Answer :
To solve the differential equation
[tex]\[ 3 \frac{d^2 y}{d x^2} + \frac{d y}{d x} - 4 y = 8, \][/tex]
with initial conditions
[tex]\[ y(0) = 0 \quad \text{and} \quad \frac{d y}{d x}\Big|_{x = 0} = 0, \][/tex]
we'll proceed with the following steps:
### Step 1: Find the general solution of the homogeneous equation
Consider the homogeneous version of the given differential equation:
[tex]\[ 3 \frac{d^2 y}{d x^2} + \frac{d y}{d x} - 4 y = 0. \][/tex]
To solve this, we assume a solution of the form [tex]\( y = e^{mx} \)[/tex]. Substituting [tex]\( y = e^{mx} \)[/tex] into the homogeneous equation, we get:
[tex]\[ 3m^2 e^{mx} + m e^{mx} - 4e^{mx} = 0. \][/tex]
Dividing through by [tex]\( e^{mx} \)[/tex] (which is never zero), we obtain:
[tex]\[ 3m^2 + m - 4 = 0. \][/tex]
This is a quadratic equation in terms of [tex]\( m \)[/tex]. To solve for [tex]\( m \)[/tex], we use the quadratic formula [tex]\( m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 3 \)[/tex], [tex]\( b = 1 \)[/tex], and [tex]\( c = -4 \)[/tex]:
[tex]\[ m = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 3 \cdot (-4)}}{2 \cdot 3} = \frac{-1 \pm \sqrt{1 + 48}}{6} = \frac{-1 \pm \sqrt{49}}{6} = \frac{-1 \pm 7}{6}. \][/tex]
Thus, the roots are:
[tex]\[ m = \frac{6}{6} = 1 \quad \text{and} \quad m = \frac{-8}{6} = -\frac{4}{3}. \][/tex]
The general solution to the homogeneous equation is then:
[tex]\[ y_h(x) = C_1 e^{x} + C_2 e^{-4x/3}. \][/tex]
### Step 2: Find a particular solution to the inhomogeneous equation
To find a particular solution [tex]\( y_p(x) \)[/tex] of the inhomogeneous equation
[tex]\[ 3 \frac{d^2 y}{d x^2} + \frac{d y}{d x} - 4 y = 8, \][/tex]
we can try a constant solution [tex]\( y_p = k \)[/tex]. Substituting [tex]\( y_p = k \)[/tex] into the differential equation, we get:
[tex]\[ 3 \cdot 0 + 0 - 4k = 8, \][/tex]
which simplifies to:
[tex]\[ -4k = 8 \quad \Rightarrow \quad k = -2. \][/tex]
Thus, the particular solution is:
[tex]\[ y_p(x) = -2. \][/tex]
### Step 3: Form the general solution of the full differential equation
The general solution of the full differential equation is then the sum of the homogeneous solution and the particular solution:
[tex]\[ y(x) = C_1 e^{x} + C_2 e^{-4x/3} - 2. \][/tex]
### Step 4: Apply the initial conditions to find [tex]\( C_1 \)[/tex] and [tex]\( C_2 \)[/tex]
We use the initial conditions [tex]\( y(0) = 0 \)[/tex] and [tex]\( y'(0) = 0 \)[/tex].
1. Applying [tex]\( y(0) = 0 \)[/tex]:
[tex]\[ 0 = C_1 e^{0} + C_2 e^{0} - 2 = C_1 + C_2 - 2. \][/tex]
[tex]\[ C_1 + C_2 = 2. \][/tex]
2. Applying [tex]\( y'(0) = 0 \)[/tex]:
First, find [tex]\( y'(x) \)[/tex]:
[tex]\[ y'(x) = C_1 e^{x} - \frac{4}{3} C_2 e^{-4x/3}. \][/tex]
Then:
[tex]\[ 0 = C_1 e^{0} - \frac{4}{3} C_2 e^{0} = C_1 - \frac{4}{3} C_2. \][/tex]
Solving these two equations simultaneously:
[tex]\[ C_1 + C_2 = 2, \][/tex]
[tex]\[ C_1 - \frac{4}{3} C_2 = 0. \][/tex]
From the second equation:
[tex]\[ C_1 = \frac{4}{3} C_2. \][/tex]
Substitute this into the first equation:
[tex]\[ \frac{4}{3} C_2 + C_2 = 2, \][/tex]
[tex]\[ \frac{7}{3} C_2 = 2, \][/tex]
[tex]\[ C_2 = \frac{6}{7}. \][/tex]
Then:
[tex]\[ C_1 = \frac{4}{3} \cdot \frac{6}{7} = \frac{8}{7}. \][/tex]
### Step 5: Write the particular solution
Substituting [tex]\( C_1 \)[/tex] and [tex]\( C_2 \)[/tex] back into the general solution, we obtain the particular solution:
[tex]\[ y(x) = \frac{8}{7} e^{x} + \frac{6}{7} e^{-4x/3} - 2. \][/tex]
So, the particular solution to the differential equation with the given initial conditions is:
[tex]\[ y(x) = \frac{8}{7} e^{x} + \frac{6}{7} e^{-4x/3} - 2. \][/tex]
[tex]\[ 3 \frac{d^2 y}{d x^2} + \frac{d y}{d x} - 4 y = 8, \][/tex]
with initial conditions
[tex]\[ y(0) = 0 \quad \text{and} \quad \frac{d y}{d x}\Big|_{x = 0} = 0, \][/tex]
we'll proceed with the following steps:
### Step 1: Find the general solution of the homogeneous equation
Consider the homogeneous version of the given differential equation:
[tex]\[ 3 \frac{d^2 y}{d x^2} + \frac{d y}{d x} - 4 y = 0. \][/tex]
To solve this, we assume a solution of the form [tex]\( y = e^{mx} \)[/tex]. Substituting [tex]\( y = e^{mx} \)[/tex] into the homogeneous equation, we get:
[tex]\[ 3m^2 e^{mx} + m e^{mx} - 4e^{mx} = 0. \][/tex]
Dividing through by [tex]\( e^{mx} \)[/tex] (which is never zero), we obtain:
[tex]\[ 3m^2 + m - 4 = 0. \][/tex]
This is a quadratic equation in terms of [tex]\( m \)[/tex]. To solve for [tex]\( m \)[/tex], we use the quadratic formula [tex]\( m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 3 \)[/tex], [tex]\( b = 1 \)[/tex], and [tex]\( c = -4 \)[/tex]:
[tex]\[ m = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 3 \cdot (-4)}}{2 \cdot 3} = \frac{-1 \pm \sqrt{1 + 48}}{6} = \frac{-1 \pm \sqrt{49}}{6} = \frac{-1 \pm 7}{6}. \][/tex]
Thus, the roots are:
[tex]\[ m = \frac{6}{6} = 1 \quad \text{and} \quad m = \frac{-8}{6} = -\frac{4}{3}. \][/tex]
The general solution to the homogeneous equation is then:
[tex]\[ y_h(x) = C_1 e^{x} + C_2 e^{-4x/3}. \][/tex]
### Step 2: Find a particular solution to the inhomogeneous equation
To find a particular solution [tex]\( y_p(x) \)[/tex] of the inhomogeneous equation
[tex]\[ 3 \frac{d^2 y}{d x^2} + \frac{d y}{d x} - 4 y = 8, \][/tex]
we can try a constant solution [tex]\( y_p = k \)[/tex]. Substituting [tex]\( y_p = k \)[/tex] into the differential equation, we get:
[tex]\[ 3 \cdot 0 + 0 - 4k = 8, \][/tex]
which simplifies to:
[tex]\[ -4k = 8 \quad \Rightarrow \quad k = -2. \][/tex]
Thus, the particular solution is:
[tex]\[ y_p(x) = -2. \][/tex]
### Step 3: Form the general solution of the full differential equation
The general solution of the full differential equation is then the sum of the homogeneous solution and the particular solution:
[tex]\[ y(x) = C_1 e^{x} + C_2 e^{-4x/3} - 2. \][/tex]
### Step 4: Apply the initial conditions to find [tex]\( C_1 \)[/tex] and [tex]\( C_2 \)[/tex]
We use the initial conditions [tex]\( y(0) = 0 \)[/tex] and [tex]\( y'(0) = 0 \)[/tex].
1. Applying [tex]\( y(0) = 0 \)[/tex]:
[tex]\[ 0 = C_1 e^{0} + C_2 e^{0} - 2 = C_1 + C_2 - 2. \][/tex]
[tex]\[ C_1 + C_2 = 2. \][/tex]
2. Applying [tex]\( y'(0) = 0 \)[/tex]:
First, find [tex]\( y'(x) \)[/tex]:
[tex]\[ y'(x) = C_1 e^{x} - \frac{4}{3} C_2 e^{-4x/3}. \][/tex]
Then:
[tex]\[ 0 = C_1 e^{0} - \frac{4}{3} C_2 e^{0} = C_1 - \frac{4}{3} C_2. \][/tex]
Solving these two equations simultaneously:
[tex]\[ C_1 + C_2 = 2, \][/tex]
[tex]\[ C_1 - \frac{4}{3} C_2 = 0. \][/tex]
From the second equation:
[tex]\[ C_1 = \frac{4}{3} C_2. \][/tex]
Substitute this into the first equation:
[tex]\[ \frac{4}{3} C_2 + C_2 = 2, \][/tex]
[tex]\[ \frac{7}{3} C_2 = 2, \][/tex]
[tex]\[ C_2 = \frac{6}{7}. \][/tex]
Then:
[tex]\[ C_1 = \frac{4}{3} \cdot \frac{6}{7} = \frac{8}{7}. \][/tex]
### Step 5: Write the particular solution
Substituting [tex]\( C_1 \)[/tex] and [tex]\( C_2 \)[/tex] back into the general solution, we obtain the particular solution:
[tex]\[ y(x) = \frac{8}{7} e^{x} + \frac{6}{7} e^{-4x/3} - 2. \][/tex]
So, the particular solution to the differential equation with the given initial conditions is:
[tex]\[ y(x) = \frac{8}{7} e^{x} + \frac{6}{7} e^{-4x/3} - 2. \][/tex]
