The solution of frac{dy}{dx} + frac{y}{3} = 1 | 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 $Q = 1$.First,w

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 $Q = 1$.First,we find the integrating factor $(I.F.)$:$I.F. = e^{\int P dx} = e^{\int \frac{1}{3} dx} = e^{x/3}$.The general solution is given by $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + c$.Substituting the values:$y \cdot e^{x/3} = \int 1 \cdot e^{x/3} dx + c$.Integrating the right side:$y \cdot e^{x/3} = 3e^{x/3} + c$.Dividing both sides by $e^{x/3}$:$y = 3 + c e^{-x/3}$.

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

Similar Questions

An integrating factor of the differential equation $\left(1-x^2\right) \frac{d y}{d x}+x y=\frac{x^4}{\left(1+x^5\right)}\left(\sqrt{1-x^2}\right)^3$ is

The general solution of the differential equation $(x + y) \frac{dy}{dx} = 1$ is

The solution of $\frac{dy}{dx} + p(x)y = 0$ is

The equation of the curve passing through $(1,2)$ and whose tangent at any point $(x, y)$ makes an angle $\tan ^{-1}(2 x+3 y)$ with the $X$-axis is .........

Let $y=y(x)$ be the solution of the differential equation $\frac{dy}{dx}=2(y+2 \sin x-5)x-2 \cos x$ such that $y(0)=7$. Then $y(\pi)$ is equal to :

Vedclass Test Series

Exam Paper Generator

Online Exam Module