Problem 16 - Advanced Engineering Mathematics

Question

In Problems 1-18, solve the given differential equation. $$ x^{3} y^{\prime \prime \prime}+x y^{\prime}-y=0 $$

Step-by-Step Solution

Answer

The general solution is \( y(x) = C_1 x + x^{\frac{1}{2}} (C_2 \cos(\frac{\sqrt{3}}{2} \ln x) + C_3 \sin(\frac{\sqrt{3}}{2} \ln x)) \).

Step 1: Identify the type of differential equation

The given differential equation is \( x^3 y''' + x y' - y = 0 \), a third-order linear homogeneous differential equation with variable coefficients.

Step 2: Analyze for simplification techniques

Since the equation involves terms with coefficients that are powers of \( x \), consider using methods applicable to equations with variable coefficients, such as assuming a solution of the form \( y = x^m \) and seeking a power solution.

Step 3: Assume a power series solution

Let \( y = x^m \) where \( m \) is a constant. Then the derivatives are \( y' = m x^{m-1} \) and \( y''' = m(m-1)(m-2)x^{m-3} \). Substitute these into the equation.

Step 4: Substitute the assumed solution into the original equation

By substituting the derivatives, we get: \( x^3[m(m-1)(m-2)x^{m-3}] + x[m x^{m-1}] - x^m = 0 \). Simplifying results in: \( m(m-1)(m-2)x^m + m x^m - x^m = 0 \).

Step 5: Simplify and solve the characteristic polynomial

Combine the like terms: \( [m(m-1)(m-2) + m - 1]x^m = 0 \). For this to hold for any \( x eq 0 \), the expression inside the bracket must be zero: \( m(m-1)(m-2) + m - 1 = 0 \).

Step 6: Solve the characteristic equation for \( m \)

Solve the characteristic polynomial: \( m^3 - 3m^2 + 2m + m - 1 = 0 \). Combine the terms: \( m^3 - 2m^2 + m - 1 = 0 \). First, factor the polynomial to find roots for \( m \).

Step 7: Find roots of the polynomial equation

Using synthetic division or the rational root theorem, determine possible roots for the polynomial \( m^3 - 2m^2 + m - 1 = 0 \). After checking, one finds that \( m = 1 \) is a root. Factor out \( m-1 \) to solve the remaining quadratic: \( m^2 - m + 1 = 0 \). Solve using the quadratic formula if no real roots are apparent.

Step 8: Solve remaining quadratic equation

Upon solving \( m^2 - m + 1 = 0 \) using the quadratic formula, \( m_{2,3} = \frac{1 \pm i \sqrt{3}}{2} \), revealing complex roots.

Step 9: Write the general solution using roots

Using the roots found, write the general solution for the differential equation: \( y(x) = C_1 x + C_2 x^{rac{1}{2}} e^{i \frac{\sqrt{3}}{2} \ln x} + C_3 x^{\frac{1}{2}} e^{-i \frac{\sqrt{3}}{2} \ln x} \). Use Euler's formula to rewrite the complex exponentials in terms of sines and cosines.

Step 10: Final form of the general solution

Applying Euler's formula, \( e^{i\theta} = \cos(\theta) + i\sin(\theta) \), the terms simplify to real functions. The complete solution is \( y(x) = C_1 x + x^{\frac{1}{2}}(C_2 \cos(\frac{\sqrt{3}}{2} \ln x) + C_3 \sin(\frac{\sqrt{3}}{2} \ln x)) \).

Key Concepts

Third-Order Differential Equation

A third-order differential equation is an equation involving a function and its derivatives up to the third order. In the given problem, we deal with:

• Function: \( y \)
• First derivative: \( y' \)
• Third derivative: \( y''' \)

These derivatives indicate the degree of the equation, which is crucial for understanding the complexity and the potential solutions needed. This specific problem is a **linear homogeneous** equation, meaning that all the terms depend linearly on \( y \) and its derivatives. This type of equation can be tricky, but knowing it's linear implies there could be systematic methods to find solutions.

Variable Coefficients

Differential equations can have either constant or variable coefficients. In our problem, the term \( x^3 \) in front of the third derivative \( y''' \) and \( x \) in front of the first derivative \( y' \) indicate that we're dealing with a differential equation with variable coefficients. These coefficients depend on the independent variable \( x \).
Dealing with variable coefficients often requires different techniques compared to constant coefficient equations. For this exercise, identifying a solution of the form \( y = x^m \) helps handle the variable coefficients effectively by transforming the problem into a power series form, which further simplifies into finding a characteristic equation.

Characteristic Equation

The characteristic equation is a crucial part of solving linear differential equations, especially those with constant coefficients. However, in this case, since we have variable coefficients, the characteristic equation arises from substituting a trial solution into the original differential equation.

• For \( y = x^m \), the derivatives are substituted back into the equation.
• Combining like terms leads to the development of a characteristic polynomial equation in terms of \( m \).

The characteristic polynomial here is \( m^3 - 2m^2 + m - 1 = 0 \), and by solving it, we can determine the roots of the equation, which are necessary to write the general solution of the differential equation. Finding these roots involves algebraic techniques like synthetic division or the rational root theorem.

Complex Roots

Complex roots occur when solving polynomial equations that do not intersect the real number line completely, implying the presence of imaginary components. After determining that \( m = 1 \) is a real root for the characteristic polynomial, the remaining expression \( m^2 - m + 1 = 0 \) is solved using the quadratic formula.
Solving this equation gives the complex roots \( m_{2,3} = \frac{1 \pm i \sqrt{3}}{2} \). These roots indicate the oscillatory behavior of the solutions.

• Use Euler’s formula: \( e^{i\theta} = \cos(\theta) + i\sin(\theta) \) to transform the complex exponential solutions into a real domain.
• This transformation helps express the general solution in terms of sine and cosine functions.

Understanding complex roots and how they manifest in solutions via trigonometric functions is key to solving differential equations with oscillatory behavior.