The particular solution of the differential e | vedclass

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

Question

(B) Given differential equation is $(1+e^{2x}) dy + e^x(1+y^2) dx = 0$. Rearranging the terms to separate variables,we get $\frac{dy}{1+y^2} = -\frac{

Vedclass · Accepted Answer

$	an^{-1} e^x + 	an^{-1} y = \frac{\pi}{2}$

Vedclass · Answer

(B) Given differential equation is $(1+e^{2x}) dy + e^x(1+y^2) dx = 0$. Rearranging the terms to separate variables,we get $\frac{dy}{1+y^2} = -\frac{e^x}{1+e^{2x}} dx$. Integrating both sides,$\int \frac{dy}{1+y^2} = -\int \frac{e^x}{1+(e^x)^2} dx$. Let $e^x = t$,then $e^x dx = dt$. The integral becomes $	an^{-1}(y) = -	an^{-1}(t) + C$,which is $	an^{-1}(y) = -	an^{-1}(e^x) + C$. Thus,$	an^{-1}(y) + 	an^{-1}(e^x) = C$. Given $x=0$ and $y=1$,we substitute these values: $	an^{-1}(1) + 	an^{-1}(e^0) = C$. $\frac{\pi}{4} + \frac{\pi}{4} = C$,so $C = \frac{\pi}{2}$. The particular solution is $	an^{-1}(y) + 	an^{-1}(e^x) = \frac{\pi}{2}$.

The particular solution of the differential equation $(1+e^{2x}) dy + e^x(1+y^2) dx = 0$ at $x=0$ and $y=1$ is

Similar Questions

Find the equation of a curve passing through the point $(-2, 3),$ given that the slope of the tangent to the curve at any point $(x, y)$ is $\frac{2x}{y^2}$.

Let $y=y(x)$ be the solution of the differential equation $e^{x} \sqrt{1-y^{2}} dx + \frac{y}{x} dy = 0$,with $y(1) = -1$. Then the value of $(y(3))^{2}$ is equal to:

The general solution of the equation $({e^y} + 1)\cos x \, dx + {e^y}\sin x \, dy = 0$ is

The general solution of the differential equation $\sec^2 x \tan y \, dx + \sec^2 y \tan x \, dy = 0$ is . . . . . . .

General solution of the differential equation $\frac{dy}{dx} = \frac{x+y+1}{x+y-1}$ is given by

Vedclass Test Series

Exam Paper Generator

Online Exam Module