The solution of the equation frac{dy}{dx} = f | vedclass

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

Question

(A) Given the differential equation $\frac{dy}{dx} = \frac{1}{x + y + 1}$.Taking the reciprocal,we get $\frac{dx}{dy} = x + y + 1$.Rearranging the ter

Vedclass · Accepted Answer

$x = ce^y - y - 2$

Vedclass · Answer

(A) Given the differential equation $\frac{dy}{dx} = \frac{1}{x + y + 1}$.Taking the reciprocal,we get $\frac{dx}{dy} = x + y + 1$.Rearranging the terms,we have $\frac{dx}{dy} - x = y + 1$.This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = -1$ and $Q(y) = y + 1$.The integrating factor $(I.F.)$ is $e^{\int P(y) dy} = e^{\int -1 dy} = e^{-y}$.The general solution is given by $x \cdot (I.F.) = \int Q(y) \cdot (I.F.) dy + c$.$x e^{-y} = \int (y + 1) e^{-y} dy + c$.Using integration by parts for $\int (y + 1) e^{-y} dy$:Let $u = y + 1$ and $dv = e^{-y} dy$,then $du = dy$ and $v = -e^{-y}$.$\int (y + 1) e^{-y} dy = (y + 1)(-e^{-y}) - \int (-e^{-y}) dy = -(y + 1)e^{-y} - e^{-y} + c = -(y + 2)e^{-y} + c$.Thus,$x e^{-y} = -(y + 2)e^{-y} + c$.Multiplying both sides by $e^y$,we get $x = -(y + 2) + ce^y$,which simplifies to $x = ce^y - y - 2$.

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

Similar Questions

The solution of $e^{y-x} \frac{dy}{dx} = \frac{y(\sin x + \cos x)}{1 + y \log y}$ is

The general solution of the differential equation $\frac{dy}{dx} + \frac{\sec x}{\cos x + \sin x} y = \frac{\cos x}{1 + \tan x}$ is

The solution of $dy = \cos x(2 - y \csc x)dx$ where $y = 2$ when $x = \frac{\pi}{2}$ is

The integrating factor of the differential equation $(x^2 - 1)\frac{dy}{dx} + 2xy = x$ is

If for $x \geq 0$,$y=y(x)$ is the solution of the differential equation $(x+1) dy = ((x+1)^{2} + y - 3) dx$ with $y(2) = 0$,then $y(3)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module