The general solution of the differential equa | vedclass

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

Question

(C) Given the differential equation: $\frac{dy}{dx} = e^{x+y}$.Using the property of exponents,we can write: $\frac{dy}{dx} = e^x \cdot e^y$.Separatin

Vedclass · Accepted Answer

$e^{x} + e^{-y} = C$

Vedclass · Answer

(C) Given the differential equation: $\frac{dy}{dx} = e^{x+y}$.Using the property of exponents,we can write: $\frac{dy}{dx} = e^x \cdot e^y$.Separating the variables,we get: $\frac{dy}{e^y} = e^x dx$,which is equivalent to $e^{-y} dy = e^x dx$.Integrating both sides: $\int e^{-y} dy = \int e^x dx$.This yields: $-e^{-y} = e^x + k$,where $k$ is the constant of integration.Rearranging the terms: $e^x + e^{-y} = -k$.Letting $C = -k$,we get the general solution: $e^x + e^{-y} = C$.Thus,the correct option is $C$.

The general solution of the differential equation $\frac{dy}{dx} = e^{x+y}$ is

Similar Questions

The particular solution of the differential equation $\cos \left(\frac{dy}{dx}\right) = a$,under the conditions $a \in \mathbb{R}$ and $y(0) = 2$ is

If the solution of $\frac{dy}{dx} = \frac{y^3 \cos \sqrt{x}}{\sqrt{x} e^{1/y^2}}$ with $y(0) = 1$ is $\frac{1}{y^2} = \log_e(f(x))$,then $f(x) =$

The solution of $(x - y^3)dx + 3xy^2dy = 0$ is

Show that the general solution of the differential equation $\frac{dy}{dx} + \frac{y^{2}+y+1}{x^{2}+x+1} = 0$ is given by $(x+y+1) = A(1-x-y-2xy)$,where $A$ is a parameter.

If $y=y(x)$ satisfies the differential equation $8 \sqrt{x}(\sqrt{9+\sqrt{x}}) dy = (\sqrt{4+\sqrt{9+\sqrt{x}}})^{-1} dx$ for $x>0$ and $y(0)=\sqrt{7}$,then find $y(256)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module