Answer :
To find the particular solution of the given differential equation with initial conditions, follow these steps:
### Given Problem
The differential equation is:
[tex]\[ 9 \frac{d^2 y}{d x^2} - 12 \frac{d y}{d x} + 4 y = 3 x - 1 \][/tex]
### Initial Conditions
At [tex]\( x = 0 \)[/tex]:
[tex]\[ y(0) = 0 \][/tex]
[tex]\[ \frac{d y}{d x}\bigg|_{x=0} = -\frac{4}{3} \][/tex]
### Step-by-Step Solution
1. Formulate the Homogeneous Equation:
First, solve the corresponding homogeneous equation:
[tex]\[ 9 \frac{d^2 y}{d x^2} - 12 \frac{d y}{d x} + 4 y = 0 \][/tex]
2. Solve for Characteristic Equation:
Convert the homogeneous differential equation into its characteristic equation by assuming a solution of the form [tex]\( y = e^{rx} \)[/tex]:
[tex]\[ 9r^2 - 12r + 4 = 0 \][/tex]
3. Solve the Quadratic Equation:
Solve for [tex]\( r \)[/tex]:
[tex]\[ r = \frac{12 \pm \sqrt{144 - 144}}{18} = \frac{12 \pm 0}{18} = \frac{2}{3} \][/tex]
Since both roots are [tex]\( \frac{2}{3} \)[/tex], the general solution to the homogeneous equation is:
[tex]\[ y_h(x) = (C_1 + C_2 x) e^{2x/3} \][/tex]
4. Formulate the Particular Solution:
Find a particular solution [tex]\( y_p(x) \)[/tex] to the non-homogeneous equation. Assume a particular solution of the form:
[tex]\[ y_p(x) = Ax + B \][/tex]
Substitute [tex]\( y_p \)[/tex] and its derivatives into the original differential equation:
[tex]\[ 9(0) - 12(A) + 4(Ax + B) = 3x - 1 \implies 4Ax + 4B - 12A = 3x - 1 \][/tex]
Comparing coefficients:
[tex]\[ 4A = 3 \quad \Rightarrow \quad A = \frac{3}{4} \][/tex]
[tex]\[ 4B - 12A = -1 \quad \Rightarrow \quad B = 2 \][/tex]
Thus, the particular solution is:
[tex]\[ y_p(x) = \frac{3}{4}x + 2 \][/tex]
5. Combine Solutions:
The general solution of the differential equation is:
[tex]\[ y(x) = y_h(x) + y_p(x) = (C_1 + C_2 x) e^{2x/3} + \frac{3}{4}x + 2 \][/tex]
6. Apply Initial Conditions:
Use the initial conditions to solve for [tex]\( C_1 \)[/tex] and [tex]\( C_2 \)[/tex]:
[tex]\[ y(0) = 0 \quad \Rightarrow \quad C_1 e^0 + 2 = 0 \quad \Rightarrow \quad C_1 + 2 = 0 \quad \Rightarrow \quad C_1 = -2 \][/tex]
The derivative of [tex]\( y(x) \)[/tex] is:
[tex]\[ y'(x) = \left(C_2 e^{2x/3} + \frac{2}{3}(C_1 + C_2 x)e^{2x/3}\right) + \frac{3}{4} \][/tex]
At [tex]\( x = 0 \)[/tex]:
[tex]\[ y'(0) = C_2 + \frac{2}{3}C_1 + \frac{3}{4} = -\frac{4}{3} \][/tex]
Substituting [tex]\( C_1 = -2 \)[/tex]:
[tex]\[ C_2 + \frac{2}{3}(-2) + \frac{3}{4} = -\frac{4}{3} \][/tex]
[tex]\[ C_2 - \frac{4}{3} + \frac{3}{4} = -\frac{4}{3} \quad \Rightarrow \quad C_2 - \frac{4}{3} + \frac{3}{4} = -\frac{4}{3} \quad \Rightarrow \quad C_2 - \frac{1}{12} = -\frac{4}{3} \quad \Rightarrow \quad C_2 = -\frac{4}{3} + \frac{1}{12} \][/tex]
[tex]\[ C_2 = -\frac{16}{12} + \frac{1}{12} = -\frac{15}{12} = -1.25 \][/tex]
7. Particular Solution:
Thus, the particular solution of the differential equation is:
[tex]\[ y(x) = \frac{3}{4}x + (-0.75x - 2.0)e^{2x/3} + 2 \][/tex]
So, the particular solution of the differential equation [tex]\( 9 \frac{d^2 y}{d x^2} - 12 \frac{d y}{d x} + 4 y = 3 x - 1 \)[/tex] given the initial conditions is:
[tex]\[ y(x) = \frac{3}{4}x + (-0.75x - 2.0)e^{2x/3} + 2 \][/tex]
### Given Problem
The differential equation is:
[tex]\[ 9 \frac{d^2 y}{d x^2} - 12 \frac{d y}{d x} + 4 y = 3 x - 1 \][/tex]
### Initial Conditions
At [tex]\( x = 0 \)[/tex]:
[tex]\[ y(0) = 0 \][/tex]
[tex]\[ \frac{d y}{d x}\bigg|_{x=0} = -\frac{4}{3} \][/tex]
### Step-by-Step Solution
1. Formulate the Homogeneous Equation:
First, solve the corresponding homogeneous equation:
[tex]\[ 9 \frac{d^2 y}{d x^2} - 12 \frac{d y}{d x} + 4 y = 0 \][/tex]
2. Solve for Characteristic Equation:
Convert the homogeneous differential equation into its characteristic equation by assuming a solution of the form [tex]\( y = e^{rx} \)[/tex]:
[tex]\[ 9r^2 - 12r + 4 = 0 \][/tex]
3. Solve the Quadratic Equation:
Solve for [tex]\( r \)[/tex]:
[tex]\[ r = \frac{12 \pm \sqrt{144 - 144}}{18} = \frac{12 \pm 0}{18} = \frac{2}{3} \][/tex]
Since both roots are [tex]\( \frac{2}{3} \)[/tex], the general solution to the homogeneous equation is:
[tex]\[ y_h(x) = (C_1 + C_2 x) e^{2x/3} \][/tex]
4. Formulate the Particular Solution:
Find a particular solution [tex]\( y_p(x) \)[/tex] to the non-homogeneous equation. Assume a particular solution of the form:
[tex]\[ y_p(x) = Ax + B \][/tex]
Substitute [tex]\( y_p \)[/tex] and its derivatives into the original differential equation:
[tex]\[ 9(0) - 12(A) + 4(Ax + B) = 3x - 1 \implies 4Ax + 4B - 12A = 3x - 1 \][/tex]
Comparing coefficients:
[tex]\[ 4A = 3 \quad \Rightarrow \quad A = \frac{3}{4} \][/tex]
[tex]\[ 4B - 12A = -1 \quad \Rightarrow \quad B = 2 \][/tex]
Thus, the particular solution is:
[tex]\[ y_p(x) = \frac{3}{4}x + 2 \][/tex]
5. Combine Solutions:
The general solution of the differential equation is:
[tex]\[ y(x) = y_h(x) + y_p(x) = (C_1 + C_2 x) e^{2x/3} + \frac{3}{4}x + 2 \][/tex]
6. Apply Initial Conditions:
Use the initial conditions to solve for [tex]\( C_1 \)[/tex] and [tex]\( C_2 \)[/tex]:
[tex]\[ y(0) = 0 \quad \Rightarrow \quad C_1 e^0 + 2 = 0 \quad \Rightarrow \quad C_1 + 2 = 0 \quad \Rightarrow \quad C_1 = -2 \][/tex]
The derivative of [tex]\( y(x) \)[/tex] is:
[tex]\[ y'(x) = \left(C_2 e^{2x/3} + \frac{2}{3}(C_1 + C_2 x)e^{2x/3}\right) + \frac{3}{4} \][/tex]
At [tex]\( x = 0 \)[/tex]:
[tex]\[ y'(0) = C_2 + \frac{2}{3}C_1 + \frac{3}{4} = -\frac{4}{3} \][/tex]
Substituting [tex]\( C_1 = -2 \)[/tex]:
[tex]\[ C_2 + \frac{2}{3}(-2) + \frac{3}{4} = -\frac{4}{3} \][/tex]
[tex]\[ C_2 - \frac{4}{3} + \frac{3}{4} = -\frac{4}{3} \quad \Rightarrow \quad C_2 - \frac{4}{3} + \frac{3}{4} = -\frac{4}{3} \quad \Rightarrow \quad C_2 - \frac{1}{12} = -\frac{4}{3} \quad \Rightarrow \quad C_2 = -\frac{4}{3} + \frac{1}{12} \][/tex]
[tex]\[ C_2 = -\frac{16}{12} + \frac{1}{12} = -\frac{15}{12} = -1.25 \][/tex]
7. Particular Solution:
Thus, the particular solution of the differential equation is:
[tex]\[ y(x) = \frac{3}{4}x + (-0.75x - 2.0)e^{2x/3} + 2 \][/tex]
So, the particular solution of the differential equation [tex]\( 9 \frac{d^2 y}{d x^2} - 12 \frac{d y}{d x} + 4 y = 3 x - 1 \)[/tex] given the initial conditions is:
[tex]\[ y(x) = \frac{3}{4}x + (-0.75x - 2.0)e^{2x/3} + 2 \][/tex]
