Answer :
To solve the differential equation [tex]\(\frac{d^2 y}{d t^2} + 9 \frac{d y}{d t} = 4 \sin 2 t\)[/tex] using the method of undetermined coefficients, we'll follow these steps:
1. Solve the homogeneous equation:
[tex]\(\frac{d^2 y}{d t^2} + 9 \frac{d y}{d t} = 0\)[/tex]
2. Find the particular solution of the non-homogeneous equation:
[tex]\(\frac{d^2 y}{d t^2} + 9 \frac{d y}{d t} = 4 \sin 2 t\)[/tex]
3. Combine the solutions to form the general solution.
### Step 1: Solve the Homogeneous Equation
First, solve the homogeneous equation:
[tex]\[ \frac{d^2 y}{d t^2} + 9 \frac{d y}{d t} = 0 \][/tex]
This can be written as:
[tex]\[ y'' + 9y' = 0 \][/tex]
To solve this, we'll assume a solution of the form [tex]\(y = e^{rt}\)[/tex]. Substituting [tex]\(y = e^{rt}\)[/tex] into the differential equation, we get:
[tex]\[ r^2 e^{rt} + 9r e^{rt} = 0 \][/tex]
Factor out [tex]\(e^{rt}\)[/tex]:
[tex]\[ e^{rt} (r^2 + 9r) = 0 \][/tex]
Since [tex]\(e^{rt} \neq 0\)[/tex], we have:
[tex]\[ r^2 + 9r = 0 \][/tex]
Solve for [tex]\(r\)[/tex]:
[tex]\[ r(r + 9) = 0 \][/tex]
This gives us two roots:
[tex]\[ r = 0 \quad \text{and} \quad r = -9 \][/tex]
So the general solution to the homogeneous equation is:
[tex]\[ y_h(t) = C_1 + C_2 e^{-9t} \][/tex]
### Step 2: Find the Particular Solution
Next, we need a particular solution to the non-homogeneous equation:
[tex]\[ \frac{d^2 y}{d t^2} + 9 \frac{d y}{d t} = 4 \sin 2 t \][/tex]
Since the right-hand side is [tex]\(4 \sin 2 t\)[/tex], which is a sine function, we will guess a particular solution of the form:
[tex]\[ y_p(t) = A \cos 2t + B \sin 2t \][/tex]
Now, we need the first and second derivatives of [tex]\(y_p(t)\)[/tex]:
[tex]\[ y_p'(t) = -2A \sin 2t + 2B \cos 2t \][/tex]
[tex]\[ y_p''(t) = -4A \cos 2t - 4B \sin 2t \][/tex]
Substitute [tex]\(y_p(t)\)[/tex], [tex]\(y_p'(t)\)[/tex], and [tex]\(y_p''(t)\)[/tex] into the original differential equation:
[tex]\[ (-4A \cos 2t - 4B \sin 2t) + 9(-2A \sin 2t + 2B \cos 2t) = 4 \sin 2t \][/tex]
Simplify and collect like terms:
[tex]\[ -4A \cos 2t - 4B \sin 2t - 18A \sin 2t + 18B \cos 2t = 4 \sin 2t \][/tex]
[tex]\[ (-4A + 18B) \cos 2t + (-4B - 18A) \sin 2t = 4 \sin 2t \][/tex]
To satisfy this equation for all [tex]\(t\)[/tex], the coefficients of [tex]\(\cos 2t\)[/tex] and [tex]\(\sin 2t\)[/tex] must be equal on both sides:
[tex]\[ -4A + 18B = 0 \][/tex]
[tex]\[ -4B - 18A = 4 \][/tex]
Solve these simultaneous equations:
From the first equation:
[tex]\[ 18B = 4A \][/tex]
[tex]\[ B = \frac{2A}{9} \][/tex]
Substitute [tex]\(B\)[/tex] into the second equation:
[tex]\[ -4\left(\frac{2A}{9}\right) - 18A = 4 \][/tex]
[tex]\[ -\frac{8A}{9} - 18A = 4 \][/tex]
[tex]\[ -\frac{8A + 162A}{9} = 4 \][/tex]
[tex]\[ -\frac{170A}{9} = 4 \][/tex]
[tex]\[ -170A = 36 \][/tex]
[tex]\[ A = -\frac{36}{170} = -\frac{18}{85} \][/tex]
Then, substitute [tex]\(A\)[/tex] back to find [tex]\(B\)[/tex]:
[tex]\[ B = \frac{2A}{9} = \frac{2 \left( -\frac{18}{85} \right)}{9} = -\frac{36}{85 \cdot 9} = -\frac{36}{765} = -\frac{4}{85} \][/tex]
So the particular solution is:
[tex]\[ y_p(t) = -\frac{18}{85} \cos 2t - \frac{4}{85} \sin 2t \][/tex]
### Step 3: General Solution
Combine the homogeneous solution and the particular solution:
[tex]\[ y(t) = y_h(t) + y_p(t) = C_1 + C_2 e^{-9t} - \frac{18}{85} \cos 2t - \frac{4}{85} \sin 2t \][/tex]
Thus, the general solution to the differential equation is:
[tex]\[ \boxed{y(t) = C_1 + C_2 e^{-9t} - \frac{18}{85} \cos 2t - \frac{4}{85} \sin 2t} \][/tex]
1. Solve the homogeneous equation:
[tex]\(\frac{d^2 y}{d t^2} + 9 \frac{d y}{d t} = 0\)[/tex]
2. Find the particular solution of the non-homogeneous equation:
[tex]\(\frac{d^2 y}{d t^2} + 9 \frac{d y}{d t} = 4 \sin 2 t\)[/tex]
3. Combine the solutions to form the general solution.
### Step 1: Solve the Homogeneous Equation
First, solve the homogeneous equation:
[tex]\[ \frac{d^2 y}{d t^2} + 9 \frac{d y}{d t} = 0 \][/tex]
This can be written as:
[tex]\[ y'' + 9y' = 0 \][/tex]
To solve this, we'll assume a solution of the form [tex]\(y = e^{rt}\)[/tex]. Substituting [tex]\(y = e^{rt}\)[/tex] into the differential equation, we get:
[tex]\[ r^2 e^{rt} + 9r e^{rt} = 0 \][/tex]
Factor out [tex]\(e^{rt}\)[/tex]:
[tex]\[ e^{rt} (r^2 + 9r) = 0 \][/tex]
Since [tex]\(e^{rt} \neq 0\)[/tex], we have:
[tex]\[ r^2 + 9r = 0 \][/tex]
Solve for [tex]\(r\)[/tex]:
[tex]\[ r(r + 9) = 0 \][/tex]
This gives us two roots:
[tex]\[ r = 0 \quad \text{and} \quad r = -9 \][/tex]
So the general solution to the homogeneous equation is:
[tex]\[ y_h(t) = C_1 + C_2 e^{-9t} \][/tex]
### Step 2: Find the Particular Solution
Next, we need a particular solution to the non-homogeneous equation:
[tex]\[ \frac{d^2 y}{d t^2} + 9 \frac{d y}{d t} = 4 \sin 2 t \][/tex]
Since the right-hand side is [tex]\(4 \sin 2 t\)[/tex], which is a sine function, we will guess a particular solution of the form:
[tex]\[ y_p(t) = A \cos 2t + B \sin 2t \][/tex]
Now, we need the first and second derivatives of [tex]\(y_p(t)\)[/tex]:
[tex]\[ y_p'(t) = -2A \sin 2t + 2B \cos 2t \][/tex]
[tex]\[ y_p''(t) = -4A \cos 2t - 4B \sin 2t \][/tex]
Substitute [tex]\(y_p(t)\)[/tex], [tex]\(y_p'(t)\)[/tex], and [tex]\(y_p''(t)\)[/tex] into the original differential equation:
[tex]\[ (-4A \cos 2t - 4B \sin 2t) + 9(-2A \sin 2t + 2B \cos 2t) = 4 \sin 2t \][/tex]
Simplify and collect like terms:
[tex]\[ -4A \cos 2t - 4B \sin 2t - 18A \sin 2t + 18B \cos 2t = 4 \sin 2t \][/tex]
[tex]\[ (-4A + 18B) \cos 2t + (-4B - 18A) \sin 2t = 4 \sin 2t \][/tex]
To satisfy this equation for all [tex]\(t\)[/tex], the coefficients of [tex]\(\cos 2t\)[/tex] and [tex]\(\sin 2t\)[/tex] must be equal on both sides:
[tex]\[ -4A + 18B = 0 \][/tex]
[tex]\[ -4B - 18A = 4 \][/tex]
Solve these simultaneous equations:
From the first equation:
[tex]\[ 18B = 4A \][/tex]
[tex]\[ B = \frac{2A}{9} \][/tex]
Substitute [tex]\(B\)[/tex] into the second equation:
[tex]\[ -4\left(\frac{2A}{9}\right) - 18A = 4 \][/tex]
[tex]\[ -\frac{8A}{9} - 18A = 4 \][/tex]
[tex]\[ -\frac{8A + 162A}{9} = 4 \][/tex]
[tex]\[ -\frac{170A}{9} = 4 \][/tex]
[tex]\[ -170A = 36 \][/tex]
[tex]\[ A = -\frac{36}{170} = -\frac{18}{85} \][/tex]
Then, substitute [tex]\(A\)[/tex] back to find [tex]\(B\)[/tex]:
[tex]\[ B = \frac{2A}{9} = \frac{2 \left( -\frac{18}{85} \right)}{9} = -\frac{36}{85 \cdot 9} = -\frac{36}{765} = -\frac{4}{85} \][/tex]
So the particular solution is:
[tex]\[ y_p(t) = -\frac{18}{85} \cos 2t - \frac{4}{85} \sin 2t \][/tex]
### Step 3: General Solution
Combine the homogeneous solution and the particular solution:
[tex]\[ y(t) = y_h(t) + y_p(t) = C_1 + C_2 e^{-9t} - \frac{18}{85} \cos 2t - \frac{4}{85} \sin 2t \][/tex]
Thus, the general solution to the differential equation is:
[tex]\[ \boxed{y(t) = C_1 + C_2 e^{-9t} - \frac{18}{85} \cos 2t - \frac{4}{85} \sin 2t} \][/tex]
