The general solution of the differential equa | vedclass

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

Question

(B) Given the differential equation $\frac{dy}{dx} = 2^{y-x}$.Using the property of exponents,we can write $\frac{dy}{dx} = \frac{2^y}{2^x}$.Separatin

Vedclass · Accepted Answer

$\frac{1}{2^x} - \frac{1}{2^y} = c$

Vedclass · Answer

(B) Given the differential equation $\frac{dy}{dx} = 2^{y-x}$.Using the property of exponents,we can write $\frac{dy}{dx} = \frac{2^y}{2^x}$.Separating the variables,we get $\frac{dy}{2^y} = \frac{dx}{2^x}$,which is equivalent to $2^{-y} dy = 2^{-x} dx$.Integrating both sides,we have $\int 2^{-y} dy = \int 2^{-x} dx$.Using the formula $\int a^u du = \frac{a^u}{\ln a} + C$,we get $\frac{2^{-y}}{-\ln 2} = \frac{2^{-x}}{-\ln 2} + C_1$.Multiplying by $-\ln 2$,we get $2^{-y} = 2^{-x} - C_1 \ln 2$.Rearranging the terms,$2^{-x} - 2^{-y} = C_1 \ln 2$.Letting $c = C_1 \ln 2$,we obtain $\frac{1}{2^x} - \frac{1}{2^y} = c$.

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

Similar Questions

The general solution of the differential equation $y \log y \, dx - x \, dy = 0$ is . . . . . . .

The general solution of the differential equation $\frac{dy}{dx} = \frac{2x-3y+5}{6x-9y+7}$ is

The solution of the differential equation $\sec^2 x \tan y \, dx + \sec^2 y \tan x \, dy = 0$ with the initial condition $y(\frac{\pi}{4}) = \frac{\pi}{3}$ is:

The general solution of the differential equation $(x-2y+1)dy-(3x-6y+2)dx=0$ is

Given that the slope of the tangent to a curve $y=y(x)$ at any point $(x, y)$ is $\frac{2y}{x^2}$. If the curve passes through the centre of the circle $x^2+y^2-2x-2y=0$,then its equation is

Vedclass Test Series

Exam Paper Generator

Online Exam Module