Answered: y" – y' – 6 y = sin t + 3 cos t

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Please find a particular solution to the differential equation. Thank you.

Expert Solution

Step 1

The given equation is a second order linear non homogeneous differential equation with constant coefficient.

The general solution for a(t) y’’ + b(t) y’ + c(t) y = g(t)

The general solution of the given differential equation can be written as y = y_h+ y_p

y_h is the solution to the homogeneous ODE a(t) y’’ + b(t) y’ + c(t) y = 0

and the particular solution, is any function that satisfies the non-homogeneous equation.

The complementary solution for the given equation is:

$\begin{array}{l} y^{''} - y^{'} - 6 y = \sin t + 3 \cos t \\ y^{''} - y^{'} - 6 y = 0 \\ (D^{2} - D - 6) y = 0 \\ λ^{2} - λ - 6 = 0 \\ (λ - 3) (λ + 2) = 0 \\ λ = 3, λ = - 2 \\ y = c_{1} e^{3 t} + c_{2} e^{- 2 t} \\ Two real and different root λ_{1} and λ_{2} \\ The solution is of the form: y = c_{1} e^{λ_{1} t} + c_{2} e^{λ_{2} t} \\ So the solution is of the form: \\ y_{c} = c_{1} e^{3 t} + c_{2} e^{- 2 t} \end{array}$

Step 2

And the particular solution is given as:

$\begin{matrix} (D^{2} - D - 6) y = \sin t + 3 \cos t \\ y_{p} = \frac{\sin t + 3 \cos t}{D^{2} - D - 6} = \frac{1}{(D^{2} - D - 6)} \sin t + 3 \cos t \\ = \frac{1}{(D^{2} - D - 6)} \sin t + \frac{1}{(D^{2} - D - 6)} 3 \cos t \\ \{\frac{1}{f (D^{2})} \sin (a x + b) or cos (a x + b) = \frac{\sin (a x + b)}{f (- a^{2})}, f (- a^{2}) \neq 0\} \\ = \frac{1}{((- 1) - D - 6)} \sin t + \frac{1}{((- 1) - D - 6)} 3 \cos t \\ = \frac{1}{(- D - 7)} \sin t + 3 \frac{1}{(- D - 7)} \cos t \\ = - \frac{(D - 7)}{(D + 7) (D - 7)} \sin t - 3 \frac{(D - 7)}{(D + 7) (D - 7)} \cos t \{Ratinalizing the denominator\} \\ = - \frac{(D - 7)}{(D^{2} - 49)} \sin t - 3 \frac{(D - 7)}{(D^{2} - 49)} \cos t \\ = - \frac{(D - 7)}{(- 1 - 49)} \sin t - 3 \frac{(D - 7)}{(- 1 - 49)} \cos t \\ = \frac{(D - 7)}{50} \sin t + 3 \frac{(D - 7)}{50} \cos t \\ = \frac{\cos t}{50} - \frac{7}{50} \sin t + 3 \frac{(- \sin t)}{50} - \frac{21}{50} \cos t \\ = - \frac{1}{5} \sin (t) - \frac{2}{5} \cos (t) \end{matrix}$

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