Question

The substitution which reduces the differential equation \( t^2 \frac{d^2y}{dt^2} + 5t \frac{dy}{dt} + 7y = 0 \) to \( \frac{d^2y}{dx^2} + 4 \frac{dy}{dx} + 7y = 0 \) is:

• A. \( x = e^t \)
• B. \( t = \log y \)
• C. \( t = e^x \)
• D. \( x = \log y \)

Hint:

To simplify Cauchy–Euler differential equations, use the substitution \( t = e^x \) to convert variable coefficients to constant coefficients.

Answer 1

The Correct Option is C

Step 1: Use the substitution \( t = e^x \), or \( x = \log t \).
This transformation is common for simplifying Cauchy–Euler equations into constant coefficient linear equations.
Step 2: Apply the chain rule.
\[ \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} = \frac{dy}{dx} \cdot \frac{1}{t} \Rightarrow \frac{dy}{dx} = t \frac{dy}{dt} \] Differentiate again: \[ \frac{d^2y}{dx^2} = \frac{d}{dx}(t \frac{dy}{dt}) = \frac{dt}{dx} \cdot \frac{dy}{dt} + t \cdot \frac{d}{dx} \left( \frac{dy}{dt} \right) = t \frac{dy}{dt} + t \cdot \left( \frac{d^2y}{dt^2} \cdot \frac{dt}{dx} \right) = t \frac{dy}{dt} + t^2 \frac{d^2y}{dt^2} \]
Step 3: Substitute into the new equation.
\[ \frac{d^2y}{dx^2} + 4\frac{dy}{dx} + 7y = t^2 \frac{d^2y}{dt^2} + t \frac{dy}{dt} + 4t \frac{dy}{dt} + 7y = t^2 \frac{d^2y}{dt^2} + 5t \frac{dy}{dt} + 7y \] This confirms the equation is correctly transformed.