Solve 7d2ydx2−126dydx+567y=0, y(0)=3, y′(0)=27d2ydx2−126dydx+567y=0, y(0)=3, y′(0)=2 where ′′ denotes differentiation with respect to x

The differential equation to solve is:

\[
7 \frac{d^2y}{dx^2} - 126 \frac{dy}{dx} + 567y = 0
\]

This is a second-order linear homogeneous differential equation with constant coefficients. The general approach to solving such equations involves the following steps:

### Step 1: Form the characteristic equation
Assume a solution of the form \( y(x) = e^{rx} \), where \( r \) is a constant. Substituting this into the differential equation gives:

\[
7r^2 e^{rx} - 126r e^{rx} + 567e^{rx} = 0
\]

Dividing through by \( e^{rx} \) (which is never zero), we obtain the characteristic equation:

\[
7r^2 - 126r + 567 = 0
\]

### Step 2: Solve the characteristic equation
The characteristic equation is a quadratic equation:

\[
7r^2 - 126r + 567 = 0
\]

Dividing by 7 to simplify:

\[
r^2 - 18r + 81 = 0
\]

This can be factored as:

\[
(r - 9)^2 = 0
\]

Thus, \( r = 9 \) is a repeated root.

### Step 3: Write the general solution
Since we have a repeated root \( r = 9 \), the general solution of the differential equation is:

\[
y(x) = (C_1 + C_2x)e^{9x}
\]

where \( C_1 \) and \( C_2 \) are constants to be determined by the initial conditions.

### Step 4: Apply the initial conditions
The initial conditions are given as \( y(0) = 3 \) and \( y'(0) = 2 \).

1. **Apply \( y(0) = 3 \):**

\[
y(0) = (C_1 + C_2 \cdot 0)e^{0} = C_1 = 3
\]

So, \( C_1 = 3 \).

2. **Apply \( y'(0) = 2 \):**

First, compute the derivative of \( y(x) \):

\[
y'(x) = \frac{d}{dx} \left[(C_1 + C_2x)e^{9x}\right]
\]

Using the product rule:

\[
y'(x) = C_2 e^{9x} + (C_1 + C_2x) \cdot 9e^{9x}
\]

At \( x = 0 \):

\[
y'(0) = C_2 \cdot e^{0} + (C_1 + C_2 \cdot 0) \cdot 9e^{0} = C_2 + 9C_1
\]

Substitute the values \( y'(0) = 2 \) and \( C_1 = 3 \):

\[
2 = C_2 + 9 \times 3
\]

\[
2 = C_2 + 27
\]

\[
C_2 = 2 - 27 = -25
\]

### Final Solution
Substitute the values of \( C_1 \) and \( C_2 \) back into the general solution:

\[
y(x) = (3 - 25x)e^{9x}
\]

This is the solution to the differential equation with the given initial conditions.

Would you like further details or have any questions?

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Here are five related questions to explore further:

1. What are the general forms of solutions for different types of characteristic equations (real, complex, repeated roots)?
2. How does the solution change if we have complex roots instead of real roots?
3. What is the significance of the initial conditions in determining the constants \( C_1 \) and \( C_2 \)?
4. Can this method be extended to non-homogeneous differential equations?
5. How do the solutions of second-order differential equations behave asymptotically for large \( x \)?

**Tip:** When solving second-order differential equations, always check for the type of roots (real, repeated, complex) to determine the appropriate form of the solution.

Learn how to solve the second-order linear differential equation 7d^2y/dx^2 - 126dy/dx + 567y = 0 with detailed step-by-step solutions. This problem covers differential equations, characteristic equations, and initial value problems. Suitable for advanced college-level students.

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