Problem 35 - Calculus Volume 3

Question

Solve the initial-value problem. $$ y^{\prime \prime}-y^{\prime}-30 y=0, \quad y(0)=1, \quad y^{\prime}(0)=-16 $$

Step-by-Step Solution

Answer

The solution is \( y(t) = -e^{6t} + 2e^{-5t} \).

Step 1: Write the Characteristic Equation

To solve the homogeneous differential equation, write the characteristic equation. Use the coefficients from the differential equation.\[ r^2 - r - 30 = 0 \]

Step 2: Solve the Characteristic Equation

Factor the quadratic equation to find the roots.\[ r^2 - r - 30 = (r - 6)(r + 5) = 0 \] The roots are \( r = 6 \) and \( r = -5 \).

Step 3: Write the General Solution

Using the roots from the characteristic equation, write the general solution for the differential equation.\[ y(t) = C_1 e^{6t} + C_2 e^{-5t} \]

Step 4: Solve for Constants Using Initial Conditions

Apply the initial conditions to find \( C_1 \) and \( C_2 \). Start with \( y(0) = 1 \).\[ 1 = C_1 e^{0} + C_2 e^{0} = C_1 + C_2 \]Thus, \( C_1 + C_2 = 1 \).

Step 5: Use the First Derivative

Differentiate the general solution to apply the second initial condition.\[ y'(t) = 6C_1 e^{6t} - 5C_2 e^{-5t} \]Apply \( y'(0) = -16 \).\[ -16 = 6C_1 - 5C_2 \]

Step 6: Solve the System of Equations

Use the equations from Steps 4 and 5:1. \( C_1 + C_2 = 1 \)2. \( 6C_1 - 5C_2 = -16 \)Solve this system using substitution or elimination. From equation 1, \( C_2 = 1 - C_1 \). Substitute into equation 2:\[ 6C_1 - 5(1 - C_1) = -16 \]\[ 6C_1 - 5 + 5C_1 = -16 \]\[ 11C_1 = -11 \]\[ C_1 = -1 \]Now find \( C_2 \):\( C_2 = 1 - (-1) = 2 \).

Step 7: Write the Particular Solution

Substitute \( C_1 \) and \( C_2 \) back into the general solution:\[ y(t) = -e^{6t} + 2e^{-5t} \]

Key Concepts

Characteristic Equation

When dealing with a linear differential equation, a powerful technique to find solutions is through the characteristic equation. This equation is formed using the coefficients of the differential equation. For instance, consider our differential equation: \[ y'' - y' - 30y = 0 \] Here, the characteristic equation is derived by replacing the derivatives with powers of a variable, typically \( r \), leading to: \[ r^2 - r - 30 = 0 \] This characteristic equation is essentially a quadratic equation. Solving it gives the roots, which play a crucial role in forming the general solution of the differential equation.

• The characteristic equation transforms a differential problem into an algebraic one.
• The roots of the characteristic polynomial indicate the nature of the general solution.

By factoring the quadratic equation or using other methods like the quadratic formula, you can determine these roots, guiding you to the solution's form.

Initial Value Problem

An initial value problem is a differential equation that comes with additional conditions, called initial conditions. These specify the value of the function and perhaps its derivatives at a particular point. In our case, the initial value problem is given by:

• The differential equation: \( y'' - y' - 30y = 0 \)
• Initial conditions: \( y(0) = 1 \) and \( y'(0) = -16 \)

Initial conditions are crucial as they help identify a unique solution from the family of solutions represented by the general solution. Without these conditions, the solution contains undetermined constants. Using these initial conditions, we input values into the general solution and its derivative. Doing this helps us solve for any unknown constants. This process ensures that the solution fits precisely to the given initial constraints.

General Solution

The general solution of a differential equation represents all possible solutions of the equation. It's expressed in terms of arbitrary constants. After solving the characteristic equation, the resulting roots are used to construct this solution. For our problem, the characteristic roots were \( r = 6 \) and \( r = -5 \). Thus, the general solution is: \[ y(t) = C_1 e^{6t} + C_2 e^{-5t} \] Here, \( C_1 \) and \( C_2 \) are constants that need to be determined.

• Each root of the characteristic equation corresponds to an exponential term in the solution.
• The number of undetermined constants, \( C_1, C_2, \ldots \), equals the order of the differential equation.

Ultimately, the purpose of forming a general solution is to cover every possibility. You apply the initial conditions to determine specific values for these constants, transforming the general solution into the particular solution that uniquely fits the specified initial values.