Question

Consider the differential equation \( x^2 \frac{d^2y}{dx^2} = 6y \). The general solution of the above equation is

Hint:

Recognize the \(x^2y'', xy', y\) pattern as a Cauchy-Euler equation. The substitution \(y=x^m\) quickly converts it into an algebraic auxiliary equation, which is easy to solve.

Answer 1

Step 1: Understanding the Question:
The given differential equation, \( x^2 y" - 6y = 0 \), is a homogeneous second-order linear differential equation with variable coefficients. This specific form is known as a Cauchy-Euler (or equidimensional) equation.
Step 2: Key Formula or Approach:
For a Cauchy-Euler equation of the form \(ax^2y" + bxy' + cy = 0\), we assume a solution of the form \(y = x^m\). Substituting this into the equation yields an auxiliary (or characteristic) equation in \(m\).
Step 3: Detailed Explanation:
Assume the solution is \(y = x^m\). Find the derivatives: \[ y' = mx^{m-1} \] \[ y" = m(m-1)x^{m-2} \] Substitute these into the differential equation: \[ x^2 [m(m-1)x^{m-2}] - 6(x^m) = 0 \] \[ m(m-1)x^m - 6x^m = 0 \] Since \(x^m \neq 0\), we can divide by it to get the auxiliary equation: \[ m(m-1) - 6 = 0 \] \[ m^2 - m - 6 = 0 \] This is a quadratic equation for \(m\). We can factor it: \[ (m-3)(m+2) = 0 \] The roots are \(m_1 = 3\) and \(m_2 = -2\).
Since the roots are real and distinct, the general solution is a linear combination of the two corresponding solutions: \[ y = c_1 x^{m_1} + c_2 x^{m_2} \] \[ y = c_1 x^3 + c_2 x^{-2} \] Using the constants \(a\) and \(b\) as in the options, the solution is: \[ y = ax^3 + \frac{b}{x^2} \] Step 4: Final Answer:
The general solution matches option (a).