Find a particular solution to the differential equation.
$y^{″} - y^{'} - 6 y = \sin t + 3 \cos t$

Name: Find a particular solution to the differential equation.y″−y′−6y=sin⁡t+3cos⁡t
Uploaded: 2022-02-23T17:22:57+00:00
Description: Find a particular solution to the differential equation.y″−y′−6y=sin⁡t+3cos⁡t

Question

Find a particular solution to the differential equation.y″−y′−6y=sin⁡t+3cos⁡t

Alannah Farmer · Accepted Answer

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:

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} \cdot t} + c_{2} e^{λ_{2} \cdot t}

So the solution is of the form:

y_{c} = c_{1} e^{3 t} + c_{2} e^{- 2 t}

And the particular solution is given as:

(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} \cdot 3 \cos t

= \frac{1}{(- 1) - D - 6} \sin t + \frac{1}{(- 1) - D - 6} \cdot 3 \cos t

= \frac{1}{- D - 7} \sin t + 3 \cdot \frac{1}{- D - 7} \cos t

= - \frac{D - 7}{(D + 7) (D - 7)} \sin t - 3 \cdot \frac{D - 7}{(D - 7) (D + 7)} \cos t

{Ratinalizing the denominator}

= - \frac{D - 7}{D^{2} - 49} \sin t - 3 \cdot \frac{D - 7}{D^{2} - 49} \cos t

= - \frac{D - 7}{- 1 - 49} \sin t - 3 \cdot \frac{D - 7}{- 1 - 49} \cos t

= \frac{D - 7}{50} \sin t + 3 \cdot \frac{D - 7}{50} \cos t

= \frac{\cos (t)}{50} - \frac{7}{50} \sin t + \frac{3 (- \sin t)}{50} - \frac{21}{50} \cos t

= - \frac{1}{5} \sin (t) - \frac{2}{5} \cos (t)

Now, the solution of the given ODE is:

y = y_{p} + y_{c}

y (x) = c_{1} e^{3 t} + c_{2} e^{- 2 t} - \frac{1}{5} \sin (t) - \frac{2}{5} \cos (t)

Jeffrey Jordon · Accepted Answer

Answer is given below (on video)

Find a particular solution to the differential equation.

Answered question

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