Find the solution(s) to y''+9y=27cos(6t)(If it exists) satisfying the boundary conditions.role="math" localid="1655127897158" y(0)=-1, y(π6)=3role="math" localid="1655127945613" y(0)=-1, y(π3)=5role="math" localid="1655127991356" y(0)=-1, y(π3)=-1

Q47E - Fundamentals Of Differential Equations And Boundary Value Problems

Q47E

Question

Find the solution(s) to $y'' + 9 y = 27 \cos (6 t)$

(If it exists) satisfying the boundary conditions.

$y (0) = - 1, y (\frac{π}{6}) = 3$
$y (0) = - 1, y (\frac{π}{3}) = 5$
$y (0) = - 1, y (\frac{π}{3}) = - 1$

Step-by-Step Solution

Verified

Answer

The solution to the given differential equations are:

a. $y = 2 \sin (3 t) - \cos (6 t)$

b. No solution

c. $y = c_{2} \sin (3 t) - \cos (6 t)$

1Step 1: Use the given information to write the homogeneous differential equation.

The differential equation is,

$y'' + 9 y = 27 \cos (6 t) . ..... (1)$

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

$y'' + 9 y = 0$

2Step 2: Now find the complementary solution of the given equation.

The auxiliary equation for the above equation,

$\begin{matrix} m^{2} + 3^{2} = 0 \\ m = \pm 3 i \end{matrix}$

The roots of the auxiliary equation are,

$m_{1} = 3 i, m_{2} = - 3 i$

The complementary solution of the given equation is,

$y_{c} = c_{1} \cos (3 t) + c_{2} \sin (3 t)$

3Step 3: Find the particular solution to find a general solution for the equation.

Assume, the particular solution of equation (1),

$y_{p} (t) = Acos (6 t) + Bsin (6 t) . ..... (2)$

Now find the first and second derivatives of the above equation,

$\begin{matrix} y_{p}' (t) = - 6 Asin (6 t) + 6 Bcos (6 t) \\ y_{p}'' (t) = - 36 Acos (6 t) - 36 Bsin (6 t) \end{matrix}$

Substitute the value of $y_{p} (t)$ and $y_{p}'' (t)$ in the equation (1),

$\begin{matrix} y'' + 9 y = 27 \cos (6 t) \\ - 36 Acos (6 t) - 36 Bsin (6 t) + 9 [Acos (6 t) + Bsin (6 t)] = 27 \cos (6 t) \\ - 27 Acos (6 t) - 27 Bsin (6 t) = 27 \cos (6 t) \end{matrix}$

Comparing all coefficients of the above equation,

$\begin{matrix} - 27 A = 27 \Rightarrow A = - 1 \\ - 27 B = 0 \Rightarrow B = 0 \end{matrix}$

Substitute the value of A and B in the equation (2),

$\begin{matrix} y_{p} (t) = Acos (6 t) + Bsin (6 t) \\ y_{p} (t) = (- 1) \cos (6 t) + (0) \sin (6 t) \\ y_{p} (t) = - \cos (6 t) \end{matrix}$

Therefore, the particular solution of equation (1),

$y_{p} (t) = - \cos (6 t)$

Therefore, the general solution is,

$\begin{array}{l} y = y_{c} (t) + y_{p} (t) \\ y = c_{1} \cos (3 t) + c_{2} \sin (3 t) - \cos (6 t) . ..... (3) \end{array}$

4Step 4: Use the given boundary conditions.

Given boundary conditions,

$y (0) = - 1, y (\frac{π}{6}) = 3$

Substitute the value of y = -1 and t = 0 in the equation (3),

$\begin{matrix} - 1 = c_{1} \cos (0) + c_{2} \sin (0) - \cos (0) \\ - 1 = c_{1} - 1 \\ c_{1} = 0 \end{matrix}$

Substitute the value of y= 3 and $t = \frac{π}{6}$ in the equation (3),

$\begin{matrix} 3 = c_{1} \cos (3 \frac{π}{6}) + c_{2} \sin (3 \frac{π}{6}) - \cos (6 \frac{π}{6}) \\ 3 = c_{1} \cos (\frac{π}{2}) + c_{2} \sin (\frac{π}{2}) - \cos (π) \\ 3 = c_{2} + 1 \\ c_{2} = 2 \end{matrix}$

Substitute the value of $c_{1}$ and $c_{2}$ in the equation (3),

$\begin{matrix} y = (0) \cos (3 t) + (2) \sin (3 t) - \cos (6 t) \\ y = 2 \sin (3 t) - \cos (6 t) \end{matrix}$

5Step 5: Use the given boundary conditions.

Given boundary conditions,

$y (0) = - 1, y (\frac{π}{3}) = 5$

Substitute the value of y = -1 and t = 0 in the equation (3),

$\begin{matrix} - 1 = c_{1} \cos (0) + c_{2} \sin (0) - \cos (0) \\ - 1 = c_{1} - 1 \\ c_{1} = 0 \end{matrix}$

Substitute the value of y = 5 and $t = \frac{π}{3}$ in the equation (3),

$\begin{matrix} 5 = c_{1} \cos (3 \frac{π}{3}) + c_{2} \sin (3 \frac{π}{3}) - \cos (6 \frac{π}{3}) \\ 5 = c_{1} \cos (π) + c_{2} \sin (π) - \cos (2 π) \\ 5 = - c_{1} - 1 \\ c_{1} = - 6 \end{matrix}$

Here $c_{1}$ have two different values, so this is an absurd case.

Therefore, no solution to this boundary value problem,

6Step 6: Use the given boundary conditions.

Given boundary conditions,

$y (0) = - 1, y (\frac{π}{3}) = - 1$

Substitute the value of y = -1 and t = 0 in the equation (3),

$\begin{matrix} - 1 = c_{1} \cos (0) + c_{2} \sin (0) - \cos (0) \\ - 1 = c_{1} - 1 \\ c_{1} = 0 \end{matrix}$

Substitute the value of y = -1 and $t = \frac{π}{3}$ in the equation (3),

$\begin{matrix} - 1 = c_{1} \cos (3 \frac{π}{3}) + c_{2} \sin (3 \frac{π}{3}) - \cos (6 \frac{π}{3}) \\ - 1 = c_{1} \cos (π) + c_{2} \sin (π) - \cos (2 π) \\ - 1 = - c_{1} - 1 \\ c_{1} = 0 \end{matrix}$

Substitute the value of $c_{1}$ in the equation (3),

$\begin{matrix} y = (0) \cos (3 t) + c_{2} \sin (3 t) - \cos (6 t) \\ y = c_{2} \sin (3 t) - \cos (6 t) \end{matrix}$

Thus, the solution is $y = c_{2} \sin (3 t) - \cos (6 t)$

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