Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Differential Equations

Question:

The differential equation representing family of curves y = aemx + benx, where a and b are arbitrary constants, is

Options:

\(\frac{d^2y}{dx^2}\)+(m + n)\(\frac{dy}{dx}\) + y =0

\(\frac{d^2y}{dx^2}\)+(mn)\(\frac{dy}{dx}\) +(m + n) y =0

\(\frac{d^2y}{dx^2}\)-(m + n)\(\frac{dy}{dx}\) + mny =0

\(\frac{d^2y}{dx^2}\)+(m + n)\(\frac{dy}{dx}\) - mny =0

Correct Answer:

\(\frac{d^2y}{dx^2}\)-(m + n)\(\frac{dy}{dx}\) + mny =0

Explanation:

y = aemx + benx

on differentiating both side, we get

$\frac{dy}{dx}=ame^{mx}+bne^{nx}$

Again differentiating, we get

$\frac{d^2y}{dx^2}=am^2e^{mx}+bn^2e^{nx}$

$∴\frac{d^2y}{dx^2}-(m+n)\frac{dy}{dx}+mny=am^2e^{mx}+bn^2e^{nx}$

$-(m+n)[ame^{mx}+bne^{nx}]+mn[ae^{mx}+be^{nx}]$

$=0$