Q25E

Question

Find a particular solution to the differential equation.

$y'' + 2 y' + 4 y = 111 e^{2 t} \cos 3 t$

Step-by-Step Solution

Verified

Answer

The particular solution is $y_{p} = e^{2 t} (\cos 3 t + 6 \sin 3 t)$ .

1Step 1: Firstly, write the auxiliary equation of the above differential equation.

The differential equation is:

$y'' + 2 y' + 4 y = 111 e^{2 t} \cos 3 t \dots (1)$

Write the homogeneous differential equation of the equation (1),

$y'' + 2 y' + 4 y = 0$

The auxiliary equation for the above equation,

$m^{2} + 2 m + 4 = 0$

2Step 2: Now find the roots of the auxiliary equation

Solve the auxiliary equation,

$\begin{matrix} m^{2} + 2 m + 4 = 0 \\ m = \frac{- 2 \pm \sqrt{2^{2} - 4 (1) (4)}}{2 (1)} \\ m = \frac{- 2 \pm \sqrt{- 12}}{2} \\ m = - 1 \pm i \sqrt{3} \end{matrix}$

The roots of the auxiliary equation are,

$m_{1} = - 1 + i \sqrt{3}, & m_{2} = - 1 - i \sqrt{3}$

The complementary solution of the given equation is,

$y_{c} = e^{- x} (c_{1} \cos \sqrt{3} x + c_{2} \sin \sqrt{3} x)$

3Step 3: Use the method of undetermined coefficients to find a particular solution to the differential equation.

According to the method of undetermined coefficients, assume, the particular solution of equation (1),

$y_{p} = e^{2 t} (A_{0} \cos 3 t + B_{0} \sin 3 t) ...... (2)$

Now find the derivative of the above equation,

$\begin{matrix} y_{p}' = 2 e^{2 t} (A_{0} \cos 3 t + B_{0} \sin 3 t) + e^{2 t} (- 3 A_{0} \sin 3 t + 3 B_{0} \cos 3 t) \\ y_{p}'' = 4 e^{2 t} (A_{0} \cos 3 t + B_{0} \sin 3 t) + 2 e^{2 t} (- 3 A_{0} \sin 3 t + 3 B_{0} \cos 3 t) \\ + 2 e^{2 t} (- 3 A_{0} \sin 3 t + 3 B_{0} \cos 3 t) + e^{2 t} (- 9 A_{0} \cos 3 t - 9 B_{0} \sin 3 t) \\ y_{p}'' = e^{2 t} (- 5 A_{0} \cos 3 t - 5 B_{0} \sin 3 t) + e^{2 t} (- 12 A_{0} \sin 3 t + 12 B_{0} \cos 3 t) \end{matrix}$

From the equation (1), Substitute the value of $y_{p}'', y_{p}'$ and $y_{p}$ in the equation (1),

$\begin{array}{l} \Rightarrow y_{p}'' + 2 y_{p}' + 4 y_{p} = 111 e^{2 t} \cos 3 t \\ \Rightarrow e^{2 t} (- 5 A_{0} \cos 3 t - 5 B_{0} \sin 3 t) + e^{2 t} (- 12 A_{0} \sin 3 t + 12 B_{0} \cos 3 t) \\ + 2 [2 e^{2 t} (A_{0} \cos 3 t + B_{0} \sin 3 t) + e^{2 t} (- 3 A_{0} \sin 3 t + 3 B_{0} \cos 3 t)] \\ + 4 e^{2 t} (A_{0} \cos 3 t + B_{0} \sin 3 t) = 111 e^{2 t} \cos 3 t \\ \Rightarrow e^{2 t} (3 A_{0} \cos 3 t + 3 B_{0} \sin 3 t) + e^{2 t} (- 18 A_{0} \sin 3 t + 18 B_{0} \cos 3 t) = 111 e^{2 t} \cos 3 t \\ \Rightarrow e^{2 t} \cos 3 t (3 A_{0} + 18 B_{0}) + e^{2 t} \sin 3 t (3 B_{0} - 18 A_{0}) = 111 e^{2 t} \cos 3 t \end{array}$

4Step 4: Final conclusion:

Comparing all coefficients of the above equation;

$\begin{matrix} 3 A_{0} + 18 B_{0} = 111 . ..... (3) \\ 3 B_{0} - 18 A_{0} = 0 . ..... (4) \end{matrix}$

Solve the above equations,

$\begin{matrix} (3 A_{0} + 18 B_{0}) = 111 \times 6 \\ 18 A_{0} + 108 B_{0} = 666 \\ - 18 A_{0} + 3 B_{0} = 0 \\ \bar{111 B_{0} = 666} \\ B_{0} = 6 \end{matrix}$