Problem 29 - Calculus Volume 3

Question

Find the general solution to the linear differential equation. $$ \frac{d^{2} y}{d x^{2}}-9 \frac{d y}{d x}=0 $$

Step-by-Step Solution

Answer

The general solution is \( y(x) = C_1 + C_2 e^{9x} \).

Step 1: Write the differential equation in standard form

Start by writing the differential equation \[ \frac{d^{2} y}{d x^{2}} - 9 \frac{d y}{d x} = 0 \]in standard form. The differential equation is already in standard form, with the highest derivative first.

Step 2: Substitute y with exponential form

Assume a solution of the form \( y = e^{mx} \), where \( m \) is a constant to be determined. Substitute this into the differential equation.

Step 3: Calculate the derivatives

Calculate the first and second derivatives of \( y \):1. \( \frac{dy}{dx} = me^{mx} \) 2. \( \frac{d^2y}{dx^2} = m^2e^{mx} \).

Step 4: Substitute derivatives into the equation

Substitute the calculated derivatives into the differential equation:\[ m^2e^{mx} - 9me^{mx} = 0 \].

Step 5: Factor out the common terms

Factor \( e^{mx} \) out of the equation:\[ e^{mx} (m^2 - 9m) = 0 \].Since \( e^{mx} eq 0 \), we consider the equation \( m^2 - 9m = 0 \).

Step 6: Solve the characteristic equation

Solve \( m^2 - 9m = 0 \):1. Factor to get \( m(m - 9) = 0 \).2. Solve for \( m \), giving \( m = 0 \) and \( m = 9 \).

Step 7: Write the general solution

The general solution is a linear combination of the solutions corresponding to \( m = 0 \) and \( m = 9 \):\[ y(x) = C_1 e^{0 \, x} + C_2 e^{9 \, x} = C_1 + C_2 e^{9x} \],where \( C_1 \) and \( C_2 \) are arbitrary constants.

Key Concepts

Characteristic Equation

The characteristic equation is a crucial step in solving linear differential equations. It involves solving an algebraic equation derived from the original differential equation.
For equations like the one in our exercise—\( \frac{d^{2} y}{d x^{2}} - 9 \frac{d y}{d x} = 0 \)—we look for solutions of the form \( y = e^{mx} \), where \( m \) is a constant that we need to determine.
By substituting \( y = e^{mx} \) and its derivatives into the original equation, we form a new equation in terms of \( m \):

• \( \frac{dy}{dx} = me^{mx} \)
• \( \frac{d^2y}{dx^2} = m^2e^{mx} \)

Substituting these into the original differential equation and simplifying usually gives us a polynomial in \( m \)—the characteristic equation. In this problem, it simplifies to \( m^2 - 9m = 0 \). Solving this equation helps us find the values of \( m \) that lead to our general solution. These values are known as the roots of the characteristic equation.

General Solution

The general solution of a differential equation encompasses all possible solutions that satisfy the equation.
Once the characteristic equation is solved, we find roots that allow us to write the general solution. In our exercise, the characteristic equation \( m^2 - 9m = 0 \) has roots \( m = 0 \) and \( m = 9 \).
The solution to the differential equation is expressed as a linear combination of exponentials tied to these roots:

• For \( m = 0 \), the solution is \( e^{0 \cdot x} = 1 \), a constant function.
• For \( m = 9 \), it is \( e^{9x} \).

Combining these, we write the general solution as:\[y(x) = C_1 + C_2 e^{9x}\]Here, \( C_1 \) and \( C_2 \) are arbitrary constants that can be determined if initial conditions are given. They reflect the infinite variety of solutions that the linear differential equation can have.

Exponential Form Solution

Using an exponential form \( y = e^{mx} \) is a smart choice for solving linear differential equations.
This assumption greatly simplifies calculations and helps derive the characteristic equation. The beauty of exponentials lies in how they transform differentiation—each derivative of \( y = e^{mx} \) results in a simple multiplication:

• The first derivative is \( \frac{dy}{dx} = me^{mx} \).
• The second derivative is \( \frac{d^2y}{dx^2} = m^2e^{mx} \).

In our case, substituting these into \( \frac{d^{2} y}{d x^{2}} - 9 \frac{d y}{d x} = 0 \) yields the reduced form: \( e^{mx} (m^2 - 9m) = 0 \).
This nicely leverages the properties of exponentials to factor out \( e^{mx} \) and focus on solving the algebraic part, which is \( m^2 - 9m = 0 \).
Such manipulation underscores why the exponential approach is not just a convenient shortcut but a deeply integrated method in tackling differential equations.