Question

In this problem you will use undetermined coefficients to solve the nonhomogeneous equation y′′+4y′+3y=8te^−t+6e^−t−(9t+6)

Answer 1

We're given the ODE,

y'' + 4y' + 3y = 8t exp(-t ) + 6 exp(-t ) - (9t + 6)

(where I denote exp(x) = eˣ )

First determine the characteristic solution:

y'' + 4y' + 3y = 0

has characteristic equation

r ² + 4r + 3 = (r + 1) (r + 3) = 0

with roots at r = -1 and r = -3, so the characteristic solution is

y = C₁ exp(-t ) + C₂ exp(-3t )

For the non-homogeneous equation, assume two ansatz solutions

y₁ = (at ² + bt + c) exp(-t )

and

y₂ = at + b

• y'' + 4y' + 3y = 8t exp(-t ) + 6 exp(-t ) … … … [1]

Compute the derivatives of y₁ :

y₁ = (at ² + bt + c) exp(-t )

y₁' = (2at + b) exp(-t ) - (at ² + bt + c) exp(-t )

… = (-at ² + (2a - b) t + b - c) exp(-t )

y₁'' = (-2at + 2a - b) exp(-t ) - (-at ² + (2a - b) t + b - c) exp(-t )

… = (at ² + (b - 4a) t + 2a - 2b + c) exp(-t )

Substitute them into the ODE [1] to get

→   [(at ² + (b - 4a) t + 2a - 2b + c) + 4 (-at ² + (2a - b) t + b - c) + 3 (at ² + bt + c)] exp(-t ) = 8t exp(-t ) + 6 exp(-t )

(at ² + (b - 4a) t + 2a - 2b + c) + 4 (-at ² + (2a - b) t + b - c) + 3 (at ² + bt + c) = 8t + 6

4at + 2a + 2b = 8t + 6

→   4a = 8   and   2a + 2b = 6

→   a = 2   and   b = 1

→   y₁ = (2t ² + t ) exp(-t )

(Note that we don't find out anything about c, but that's okay since it would have gotten absorbed into the first characteristic solution exp(-t ) anyway.)

• y'' + 4y' + 3y = -(9t + 6) … … … [2]

Compute the derivatives of y₂ :

y₂ = at + b

y₂' = a

y₂'' = 0

Substitute these into [2] :

4a + 3 (at + b) = -9t - 6

3at + 4a + 3b = -9t - 6

→   3a = -9   and   4a + 3b = -6

→   a = -3   and   b = 2

→   y₂ = -3t + 2

Then the general solution to the original ODE is

y(t) = C₁ exp(-t ) + C₂ exp(-3t ) + (2t ² + t ) exp(-t ) - 3t + 2

Use the initial conditions y (0) = 2 and y' (0) = 2 to solve for C₁ and C₂ :

y (0) = C₁ + C₂ + 2 = 2

→   C₁ + C₂ = 0 … … … [3]

y'(t) = -C₁ exp(-t ) - 3C₂ exp(-3t ) + (-2t ² + 3t + 1) exp(-t ) - 3

y' (0) = -C₁ - 3C₂ + 1 - 3 = 2

→   C₁ + 3C₂ = -4 … … … [4]

Solve equations [3] and [4] to get C₁ = 2 and C₂ = -2. Then the particular solution to the initial value problem is

y(t) = -2 exp(-3t ) + (2t ² + t + 2) exp(-t ) - 3t + 2