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: Use the substitution $x = e^t$, that is $t = \ln x$.
This is the standard trick for a Cauchy-Euler equation: it turns the variable coefficients (powers of $x$) into constant coefficients in $t$.
With this substitution, it can be shown that $x^2\dfrac{d^2y}{dx^2} = \dfrac{d^2y}{dt^2} - \dfrac{dy}{dt}$.

Step 2: Rewrite the equation in terms of t.
The original equation $x^2y'' = 6y$ becomes
\[ \frac{d^2y}{dt^2} - \frac{dy}{dt} = 6y \]
\[ \frac{d^2y}{dt^2} - \frac{dy}{dt} - 6y = 0 \]
This is now a standard linear equation with constant coefficients.

Step 3: Solve using the characteristic equation.
Let $y = e^{kt}$. Substituting gives the characteristic equation
\[ k^2 - k - 6 = 0 \]
Factor: $(k-3)(k+2) = 0$, so $k = 3$ or $k = -2$.

Step 4: Write the solution in t, then convert back to x.
The solution in $t$ is $y(t) = ae^{3t} + be^{-2t}$.
Since $t = \ln x$, $e^{3t} = e^{3\ln x} = x^3$ and $e^{-2t} = e^{-2\ln x} = x^{-2}$.

Step 5: Substitute back to get the final answer.
\[ y(x) = ax^3 + bx^{-2} = ax^3 + \frac{b}{x^2} \]
This confirms option (A), reached here by a change-of-variable method instead of the direct trial-solution method used earlier. It also confirms no logarithmic term appears, since the roots $k=3,-2$ of the characteristic equation are real and distinct, not repeated.

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