Problem 5 - Advanced Engineering Mathematics

Question

In Problems 1-18, solve the given differential equation. $$ x^{2} y^{\prime \prime}+x y^{\prime}+4 y=0 $$

Step-by-Step Solution

Answer

The general solution is \(y(x) = x^{-\frac{3}{2}}(C_{1} \cos(\frac{\sqrt{7}}{2} \ln x) + C_{2} \sin(\frac{\sqrt{7}}{2} \ln x))\).

Step 1: Identify the Equation Type

The given differential equation is \(x^{2} y^{\prime \prime} + x y^{\prime} + 4y = 0\), indicating it's a second-order linear differential equation with variable coefficients. This suggests using methods suitable for Cauchy-Euler equations.

Step 2: Assume a Power Solution

For Cauchy-Euler equations, we assume a solution of the form \(y = x^{m}\). Calculate the derivatives: \(y' = mx^{m-1}\) and \(y'' = m(m-1)x^{m-2}\).

Step 3: Substitute into the Differential Equation

Substitute \(y = x^{m}\), \(y' = mx^{m-1}\), and \(y'' = m(m-1)x^{m-2}\) into the original equation: \[x^{2}(m(m-1)x^{m-2}) + x(mx^{m-1}) + 4x^{m} = 0\].

Step 4: Simplify the Substitution

Simplify the expression:- \(x^{2} m(m-1) x^{m-2} = m(m-1)x^{m}\)- \(x mx^{m-1} = mx^{m}\)Thus, the equation becomes: \[ m(m-1)x^{m} + mx^{m} + 4x^{m} = 0\].

Step 5: Factor and Solve for the Roots

Factor out \(x^{m}\): \[ x^{m}(m(m-1) + m + 4) = 0\].The equation inside the parentheses becomes \[ m^{2} + 3m + 4 = 0\].Solve this quadratic equation for \(m\).

Step 6: Use the Quadratic Formula

The quadratic formula is \(m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Plugging in coefficients: \(a = 1\), \(b = 3\), \(c = 4\), we get:\[ m = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot 4}}{2 \cdot 1}\].

Step 7: Calculate the Discriminant and Roots

Calculate the discriminant: \( b^{2} - 4ac = 9 - 16 = -7 \), indicating complex roots.Compute the roots:\[ m = \frac{-3 \pm \sqrt{-7}}{2}\]\[ m = \frac{-3}{2} \pm \frac{i \sqrt{7}}{2} \].

Step 8: Form the General Solution

With complex roots \(m = \alpha \pm \beta i\) where \(\alpha = -\frac{3}{2}\) and \(\beta = \frac{\sqrt{7}}{2}\), the general solution is:\[ y(x) = x^{\alpha}(C_{1} \cos(\beta \ln x) + C_{2} \sin(\beta \ln x)) \]Substitute \(\alpha\) and \(\beta\):\[ y(x) = x^{-\frac{3}{2}}(C_{1} \cos(\frac{\sqrt{7}}{2} \ln x) + C_{2} \sin(\frac{\sqrt{7}}{2} \ln x)) \]

Key Concepts

Second-Order Linear Differential Equations

Second-order linear differential equations are equations that involve the second derivative of a function. They are called linear because they maintain a linear relationship with the function and its derivatives. In mathematical terms, they have the form:\( y'' + p(x) y' + q(x) y = g(x) \).
In these equations, \(y\) is the function of \(x\) we want to determine. The coefficients \(p(x)\) and \(q(x)\) influence the behavior of the solutions. The function \(g(x)\) represents any non-homogeneous part, and if it equals zero, the equation is homogeneous.

• Second-order: This implies the highest derivative involved is the second derivative.
• Linear: No products or powers of \(y, y', y''\) are involved other than their first power.

Recognizing these characteristics helps determine the approach to solve, often involving specific techniques like assuming certain forms for the solution.

Variable Coefficients

Variable coefficients in a differential equation mean that the coefficients \(p(x)\) and \(q(x)\), which multiply the derivatives and function, are variables, often functions of \(x\) instead of constants. In simpler terms, they change as \(x\) changes.

• Impact on Solving Techniques: Variable coefficients generally make the equation more challenging to solve because standard methods for constant-coefficient equations may not apply directly. Thus, equations with variable coefficients usually require specific methods or assumptions for their solutions.
• Cauchy-Euler Equation: A special case of variable coefficient differential equations is the Cauchy-Euler equation, characterized by its coefficients being powers of \(x\), much like our given example \(x^2y''+xy'+4y=0\).

The understanding of variable coefficients is essential to identify and utilize appropriate methods such as reducing the equation using substitutions or transformations.

Complex Roots

Complex roots emerge when solving equations, such as characteristic equations, where the discriminant is negative. This happens quite often with Cauchy-Euler differential equations.
When we solve the quadratic equation \(m^2 + 3m + 4 = 0\), the discriminant \(b^2 - 4ac\) becomes negative, leading to complex roots:

• \( m = \frac{-3}{2} \pm \frac{i\sqrt{7}}{2} \)

These complex roots indicate solutions involving exponential and trigonometric functions:

• Real Part: The real part \(-\frac{3}{2}\) of the root contributes to the power of \(x\) in the solution.
• Imaginary Part: The imaginary part \(\frac{\sqrt{7}}{2}\) leads to sinusoidal functions like sine and cosine, influenced by logarithmic arguments in this context.

Complex roots guide us to general solutions showcasing exponential decays or oscillations, key in systems exhibiting cyclical or wave-like behavior.

General Solution

The general solution of a differential equation encompasses all possible solutions, derived from the roots of its characteristic equation. For equations like the Cauchy-Euler, the presence of complex roots dictates a distinct form for the solution.
With roots \( m = \frac{-3}{2} \pm \frac{i \sqrt{7}}{2} \), the general solution becomes:\[ y(x) = x^{\alpha} (C_1 \cos(\beta \ln x) + C_2 \sin(\beta \ln x)) \]where:

• \(\alpha = -\frac{3}{2}\), contributing to the polynomial part/asymptotic behavior.
• \(\beta = \frac{\sqrt{7}}{2}\), influencing the oscillatory nature through sine and cosine functions.
• \(C_1\) and \(C_2\) are constants determined by initial conditions or boundary values.

These constants allow the solution to adapt to specific situational requirements, ensuring that the general solution can cater to a wide array of scenarios typically modeled by the same differential structure.