Question

The differential equation of the family of curves \(y = Ae^{3x} + Be^{5x}\), where \(A\) and \(B\) are arbitrary constants, is

• A. \(\frac{d^2y}{dx^2} + 8\frac{dy}{dx} + 15y = 0\)
• B. \(\frac{d^2y}{dx^2} - \frac{dy}{dx} + y = 0\)
• C. \(\frac{d^2y}{dx^2} - 8\frac{dy}{dx} + 15y = 0\)
• D. None of the above

Hint:

For exponential solutions, the auxiliary equation roots are the exponents.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The given solution corresponds to the roots of the auxiliary equation being 3 and 5.

Step 2: Detailed Explanation:
For a linear homogeneous DE with constant coefficients, if the roots are \(m_1\) and \(m_2\), the DE is \(\frac{d^2y}{dx^2} - (m_1+m_2)\frac{dy}{dx} + m_1m_2 y = 0\). Here \(m_1 = 3\), \(m_2 = 5\), so sum = 8, product = 15. Thus DE: \(y'' - 8y' + 15y = 0\).

Step 3: Final Answer:
\(\frac{d^2y}{dx^2} - 8\frac{dy}{dx} + 15y = 0\), which corresponds to option (C).