Question

The differential equation among the following, whose general solution is \( y=A e^{5x}+B e^{-4x} \) is

• A. \( 5\frac{dy}{dx}+4\frac{dx}{dy}=0 \)
• B. \( \frac{d^{2}y}{dx^{2}}+\frac{dy}{dx}+20y=0 \)
• C. \( \left(\frac{dy}{dx}\right)^{2}-\frac{dy}{dx}-20y=0 \)
• D. \( \frac{d^{2}y}{dx^{2}}-\frac{dy}{dx}-20y=0 \)

Hint:

The middle term coefficient of the second-order differential equation is always equal to \( -(\text{sum of roots}) \), and the final constant multiplier is equal to the \( (\text{product of roots}) \). This simple rule allows you to check options mentally in seconds!

Answer 1

The Correct Option is D

Concept: For a second-order linear homogeneous differential equation with constant coefficients, if the general solution is \( y = A e^{m_1 x} + B e^{m_2 x} \), then \( m_1, m_2 \) are the roots of the corresponding characteristic equation: \[ (m - m_1)(m - m_2) = 0 \implies m^2 - (m_1 + m_2)m + m_1 m_2 = 0 \]

Step 1: Identifying the roots from the given solution string.
From the given expression \( y = A e^{5x} + B e^{-4x} \):

• \( m_1 = 5 \)

• \( m_2 = -4 \)

Step 2: Formulating the characteristic equation.
Using the sum and product of the roots:

• Sum of roots \( = 5 + (-4) = 1 \)

• Product of roots \( = 5 \times (-4) = -20 \)
Substituting these into the quadratic characteristic form: \[ m^2 - (1)m + (-20) = 0 \implies m^2 - m - 20 = 0 \]

Step 3: Converting back to differential operator notation.
Replacing \( m^2 _ with _ \frac{d^2 y}{dx^2} \) and \( m _ with _ \frac{dy}{dx} \): \[ \frac{d^2y}{dx^2} - \frac{dy}{dx} - 20y = 0 \] This matches option (D) perfectly.