Problem 39 - Calculus Volume 3

Question

Solve the boundary-value problem, if possible. $$ y^{\prime \prime}+y^{\prime}-42 y=0, \quad y(0)=0, \quad y(1)=2 $$

Step-by-Step Solution

Answer

The particular solution is \(y(x) = \frac{2}{e^6 - e^{-7}}e^{6x} - \frac{2}{e^6 - e^{-7}}e^{-7x}\).

Step 1: Identify the Differential Equation Type

We have a second-order linear homogeneous differential equation: \[y'' + y' - 42y = 0\] with boundary conditions \( y(0) = 0 \) and \( y(1) = 2 \).

Step 2: Write the Characteristic Equation

For a second-order linear differential equation with constant coefficients like \(y'' + y' - 42y = 0\), the characteristic equation is obtained by replacing \(y''\) with \(r^2\), \(y'\) with \(r\), and \(y\) with \(1\).Thus, the characteristic equation is:\[r^2 + r - 42 = 0\]

Step 3: Solve the Characteristic Equation

Solve the quadratic equation \(r^2 + r - 42 = 0\) using the quadratic formula:\[r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Substitute \(a = 1\), \(b = 1\), \(c = -42\):\[r = \frac{-1 \pm \sqrt{1^2 - 4 \times 1 \times (-42)}}{2 \times 1}\]\[r = \frac{-1 \pm \sqrt{1 + 168}}{2}\]\[r = \frac{-1 \pm \sqrt{169}}{2}\]\[r = \frac{-1 \pm 13}{2}\]Thus, the solutions are \(r_1 = 6\) and \(r_2 = -7\).

Step 4: Construct the General Solution

The general solution of the differential equation is derived from the roots of the characteristic equation:\[y(x) = C_1e^{6x} + C_2e^{-7x}\] where \(C_1\) and \(C_2\) are constants to be determined by the boundary conditions.

Step 5: Apply Boundary Condition y(0) = 0

Substitute \(x = 0\) into the general solution:\[0 = C_1e^{0} + C_2e^{0}\]\[0 = C_1 + C_2\]This gives us the equation: \(C_1 + C_2 = 0\). So, \(C_2 = -C_1\).

Step 6: Apply Boundary Condition y(1) = 2

Substitute \(x = 1\) into the general solution:\[2 = C_1e^{6} + C_2e^{-7}\]Using \(C_2 = -C_1\), substitute in:\[2 = C_1(e^6 - e^{-7})\]Solve for \(C_1\):\[C_1 = \frac{2}{e^6 - e^{-7}}\]

Step 7: Solve for C_2

Since \(C_2 = -C_1\), we have:\[C_2 = -\frac{2}{e^6 - e^{-7}}\]

Step 8: Write the Particular Solution

Substitute \(C_1\) and \(C_2\) back into the general solution:\[y(x) = \frac{2}{e^6 - e^{-7}}e^{6x} - \frac{2}{e^6 - e^{-7}}e^{-7x}\]

Key Concepts

Second-order differential equation

A second-order differential equation is a type of differential equation that involves derivatives of a function up to the second order. These equations play a crucial role in mathematical modeling of physical systems because they can describe phenomena such as motion, heat, and waves.

In a second-order equation, the highest derivative is the second derivative, commonly denoted as \(y''\) or \(f''(x)\). The general form is typically:\[ay'' + by' + cy = 0\] where \(a\), \(b\), and \(c\) are constants, and \(y\) is the function of a variable, usually \(x\).

Identifying the order of the differential equation is the first step in solving it and can tell us about the nature of the problem and suggest an appropriate solution method.

Characteristic equation

The characteristic equation is a crucial step in solving linear differential equations with constant coefficients. For a second-order differential equation, the characteristic equation is derived by replacing \(y''\), \(y'\), and \(y\) with polynomials \(r^2\), \(r\), and \(1\), respectively.

For instance, given a differential equation \(y'' + y' - 42y = 0\), the characteristic equation is formed as follows:\[r^2 + r - 42 = 0\]

• The characteristic equation is a quadratic equation.
• Finding its roots will help find the solution to the differential equation.

The roots obtained from solving the characteristic equation are key to constructing the general solution of the differential equation.

Homogeneous differential equation

A homogeneous differential equation is a type where all terms are dependent on the unknown function and its derivatives. This means every term in the equation either contains the dependent variable \(y\) or its derivatives (no standalone constants or terms independent of \(y\)).

The form of a linear homogeneous equation is typically non-zero if \(y\) itself or its derivatives are non-zero. For instance, in our given equation \(y'' + y' - 42y = 0\), the equation is homogeneous as each term is a multiple of \(y\) and its derivatives.

• These equations assume that any combination of solutions that satisfies them does so linearly.

Homogeneous equations are easier to solve as they often have neat computational properties.

Quadratic formula

The quadratic formula is a powerful tool for finding the roots of a quadratic equation of the form \(ax^2 + bx + c = 0\). It is given by:\[r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

This formula is applied when you derive a characteristic equation from a second-order differential equation. For our example, \(r^2 + r - 42 = 0\), applying the quadratic formula involves substituting \(a = 1\), \(b = 1\), and \(c = -42\) into the formula to find the roots.

• The roots, \(r_1 = 6\) and \(r_2 = -7\), are found efficiently and determine the exponential solution form.

The quadratic formula saves time and ensures accuracy when dealing with these types of problems.

Exponential function solution

When solving differential equations, especially with constant coefficients, exponential functions often form the basis of the solution. This is due in part to the nature of derivatives of exponential functions, which preserve their form.

For a second-order equation like \(y'' + y' - 42y = 0\), the general solution after finding roots \(r_1\) and \(r_2\) is expressed as:\[y(x) = C_1e^{r_1x} + C_2e^{r_2x}\]

• Here, \(C_1\) and \(C_2\) are constants determined by boundary conditions.

The exponential solutions reflect how solutions change and behave over time or space, representing decay, growth, or oscillation based on the sign and magnitude of the roots.