Question

If $y=k{e}^{\lambda x}$ then its differential equation is (where $k$ is arbitrary constant):

• A. $\frac{dy}{dx}=\lambda y$
• B. $\frac{dy}{dx}=ky$
• C. $\frac{dy}{dx}+ky=0$
• D. $\frac{dy}{dx}=e^{\lambda x}$

Answer 1

The correct option is A $\frac{dy}{dx}=\lambda y$

Given $y=k{e}^{\lambda x}$ where $k$ is arbitrary constant.

Consider $y=k{e}^{\lambda x}$

Differentiating both sides, we get

$dy=k\lambda e^{\lambda x}dx$

$\Rightarrow \frac{dy}{dx}=\lambda k{e}^{\lambda x}$

$\Rightarrow \frac{dy}{dx}=\lambda y[\because y=k{e}^{\lambda x}]$

Hence the differential equation of $y=k{e}^{\lambda x}$ is $\frac{dy}{dx}=\lambda y$