9y = 0, and state the period of every solution.

Accepted Answer

m² + 9 = 0 gives m = ±3i, so y = A cos 3x + B sin 3x. Pure oscillation at angular frequency 3, so every solution repeats with period 2π/3. M1 for m = ±3i, A1 for the general solution, B1 for the period.

Question 6

Find the general solution of y'' − 6y' + 9y = 0.

Accepted Answer

m² − 6m + 9 = (m − 3)² = 0, a repeated root. The solution needs its extra factor of x, so y = (A + Bx)e^3x. M1 for the repeated root, A1 for the extra factor of x, A1 for the general solution. Without the Bx the two constants would collapse into one and two initial conditions could not be met.

Question 7

Solve y'' − 2y' + 5y = 0 with y(0) = 2 and y'(0) = 0.

Accepted Answer

m² − 2m + 5 = 0 gives m = 1 ± 2i, so y = e^x(A cos 2x + B sin 2x). y(0) = 2 gives A = 2. Differentiating and setting x = 0 gives y'(0) = A + 2B = 0, so B = −1. The solution is y = e^x(2 cos 2x − sin 2x), an oscillation at angular frequency 2 growing inside e^x. M1 for the auxiliary equation, A1 for m = 1 ± 2i, A1 for the general solution, B1 for A = 2, M1 for differentiating and using y'(0) = 0, A1 for B = −1.

Question 8

Solve y'' − y = 0 with y(0) = 1 and y'(0) = 0, and identify the solution as a standard function.

Accepted Answer

m² − 1 = 0 gives m = ±1, so y = Ae^x + Be^−x. The conditions give A + B = 1 and A − B = 0, so A = B = 1/2 and y = (e^x + e^−x)/2 = cosh x. M1 for the auxiliary equation, A1 for the general solution, M1 for both conditions, A1 for A and B, B1 for naming cosh x.

Question 9

A student solving y'' − y' − 6y = 5e^3x tries the particular integral y = λe^3x, and finds this substitution gives 0 = 5e^3x, which has no solution for λ. Explain why this trial fails, state the correct trial to use instead, and find the particular integral.

Accepted Answer

The auxiliary equation m² − m − 6 = (m − 3)(m + 2) = 0 has roots 3 and −2, so e^3x is already part of the complementary function Ae^3x + Be^−2x: substituting a multiple of it into the left-hand side of the homogeneous equation gives zero, which is why every term but the forcing one vanishes. B1 for identifying the duplication. The correct trial is y = λxe^3x. M1 for the corrected trial. Then y' = λe^3x(1 + 3x) and y'' = λe^3x(6 + 9x), so the left side is λe^3x[(6 + 9x) − (1 + 3x) − 6x] = λe^3x(5 + 0x) = 5λe^3x. M1 for the substitution. Matching to 5e^3x gives λ = 1, so the particular integral is xe^3x. A1 for λ = 1, A1 for the particular integral.

Question 10

Solve y'' + 6y' + 9y = 0 with y(0) = 2 and y'(0) = 1.

Accepted Answer

The auxiliary equation m² + 6m + 9 = (m + 3)² = 0 has a repeated root m = −3. M1 for the auxiliary equation, A1 for the repeated root. The general solution is y = (A + Bx)e^−3x. A1. y(0) = 2 gives A = 2. B1. Differentiating, y' = Be^−3x − 3(A + Bx)e^−3x, so y'(0) = B − 3A. Setting this to 1 gives B − 6 = 1, so B = 7. M1 for differentiating and using y'(0) = 1, A1 for B = 7. The solution is y = (2 + 7x)e^−3x.

Question 11

Find the general solution of y'' + 4y' + 3y = 9x.

Accepted Answer

The auxiliary equation m² + 4m + 3 = (m + 1)(m + 3) = 0 has roots −1 and −3, so the complementary function is Ae^−x + Be^−3x. M1 for the auxiliary equation, A1 for the complementary function. The right-hand side is linear and matches nothing in the complementary function, so try y = ax + b: y' = a, y'' = 0. Substituting gives 4a + 3(ax + b) = 9x, that is 3ax + (4a + 3b) = 9x. M1 for the trial and substitution. Comparing x terms gives 3a = 9, so a = 3; the constants then give 12 + 3b = 0, so b = −4. A1 for both coefficients. The general solution is y = Ae^−x + Be^−3x + 3x − 4. A1. A polynomial trial must include every power up to the one on the right-hand side, including the constant term absent from 9x itself; dropping b here would leave one equation short.

Question 12

Find the general solution of y'' − 3y' + 2y = 4e^2x.

Accepted Answer

The auxiliary equation m² − 3m + 2 = 0 has roots 1 and 2, so the complementary function is Ae^x + Be^2x. Trying y = ke^2x for the particular integral fails, because that is already part of the complementary function and the left side collapses to zero. Multiply by x and try y = kxe^2x instead. Then y' = ke^2x(1 + 2x) and y'' = ke^2x(4 + 4x), so the left side is ke^2x[(4 + 4x) − 3(1 + 2x) + 2x] = ke^2x. Matching to 4e^2x gives k = 4, so the general solution is y = Ae^x + Be^2x + 4xe^2x. M1 for the auxiliary equation, A1 for the complementary function, B1 for spotting the duplication, M1 for trying kxe^2x, A1 for the substitution, A1 for k = 4, A1 for the general solution. When the right-hand side duplicates a term of the complementary function, an extra factor of x is needed, and a repeated auxiliary root would require x².

Question 13

Solve y'' + y' − 2y = 6 with y(0) = 2 and y'(0) = 2.

Accepted Answer

The auxiliary equation m² + m − 2 = (m + 2)(m − 1) = 0 has roots −2 and 1, so the complementary function is Ae^−2x + Be^x. M1 for the auxiliary equation, A1 for the complementary function. The right-hand side is constant, so try y = k: substituting gives −2k = 6, so k = −3. M1 A1 for the particular integral. The general solution is y = Ae^−2x + Be^x − 3. Both conditions apply here, to this full solution, not to the complementary function alone. y(0) = 2 gives A + B − 3 = 2, so A + B = 5. Differentiating, y' = −2Ae^−2x + Be^x, and y'(0) = 2 gives −2A + B = 2. Subtracting, 3A = 3, so A = 1 and B = 4. M1 for applying both conditions to the general solution, A1 for A and B. The solution is y = e^−2x + 4e^x − 3.

Question 14

Solve the recurrence relation u_n+2 − 6u_n+1 + 9u_n = 0, given u₀ = 1 and u₁ = 9.

Accepted Answer

Substituting u_n = mⁿ gives the auxiliary equation m² − 6m + 9 = (m − 3)² = 0, a repeated root m = 3. M1 for the auxiliary equation, A1 for the repeated root. As with a repeated root of a differential equation, the second solution needs an extra factor, here of n rather than x: u_n = (A + Bn)3ⁿ. B1 for the recurrence's version of the repeated-root form. u₀ = 1 gives A = 1. B1. At n = 1, u₁ = (A + B)(3) = 9, so A + B = 3 and B = 2. M1 for using u₁, A1 for B. The formula gives u₂ = (1 + 4)(9) = 45, and the recurrence gives u₂ = 6(9) − 9(1) = 45, confirming it.

Question 15

Find the general solution of y'' − 4y' + 4y = 6e^2x.

Accepted Answer

The auxiliary equation m² − 4m + 4 = (m − 2)² = 0 has a repeated root m = 2, so the complementary function is (A + Bx)e^2x. M1 for the auxiliary equation, A1 for the complementary function. Trying y = λe^2x duplicates the first term of the complementary function, and trying y = λxe^2x duplicates the second, since xe^2x is already present: a single factor of x is not enough here. B1 for identifying this. With y = λx²e^2x, y' = λe^2x(2x² + 2x) and y'' = λe^2x(4x² + 8x + 2), so the left side is λe^2x[(4x² + 8x + 2) − 4(2x² + 2x) + 4x²] = λe^2x[(4 − 8 + 4)x² + (8 − 8)x + 2] = 2λe^2x. M1 for substituting λx²e^2x and simplifying. Matching to 6e^2x gives λ = 3. A1. The general solution is y = (A + Bx)e^2x + 3x²e^2x. A1.