Q41E

Question

Discontinuous Forcing Term. In certain physical models, the nonhomogeneous term, or forcing term, g(t) in the equation

$ay'' + by' + cy = g (t)$

may not be continuous but may have a jump discontinuity. If this occurs, we can still obtain a reasonable solution using the following procedure. Consider the initial value problem;

$y'' + 2 y' + 5 y = g (t); y (0) = 0, y' (0) = 0$

Where,

$g (t) = \{\begin{array}{l} 10, if 0 \leq t \leq \frac{3 π}{2} \\ 0, if t > \frac{3 π}{2} \end{array}\}$

Find a solution to the initial value problem for $0 \leq t \leq \frac{3 π}{2}$ .
Find a general solution for $t > \frac{3 π}{2}$ .
Now choose the constants in the general solution from part (b) so that the solution from part (a) and the solution from part (b) agree, together with their first derivatives, at $t = \frac{3 π}{2}$ . This gives us a continuously differentiable function that satisfies the differential equation except at $t = \frac{3 π}{2}$ .

Step-by-Step Solution

Verified

Answer

$y = - 2 e^{- t} \cos (2 t) - e^{- t} \sin (2 t) + 2$
$y = c_{1} e^{- t} \cos (2 t) + c_{2} e^{- t} \sin (2 t)$
$c_{1} = - 2 ({\begin{matrix} 1 + e \end{matrix}}^{\frac{3 π}{2}})$ and $c_{2} = - (1 + e^{\frac{3 π}{2}})$

1Step 1: Firstly, determine the solutions for 0 ≤ t ≤ 3 π 2 .

(a).

Given that,

$0 \leq t \leq \frac{3 π}{2}$

Consider the differential equation is,

$y'' + 2 y' + 5 y = g (t) . ..... (1)$

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

$y'' + 2 y' + 5 y = 0$

2Step 2: Now find the complimentary solution of the given equation is

The auxiliary equation for the above equation, $m^{2} + 2 m + 5 = 0$

Solve the auxiliary equation,

$\begin{matrix} m^{2} + 2 m + 5 = 0 \\ m = \frac{- 2 \pm \sqrt{4 - 20}}{2} \\ m = - 1 \pm 2 i \end{matrix}$

The roots of auxiliary equation are, $m_{1} = - 1 + 2 i, m_{2} = - 1 - 2 i .$

The complimentary solution of the given equation is, $y_{c} = c_{1} e^{- t} \cos (2 t) + c_{2} e^{- t} \sin (2 t) .$

3Step 3: Use the given information, and find the particular solution to find a general solution for the equation

Given that,

$g (t) = 10$

Assume, the particular solution of equation (1),

$y_{p} (t) = k . ..... (2)$

Now find the first and second derivative of above equation,

$\begin{matrix} y_{p}' (t) = 0 \\ y_{p}'' (t) = 0 \end{matrix}$

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

$\begin{matrix} y'' + 2 y' + 5 y = g (t) \\ 0 + 2 (0) + 5 (k) = 10 \\ k = 2 \end{matrix}$

Therefore, the particular solution of equation (1),

$y_{p} (t) = 2$

4Step 4: Find the general solution and use the given initial condition,

Therefore, the general solution is,

$\begin{matrix} y = y_{c} (t) + y_{p} (t) \\ y = c_{1} e^{- t} \cos (2 t) + c_{2} e^{- t} \sin (2 t) + 2 . ..... (3) \end{matrix}$

Now find the first and second derivative of above equation,

$\begin{matrix} y' = - c_{1} e^{- t} \cos (2 t) - 2 c_{1} e^{- t} \sin (2 t) - c_{2} e^{- t} \sin (2 t) + 2 c_{2} e^{- t} \cos (2 t) \\ y' = (- c_{1} + 2 c_{2}) e^{- t} \cos (2 t) + (- 2 c_{1} - c_{2}) e^{- t} \sin (2 t) . ..... (4) \end{matrix}$

Given initial condition,

$y (0) = 0, y' (0) = 0$

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

$\begin{array}{l} 0 = c_{1} e^{- (0)} \cos (0) + c_{2} e^{- 0} \sin (0) + 2 \\ 0 = c_{1} + 2 \\ c_{1} = - 2 \end{array}$

Substitute the value of y’ = 0 and t = 0 in the equation (4),

$\begin{matrix} 0 = (- c_{1} + 2 c_{2}) e^{- (0)} \cos (0) + (- 2 c_{1} - c_{2}) e^{- (0)} \sin (0) \\ 0 = (- c_{1} + 2 c_{2}) \\ - c_{1} + 2 c_{2} = 0 . ... (5) \end{matrix}$

Substitute the value of $C_{1}$ in the equation (5),

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

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

$\begin{matrix} y = c_{1} e^{- t} \cos (2 t) + c_{2} e^{- t} \sin (2 t) + 2 \\ y_{1} = - 2 e^{- t} \cos (2 t) - e^{- t} \sin (2 t) + 2 . ..... (6) \end{matrix}$

5Step 5: Use the given information, and find the particular solution to find a general solution for the equation

(b).

Given that,

$t > \frac{3 π}{2}$ , $g (t) = 0$

Again assume, the particular solution of equation (1),

$y_{p} (t) = λ . ..... (7)$

Now find the first and second derivative of above equation,

$\begin{matrix} y_{p}' (t) = 0 \\ y_{p}'' (t) = 0 \end{matrix}$

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

$\begin{matrix} y'' + 2 y' + 5 y = g (t) \\ 0 + 2 (0) + 5 (λ) = 0 \\ λ = 0 \end{matrix}$

Substitute the value of $λ$ in the equation (7),

Therefore, the particular solution of equation (1),

$y_{p} (t) = 0$

6Step 6: Find the general solution,

Therefore, the general solution is,

$\begin{matrix} y = y_{c} (t) + y_{p} (t) \\ y = c_{1} e^{- t} \cos (2 t) + c_{2} e^{- t} \sin (2 t) + 0 \\ y_{2} = c_{1} e^{- t} \cos (2 t) + c_{2} e^{- t} \sin (2 t) . ..... (8) \end{matrix}$

7Step 7: From the part (a) and (b),

(c).

Now find the first derivative of the equation (6),

$\begin{matrix} y_{1}' = 2 e^{- t} \cos (2 t) + 4 e^{- t} \sin (2 t) + e^{- t} \sin (2 t) - 2 e^{- t} \cos (2 t) \\ y_{1}' = 5 e^{- t} \sin (2 t) \end{matrix}$

Now find the first derivative of the equation (8),

$\begin{matrix} y_{2}' = - c_{1} e^{- t} \cos (2 t) - 2 c_{1} e^{- t} \sin (2 t) - c_{2} e^{- t} \sin (2 t) + 2 c_{2} e^{- t} \cos (2 t) \\ y_{2}' = (- c_{1} + 2 c_{2}) e^{- t} \cos (2 t) + (- 2 c_{1} - c_{2}) e^{- t} \sin (2 t) \end{matrix}$

By the question $y_{1}' = y_{2}'$ at $t = \frac{3 π}{2}$

$\begin{matrix} y_{1}' (\frac{3 π}{2}) = y_{2}' (\frac{3 π}{2}) \\ 5 e^{- t} \sin (2 \frac{3 π}{2}) = (- c_{1} + 2 c_{2}) e^{- t} \cos (2 \frac{3 π}{2}) + (- 2 c_{1} - c_{2}) e^{- t} \sin (2 \frac{3 π}{2}) \\ 5 e^{- t} \sin (3 π) = (- c_{1} + 2 c_{2}) e^{- t} \cos (3 π) + (- 2 c_{1} - c_{2}) e^{- t} \sin (3 π) \\ 0 = (- c_{1} + 2 c_{2}) (- 1) \\ c_{1} = 2 c_{2} . ..... (9) \end{matrix}$

By the question $y_{1} = y_{2}$ at $t = \frac{3 π}{2}$

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

Substitute the value of $C_{1}$ in the equation (9),

$\begin{matrix} c_{1} = 2 c_{2} \\ - 2 (1 + e^{\frac{3 π}{2}}) = 2 c_{2} \\ c_{2} = - (1 + e^{\frac{3 π}{2}}) \end{matrix}$

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