Question

An ordinary differential equation is given below.
\[ x^2\frac{d^2y}{dx^2} = 6y \]
Considering \(a\) and \(b\) as arbitrary constants, the general solution of the equation is

• A. \(y(x) = ax^3 + \dfrac{b}{x^2}\)
• B. \(y(x) = ax^2 + \dfrac{b}{x^3}\)
• C. \(y(x) = ax^2 + b\ln x\)
• D. \(y(x) = ax^3 + b\ln x\)

Hint:

This is a Cauchy-Euler equation; try \(y=x^m\) or the substitution \(x=e^t\) to turn it into a constant-coefficient equation.

Answer 1

The Correct Option is A

Step 1: Identify the type of equation.
The equation \(x^2\dfrac{d^2y}{dx^2} = 6y\) can be written as \(x^2y'' - 6y = 0\).
This is a Cauchy-Euler (equidimensional) equation, where the power of \(x\) multiplying each derivative matches the order of that derivative.

Step 2: Assume a trial solution of the form \(y = x^m\).
For Cauchy-Euler equations, a solution of the form \(y = x^m\) always works for some value(s) of \(m\), because differentiating \(x^m\) always keeps it a power of \(x\), matching the \(x^2\) coefficient in the equation.
With \(y = x^m\): \(y' = mx^{m-1}\) and \(y'' = m(m-1)x^{m-2}\).

Step 3: Substitute into the equation.
\[ x^2\cdot m(m-1)x^{m-2} - 6x^m = 0 \]
\[ m(m-1)x^m - 6x^m = 0 \]
Since \(x^m \ne 0\) in general, divide through by \(x^m\):
\[ m(m-1) - 6 = 0 \]

Step 4: Solve the auxiliary (indicial) equation.
\[ m^2 - m - 6 = 0 \]
Factor: \((m-3)(m+2) = 0\), so \(m = 3\) or \(m = -2\).
These are two distinct real roots, so the two independent solutions are \(x^3\) and \(x^{-2}\).

Step 5: Write the general solution and rule out the other options.
The general solution is a linear combination of the two independent solutions:
\[ y(x) = ax^3 + \frac{b}{x^2} \]
This matches option (A). Option (B) swaps the powers (\(x^2\) and \(x^{-3}\)), which would only come from the wrong roots \(m=2,-3\), not the actual roots \(3,-2\). Options (C) and (D) both use a \(\ln x\) term, but a \(\ln x\) term only shows up in the solution when the auxiliary equation has a REPEATED root; here the roots \(3\) and \(-2\) are distinct, so no \(\ln x\) term should appear.

Final Answer:
\[ \boxed{y(x) = ax^3 + \frac{b}{x^2}} \]