Browse Questions

# If $y=ke^{\lambda x}$ then its differential equation is

$\begin{array}{1 1}(1)\frac{dy}{dx}=\lambda y&(2)\frac{dy}{dx}=ky\\(3)\frac{dy}{dx}+ky=0&(4)\frac{dy}{dx}=e^{\lambda x}\end{array}$

$y=ke^{\lambda x}$