Problem 39 - Advanced Engineering Mathematics

Question

In Problems 39-42, use the substitution \(y=\left(x-x_{0}\right)^{m}\) to solve the given equation. $$ (x+3)^{2} y^{\prime \prime}-8(x+3) y^{\prime}+14 y=0 $$

Step-by-Step Solution

Answer

The general solution is \( y = c_1(x + 3)^7 + c_2(x + 3)^2 \).

Step 1: Identify and set up substitution

We start by recognizing the substitution provided: let \( y = (x - x_0)^m \) where \( x_0 = -3 \). This means we substitute \( y = (x + 3)^m \).

Step 2: Calculate the derivatives

Now, we need to find the first and second derivatives of \( y \) with respect to \( x \). Using the substitution \( y = (x + 3)^m \), we find the derivatives:1. First derivative: \( y' = m(x + 3)^{m-1} \)2. Second derivative: \( y'' = m(m-1)(x + 3)^{m-2} \).

Step 3: Substitute derivatives into the differential equation

Substitute \( y \), \( y' \), and \( y'' \) into the original differential equation, \[ (x+3)^{2} y^{ ext{''}} - 8(x+3)y^{ ext{'}} + 14y = 0 \]Replacing each part, you get:\[(x + 3)^2 \cdot m(m-1)(x + 3)^{m-2} - 8(x + 3) \cdot m(x + 3)^{m-1} + 14(x + 3)^m = 0 \].

Step 4: Simplify the terms

Each term can be simplified as follows:1. First term: \((x + 3)^2 \cdot m(m-1)(x + 3)^{m-2} = m(m-1)(x + 3)^m\)2. Second term: \(- 8(x + 3) \cdot m(x + 3)^{m-1} = -8m(x + 3)^m\)3. Third term: \(14(x + 3)^m = 14(x + 3)^m\)Combining like terms results in:\[ m(m-1)(x + 3)^m - 8m(x + 3)^m + 14(x + 3)^m = 0 \].

Step 5: Factor and solve for m

Factor out \((x + 3)^m\):\[ (x + 3)^m [ m(m-1) - 8m + 14 ] = 0 \]The term in brackets must equal zero: \[ m^2 - m - 8m + 14 = 0 \]Simplify to\[ m^2 - 9m + 14 = 0 \].

Step 6: Solve the quadratic equation

The quadratic equation \( m^2 - 9m + 14 = 0 \) can be solved using the quadratic formula \( m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = -9, c = 14 \):\[ m = \frac{9 \pm \sqrt{81 - 56}}{2} \]\[ m = \frac{9 \pm \sqrt{25}}{2} \]\[ m = \frac{9 \pm 5}{2} \]This gives us solutions \( m = 7 \) and \( m = 2 \).

Step 7: Write the general solution

Using the values of \( m \), the general solution for the differential equation is \( y = c_1(x + 3)^7 + c_2(x + 3)^2 \), where \( c_1 \) and \( c_2 \) are constants.

Key Concepts

Quadratic Equation

Quadratic equations are fundamental in algebra and appear as a polynomial equation of the form \( ax^2 + bx + c = 0 \). The key feature of a quadratic equation is the "x squared" term. Let's see why it matters here.

In our solution, during Step 6, we encounter a quadratic equation: \( m^2 - 9m + 14 = 0 \). Quadratic equations are solved by finding values of "m" that satisfy this equation. In our exercise, this involves using a method called the quadratic formula.

The quadratic formula is a powerful tool: \( m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Where "a", "b", and "c" are the coefficients from the equation \( ax^2 + bx + c = 0 \).

• Here, \( a = 1 \), \( b = -9 \), and \( c = 14 \).
• Substitute these values into the formula to get \( m = \frac{9 \pm \sqrt{25}}{2} \).

This gives us the solutions \( m = 7 \) and \( m = 2 \), which are crucial for determining the general solution of the differential equation later on. Understanding how to solve quadratic equations is vital when tackling problems involving differential equations.

Substitution Method

In solving differential equations, substitution can simplify the process significantly. The substitution method works by substituting complex expressions with simpler terms. This approach can make solving the equation more manageable.

In our exercise, the substitution \( y = (x - x_0)^m \) helps us to deal with a complicated differential equation by reducing it to terms involving powers of \( (x + 3) \). By selecting \( x_0 = -3 \), we rewrite \( y \) as \( y = (x + 3)^m \).

• Key advantage: Simplifies the differential equation by replacing "x" expressions with "m" terms.
• Makes calculation of derivatives straightforward, as seen in Step 2: calculating \( y' \) and \( y'' \).

This substitution allows us to express the differential equation in terms of \( (x + 3)^m \), facilitating easier simplification and manipulation.

Derivatives

Derivatives are a central concept in calculus, representing the rate of change of a function. In the context of differential equations, they help us understand how a function evolves as it changes over one variable.

In this problem, we calculate derivatives based on our substitution \( y = (x + 3)^m \). The derivatives here are critical for inserting back into the original equation.

• First derivative: \( y' = m(x + 3)^{m-1} \). It represents the first level of change applied to our function.
• Second derivative: \( y'' = m(m-1)(x + 3)^{m-2} \). Derivatives of this order consider how the rate of change itself changes.

Substituting these derivatives into our differential equation allows us to manipulate and eventually solve it, contributing to finding the function's form \( y = c_1(x + 3)^7 + c_2(x + 3)^2 \). Understanding these foundational steps is crucial for advanced mathematical problem-solving.