Question

Solution of the differential equation dy/dx = $2^{y-x}$ is?

Answer 1

Solution :

Given differential equation is

$\frac{dy}{dx}=2^{y-x}$

$\implies \frac{dy}{dx}=2^{y}\times 2^{-x}$

$\implies \frac{dy}{2^{y}}=2^{-x}dx$

$\implies 2^{-y}dy=2^{-x}dx$

On integration, we get

$\int 2^{-y}dy=\int 2^{-x}dx$

$\implies \frac{2^{-y}}{-log_{e}2}=\frac{2^{-x}}{-log_{e}2}+\frac{C}{-log_{e}2}$

where $\frac{C}{-log_{e}2}$ is integral constant

$\implies \boxed{2^{-y}=2^{-x}+C}$

which is the required solution.

Rule :

$\int a^{mx}dx=\frac{a^{mx}}{m\:log_{e}a}+C$

where C is integral constant

Answer 2

hello!!!

sol:

=> dy/dx = 2^(y-x)

=> dy/dx = 2^y/2^x

=> arranging like terms , we get

=> dy/2^y = dx/2^x

integrating both sides

=> 1/2^ylog2 = 1/2^xlog2 + C

-------++-+++++

hope it helps