Arjun
asked
in Differential equations
Feb 24, 2019
recategorized
Mar 14, 2021
by Lakshman Patel RJIT

0 votes

The solution of the ordinary differential equation

$\frac{dy}{dx}+3y=1$, subjects to the initial condition $y=1$ at $x=0$, is

- $\frac{1}{3}\left ( 1+2e^{-x/3} \right )$
- $\frac{1}{3}\left ( 5-2e^{-x/3} \right )$
- $\frac{1}{3}\left ( 5-2e^{-3x} \right )$
- $\frac{1}{3}\left ( 1+2e^{-3x} \right )$