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 solution of the differential equation $({x^2} - y{x^2})\frac{{dy}}{{dx}} + {y^2} + x{y^2} = 0$ is

The general solution of $\frac{dy}{dx} = x + \sin x \cos y + x \cos y + \sin x$ is

The general solution of the differential equation $\frac{dy}{dx} = \frac{2xy - 4x + y - 2}{2xy + x - 4y - 2}$ is

The particular solution of the differential equation $\frac{dy}{dx} = e^{2y} \cos x$,when $y(\frac{\pi}{6}) = 0$ is

The solution of the differential equation $\frac{dy}{dx}=(x-y)^2$ when $y(1)=1$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module