Question

The general solution of the differential equation \( x^2 \frac{d^2y}{dx^2} - 5x \frac{d5y}{dx} + 13y = 0 \) is \dots\dots.

• A. \( y = e^{3x}(C_1 \cos 2x + C_2 \sin 2x) \)
• B. \( y = e^{3x}(C_1 \cos x + C_2 \sin x) \)
• C. \( y = x^3(C_1 \cos(2 \ln x) + C_2 \sin(2 \ln x)) \)
• D. \( y = x^3(C_1 \cos(\ln x) + C_2 \sin(\ln x)) \)

Hint:

For a Cauchy-Euler equation, if the auxiliary equation yields complex conjugate roots \(m = \alpha \pm i\beta\), the solution always involves logarithmic arguments in the trigonometric terms, take the form \(\cos(\beta \ln x)\). This allows you to eliminate options with linear exponents like \(2x\) or \(x\) immediately.

Answer 1

The Correct Option is C

Concept: The given equation is a second-order linear homogeneous differential equation with variable coefficients, known as a Cauchy-Euler equation. The general form is: \[ x^2 \frac{d^2y}{dx^2} + Px \frac{dy}{dx} + Qy = 0 \] To solve this class of differential equations, we change the independent variable from \( x \) to \( z \) using the substitution: \[ x = e^z \implies z = \ln x \] This transforms the variable-coefficient equation into a constant-coefficient linear differential equation using the operator identities: \[ x \frac{dy}{dx} = Dy \] \[ x^2 \frac{d^2y}{dx^2} = D(D-1)y \] where \( D = \frac{d}{dz} \).

Step 1: Rewrite the differential equation with constant coefficients
The given differential equation is: \[ x^2 \frac{d^2y}{dx^2} - 5x \frac{dy}{dx} + 13y = 0 \] Substituting the operator identities into the equation gives: \[ [D(D-1) - 5D + 13]y = 0 \] Expanding and collecting like terms: \[ [D^2 - D - 5D + 13]y = 0 \] \[ [D^2 - 6D + 13]y = 0 \]

Step 2: Solve the Auxiliary Equation
The auxiliary equation associated with this constant-coefficient differential equation is: \[ m^2 - 6m + 13 = 0 \] Using the quadratic formula to find the roots: \[ m = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(13)}}{2(1)} \] \[ m = \frac{6 \pm \sqrt{36 - 52}}{2} = \frac{6 \pm \sqrt{-16}}{2} \] \[ m = \frac{6 \pm 4i}{2} = 3 \pm 2i \] The roots are complex conjugates of the form \(\alpha \pm i\beta\), where \(\alpha = 3\) and \(\beta = 2\).

Step 3: Formulate the solution in terms of \( z \) and convert back to \( x \)
For complex roots \( \alpha \pm i\beta \), the solution with respect to independent variable \( z \) is written as: \[ y = e^{\alpha z} (C_1 \cos \beta z + C_2 \sin \beta z) \] Substitute \(\alpha = 3\) and \(\beta = 2\): \[ y = e^{3z} (C_1 \cos 2z + C_2 \sin 2z) \] Now substitute \( z = \ln x \) and \( e^{3z} = (e^z)^3 = x^3 \) to return to the original independent variable \( x \): \[ y = x^3 (C_1 \cos(2 \ln x) + C_2 \sin(2 \ln x)) \] This matches Option (C).