The solution of frac{dy}{dx} + frac{1}{3}y = | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$, where $P = \frac{1}{3}$ and $Q = 1$.The in

Vedclass · Accepted Answer

$y = 3 + ce^{-x/3}$

Vedclass · Answer

(B) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$, where $P = \frac{1}{3}$ and $Q = 1$.The integrating factor $(IF)$ is given by $IF = e^{\int P dx} = e^{\int \frac{1}{3} dx} = e^{x/3}$.The general solution is $y 	imes (IF) = \int (Q 	imes IF) dx + c$.Substituting the values, we get $y 	imes e^{x/3} = \int (1 	imes e^{x/3}) dx + c$.Integrating the right side, $y 	imes e^{x/3} = 3e^{x/3} + c$.Dividing both sides by $e^{x/3}$, we get $y = 3 + ce^{-x/3}$.

The solution of $\frac{dy}{dx} + \frac{1}{3}y = 1$ is

Similar Questions

Semi-water gas is a mixture of which of the following?

The potential energy of a particle of mass $1\, kg$ in motion along the $x-$axis is given by $U = 4(1 - \cos 2x)$,where $x$ is in $meters$. The period of small oscillation (in $seconds$) is

If a chord of the parabola $y^2=4x$ passes through its focus and makes an angle $\theta$ with the $X$-axis,then its length is

$A$ uniform rope of mass $0.1 ~kg$ and length $2.45 ~m$ hangs from a rigid support. The time taken by the transverse wave formed in the rope to travel through the full length of the rope is (Assume $g=9.8 ~m/s^2$) (in $~s$)

The major product obtained on interaction of phenol with sodium hydroxide and carbon dioxide is

Vedclass Test Series

Exam Paper Generator

Online Exam Module