Question

Which one of the following differential equations has the general solution

y = ae^x + be^-x?

1. \(\rm \frac{d^2y}{dx^2}+y=0\)
2. \(\rm \frac{d^2y}{dx^2}-y=0\)
3. \(\rm \frac{d^2y}{dx^2}+y=1\)
4. \(\rm \frac{dy}{dx}-y=0\)

Answer 1

Correct Answer - Option 2 : \(\rm \frac{d^2y}{dx^2}-y=0\)

Concept:

• Differential Equation: A differential equation is an equation that relates one or more functions and their derivatives. 
• e.g. \(\rm \frac{dy}{dx}\) + x = 2y + 3, etc.
• \(\rm \dfrac{d}{dx}e^{f(x)}=\dfrac{d}{dx}f(x)e^{f(x)}\)

 

Calculation:

y = aex + be-x

⇒ \(\rm \frac{dy}{dx}=\frac{d}{dx}ae^x +\frac{d}{dx}be^{-x}\)

⇒ \(\rm \frac{dy}{dx}\) = aex - be-x

⇒ \(\rm \frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}(ae^x -be^{-x})\)

⇒ \(\rm \frac{d^2y}{dx^2}\) = aex + be-x = y

∴ The general solution of y = aex + be-x is \(\rm \frac{d^2y}{dx^2}\) - y = 0.