Problem 7 - Advanced Engineering Mathematics

Question

Solve the given differential equation. $$ x^{2} y^{\prime \prime}-3 x y^{\prime}-2 y=0 $$

Step-by-Step Solution

Answer

The general solution is \( y(x) = C_1 x^{2+\sqrt{6}} + C_2 x^{2-\sqrt{6}} \).

Step 1: Identify the Type of Differential Equation

The given differential equation is \(x^2y'' - 3xy' - 2y = 0\). This is a Cauchy-Euler (or equidimensional) differential equation because it can be written in the form \( a_n x^n y^{(n)} + a_{n-1} x^{n-1} y^{(n-1)} + \, ... \, + a_0 y = 0 \).

Step 2: Assume a Solution Form

For Cauchy-Euler equations, we assume a solution of the form \( y = x^m \). This will help us convert the differential equation into an algebraic equation.

Step 3: Compute Derivatives

Calculate the first and second derivatives of \( y = x^m \). We have \( y' = mx^{m-1} \) and \( y'' = m(m-1)x^{m-2} \).

Step 4: Substitute into the Differential Equation

Substitute \( y = x^m \), \( y' = mx^{m-1} \), and \( y'' = m(m-1)x^{m-2} \) back into the original differential equation. This yields: \[ x^2(m(m-1)x^{m-2}) - 3x(mx^{m-1}) - 2(x^m) = 0. \]

Step 5: Simplify the Equation

Simplify the terms obtained from substitution: \[ m(m-1)x^m - 3mx^m - 2x^m = 0. \] Combine similar terms to get the characteristic equation: \[ m^2 - 4m - 2 = 0. \]

Step 6: Solve the Characteristic Equation

Solve the quadratic equation \( m^2 - 4m - 2 = 0 \) using the quadratic formula \( m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1 \), \( b = -4 \), and \( c = -2 \). Calculate: \[ m = \frac{4 \pm \sqrt{16 + 8}}{2} = \frac{4 \pm \sqrt{24}}{2} = \frac{4 \pm 2\sqrt{6}}{2} = 2 \pm \sqrt{6}. \]

Step 7: Write the General Solution

Since the roots are real and distinct, the general solution of the differential equation is a linear combination of the two solutions: \[ y(x) = C_1 x^{2+\sqrt{6}} + C_2 x^{2-\sqrt{6}}, \] where \( C_1 \) and \( C_2 \) are arbitrary constants.

Key Concepts

Characteristic Equation

The characteristic equation is a crucial step in solving differential equations, especially those of the Cauchy-Euler form. For a Cauchy-Euler equation like the one given, the differential equation can be transformed into an algebraic problem by assuming a solution of the form \( y = x^m \). This assumption simplifies the complicated differential equation into something more manageable. When we substitute back these expressions of derivatives, we get an expression in terms of \(x^m\). By factorizing out \(x^m\), we are left with a simpler algebraic equation involving \(m\).

This algebraic equation derived from the process is known as the characteristic equation. In our specific problem, the characteristic equation was \( m^2 - 4m - 2 = 0 \). Solving this equation gives us the values of \(m\) which correspond to the exponents in the solutions of the differential equation. These values then guide us in forming the general solution to the differential equation by using the roots to determine the power of \(x\) in the solution.

Quadratic Formula

The quadratic formula is an essential mathematical tool used to find solutions to quadratic equations. Any quadratic equation can be written in the standard form \( ax^2 + bx + c = 0 \). In the context of differential equations, when we arrive at a characteristic equation like \( m^2 - 4m - 2 = 0 \), the quadratic formula comes into play.

The quadratic formula states:

• \( m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

In our example, \( a = 1 \), \( b = -4 \), and \( c = -2 \). Plugging these values into the formula, we calculated:

• \( m = \frac{4 \pm \sqrt{24}}{2} = \frac{4 \pm 2\sqrt{6}}{2} = 2 \pm \sqrt{6} \)

Using the quadratic formula is a systematic way to find the roots of the quadratic equation, even when the numbers are not easy to factor intuitively. This step is critical in finding distinct solutions when solving higher order differential equations.

Solution of Differential Equations

Once the characteristic equation is solved using the quadratic formula, we can construct the general solution of the differential equation. For real and distinct roots like \( 2 + \sqrt{6} \) and \( 2 - \sqrt{6} \), the solution to the original differential equation is a linear combination of individual solutions. This is a common technique where the complete solution is expressed as:

• \( y(x) = C_1 x^{2+\sqrt{6}} + C_2 x^{2-\sqrt{6}} \)

Here, \( C_1 \) and \( C_2 \) are arbitrary constants, which would typically be determined if initial conditions were provided.

In summary, solving differential equations involves transforming them into algebraic ones through characteristic equations, using the quadratic formula to find the roots and formulating the general solution. This process demonstrates the interconnected nature of algebraic and differential methods, helping in solving complex equations systematically.