The solution of frac{dy}{dx} + 1 = e^{x+y} is | vedclass

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

Question

(A) Given differential equation is $\frac{dy}{dx} + 1 = e^{x+y}$.Let $x + y = z$.Differentiating both sides with respect to $x$,we get $1 + \frac{dy}{

Vedclass · Accepted Answer

$e^{-(x+y)} + x + c = 0$

Vedclass · Answer

(A) Given differential equation is $\frac{dy}{dx} + 1 = e^{x+y}$.Let $x + y = z$.Differentiating both sides with respect to $x$,we get $1 + \frac{dy}{dx} = \frac{dz}{dx}$.Substituting this into the original equation,we have $\frac{dz}{dx} = e^z$.Separating the variables,we get $e^{-z} dz = dx$.Integrating both sides,$\int e^{-z} dz = \int dx$.This gives $-e^{-z} = x + c$.Substituting $z = x + y$ back,we get $-e^{-(x+y)} = x + c$.Rearranging the terms,we get $x + e^{-(x+y)} + c = 0$.

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

Similar Questions

The solution of the differential equation $x dy - y dx = 0$ represents:

The general solution of the differential equation $2x \frac{dy}{dx} - y = 3$ is a family of

The equation of the curve passing through the point $(0, -2)$ given that at any point $(x, y)$ on the curve,the product of the slope of its tangent and the $y$-coordinate of the point is equal to the $x$-coordinate of the point,is

The solution of the differential equation ${x^2}dy = - 2xydx$ is

The solution of $x dx + y dy = x^2 y dy - x y^2 dx$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module