Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Differential Equations
0 votes

Find the solution of $\Large \frac{dy}{dx}\normalsize =2^{y-x}.$

$\begin{array}{1 1}(A)\; 2^{-y}-2^{-x}=c \\ (B)\; 2^{-y}+2^{-x}=c \\ (C)\; 2^{y}+2^{x}=c \\(D)\; 2^{-x}-2^{y}=c \end{array} $

Can you answer this question?

1 Answer

0 votes
  • A relation between the dependent and independent variable which when substituted in the differential equation reduces it to an identify is called a solution.
  • Form $\large\frac{dy}{dx}=f(x,y),$ when variable are seperable we express it in the form $f(x)dx=g(y) dy$
  • Then $\int g(y)dx=\int f(x) dx+c$
  • $\int a^xdx=(a^x/\log\;a)+c$
Given $\large\frac{dy}{dx}$$=2^{y-x}$
Clearly here the variables can be seperated to the form $f(x)dx=g(y)dy$
(ie) $\Large\frac{dy}{dx}=\Large\frac{2^y}{2^x}$
On seperating the variables we get,
=>$2^{-y} dy =2^{-x} dx$
Integrating on both sides we get
$\int 2^{-y}dy=\int 2^{-x} dx$
$\int a^xdx=\Large\frac{a^x}{\log\;a}$$+c$
$\Large\frac{2^{-y}}{\log 2}=\frac{2^{-x}}{\log 2}$$+c$
=>$ 2^{-y}=2^{-x}+c$
or $ 2^{-y}-2^{-x}=c$
answered May 7, 2013 by meena.p
Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App