Problem 42 - Calculus Volume 3

Question

Solve the boundary-value problem, if possible. $$ y^{\prime \prime}+7 y^{\prime}-60 y=0, \quad y(0)=4, \quad y(2)=0 $$

Step-by-Step Solution

Answer

The boundary-value problem solution is \( y(x) = \frac{4e^{-24}e^{5x}}{e^{10} - e^{-24}} - \frac{4e^{10}e^{-12x}}{e^{10} - e^{-24}} \).

Step 1: Identify the Differential Equation Type

The given differential equation is \( y'' + 7y' - 60y = 0 \). This is a second-order linear homogeneous differential equation with constant coefficients.

Step 2: Write the Characteristic Equation

For a second-order linear homogeneous differential equation with constant coefficients, we start by writing the characteristic equation. This reduces to solving for \( r \) in the equation: \( r^2 + 7r - 60 = 0 \).

Step 3: Solve the Characteristic Equation

Solve the quadratic characteristic equation \( r^2 + 7r - 60 = 0 \) using the quadratic formula: \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1 \), \( b = 7 \), and \( c = -60 \).

Step 4: Simplify to Find the Roots

Calculate the discriminant: \( b^2 - 4ac = 49 + 240 = 289 \). Therefore, the roots are \( r = \frac{-7 \pm \sqrt{289}}{2} \), giving \( r = \frac{-7 \pm 17}{2} \). Thus, the roots are \( r_1 = 5 \) and \( r_2 = -12 \).

Step 5: Write the General Solution

Since the equation has two distinct real roots, the general solution is given by \( y(x) = C_1e^{5x} + C_2e^{-12x} \), where \( C_1 \) and \( C_2 \) are constants to be determined.

Step 6: Apply the Initial Condition \( y(0) = 4 \)

Substitute \( y(0) = 4 \) into the general solution: \( 4 = C_1e^{0} + C_2e^{0} = C_1 + C_2 \). Thus, \( C_1 + C_2 = 4 \).

Step 7: Apply the Boundary Condition \( y(2) = 0 \)

Substitute \( y(2) = 0 \) into the general solution: \( 0 = C_1e^{10} + C_2e^{-24} \). This provides a second equation to solve for \( C_1 \) and \( C_2 \).

Step 8: Solve the System of Equations

We have the system of equations: \( C_1 + C_2 = 4 \) and \( C_1e^{10} + C_2e^{-24} = 0 \). Using substitution or elimination, solve the system to find \( C_1 =\frac{4e^{-24}}{e^{10} - e^{-24}} \) and \( C_2 =\frac{-4e^{10}}{e^{10} - e^{-24}} \).

Step 9: Final Solution

Substituting \( C_1 \) and \( C_2 \) back into the general solution provides the final function: \( y(x) = \frac{4e^{-24}e^{5x}}{e^{10} - e^{-24}} + \frac{-4e^{10}e^{-12x}}{e^{10} - e^{-24}} \). This function satisfies the given boundary conditions.

Key Concepts

Second-Order Differential Equation

A second-order differential equation is an equation involving the second derivative of a function. In many physical systems, second-order differential equations appear when expressing the rate of change of a rate of change, like acceleration in physics. For our particular problem, the equation is \( y'' + 7y' - 60y = 0 \). Here, \( y'' \) is the second derivative of \( y \), reflecting how the acceleration of a system changes concerning \( y \). The coefficients in the equation, such as 7 and -60, play a role in determining the behavior and dynamics of the solution.
The equation is linear, meaning that \( y \), \( y' \), and \( y'' \) are combined linearly (no powers or products between them). The equation is called homogeneous because the right side is zero. Homogeneous equations have solutions that involve exponential functions, as we will discuss later.
Learning to identify and solve these equations is essential as they model many natural phenomena, from oscillations to population dynamics.

Characteristic Equation

To solve a second-order linear homogeneous differential equation, we rely heavily on its characteristic equation. This is an algebraic equation obtained by replacing derivatives in the original differential equation with powers of \( r \). For our equation, the characteristic equation is \( r^2 + 7r - 60 = 0 \).

Discriminant and Roots

The roots of this quadratic equation tell us about the behavior of the differential equation's solutions. We find these roots using the quadratic formula:

• \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
• For our example: \( a = 1, b = 7, c = -60 \)

Calculating the discriminant, \( b^2 - 4ac \), helps determine the type of roots (real or complex). In our case, it is 289, a perfect square, giving two distinct real roots, 5 and -12.
The nature of these roots (real, distinct, repeated, or complex) influences the form of the solution we construct for the differential equation.

Homogeneous Differential Equation

A homogeneous differential equation has all terms involving the unknown function or its derivatives, set equal to zero, as in \( y'' + 7y' - 60y = 0 \). These equations inherently imply that the solutions decay or grow exponentially, commonly involving terms like \( e^{rx} \).

General Solution

Since we have two distinct real roots from the characteristic equation, the general solution takes the form:

• \( y(x) = C_1e^{r_1x} + C_2e^{r_2x} \)
• For our example: \( y(x) = C_1e^{5x} + C_2e^{-12x} \)

Here, \( C_1 \) and \( C_2 \) are constants. The exponential terms reflect the effect of each root on the solution. The entire approach benefits from the Zero Function: if you insert the solution into the equation, it still holds true because the original equation equals zero.

Initial Conditions

Initial conditions help pinpoint a unique solution from the family of possible solutions derived from the general solution of a differential equation. These conditions specify the value of the function at certain points, aiding in the determination of the constants involved.
For our problem, we have:

• \( y(0) = 4 \)
• \( y(2) = 0 \)

These boundary conditions allow us to formulate a system of equations concerning \( C_1 \) and \( C_2 \). By substituting the conditions into the general solution, we can solve for the constants. For example, \( y(0) = 4 \) gives us \( C_1 + C_2 = 4 \). Similarly, substituting \( y(2) = 0 \) helps derive another equation to solve for these constants.
Once solved, these conditions ensure that our solution matches the specific behavior outlined by the problem at defined points, thus providing a precise intervention of the general solution to address specific scenarios.