The differential equation representing family of curves \(y = ae^{mx} + be^{nx}\), where a and b are arbitrary constants, is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{d^2y}{dx^2}-(m+n)\frac{dy}{dx}+mny=0\)
There are two arbitrary constants, so differentiate twice and eliminate a and b.
\(y = ae^{mx} + be^{nx}\)
\( \displaystyle \frac{dy}{dx} = ame^{mx} + bne^{nx} \)
\( \displaystyle \frac{d^2y}{dx^2} = am^2e^{mx} + bn^2e^{nx} \)
Now form \(\frac{d^2y}{dx^2} - (m+n)\frac{dy}{dx} + mny\) and collect the terms:
\(\text{coefficient of } ae^{mx}:\ m^2 - (m+n)m + mn = 0\)
\(\text{coefficient of } be^{nx}:\ n^2 - (m+n)n + mn = 0\)
Hence
\( \displaystyle \frac{d^2y}{dx^2} - (m+n)\frac{dy}{dx} + mny = 0 \)