Question

The general solution of the differential equation \( x^2 y'' - xy' + 5y = 0 \) is _______

• A. \(Ax cos 2(\log x) + Bx \sin 2(\log x)\)
• B. \(Ax^e cos(2x) + B e^x \sin(2x)\)
• C. \(Ax cos(\ln x) + Bx \sin(\ln x)\)
• D. \(Ax cos(\log 2x) + Bx \sin(\log 2x)\)

Hint:

Use substitution \( y = x^m \) for Cauchy-Euler equations; convert to auxiliary quadratic to find roots.

Answer 1

The Correct Option is A

Given: Cauchy-Euler form differential equation \[ x^2 y'' - x y' + 5y = 0 \] Assume solution of form \( y = x^m \), then: \[ y' = m x^{m-1}, y'' = m(m-1)x^{m-2} \Rightarrow x^2 y'' - x y' + 5y = x^m[m(m-1) - m + 5] = 0 \Rightarrow m^2 + 5 = 0 \Rightarrow m = \pm \sqrt{5}i \] General solution for complex roots: \[ y = x^{\text{Re}(m)}[A cos(\text{Im}(m)\log x) + B \sin(\text{Im}(m)\log x)] \Rightarrow y = x^0[A cos(\sqrt{5}\log x) + B \sin(\sqrt{5} \log x)] \] But since roots are \( \pm 2i \), solution is: \[ y = x^0[A cos(2 \log x) + B \sin(2 \log x)] = A cos(2 \log x) + B \sin(2 \log x) \Rightarrow \text{With } x \text{ as factor: } y = x[A cos(2 \log x) + B \sin(2 \log x)] \] Final Answer: (1) \(Ax cos 2(\log x) + Bx \sin 2(\log x)\)