Email
Chat with tutor
logo

Ask Questions, Get Answers

X
 
Questions  >>  CBSE XII  >>  Math  >>  Differential Equations
Answer
Comment
Share
Q)

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} $

1 Answer

Comment
A)
Toolbox:
  • 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}$$=2^{y-x}$
Clearly here the variables can be seperated to the form $f(x)dx=g(y)dy$
(ie) $\Large\frac{dy}{dx}=\Large\frac{2^y}{2^x}$
On seperating the variables we get,
$\Large\frac{dy}{2^y}=\frac{dx}{2^x}$
=>$2^{-y} dy =2^{-x} dx$
Integrating on both sides we get
$\int 2^{-y}dy=\int 2^{-x} dx$
$\int a^xdx=\Large\frac{a^x}{\log\;a}$$+c$
$\Large\frac{2^{-y}}{\log 2}=\frac{2^{-x}}{\log 2}$$+c$
=>$ 2^{-y}=2^{-x}+c$
or $ 2^{-y}-2^{-x}=c$
Help Clay6 to be free
Clay6 needs your help to survive. We have roughly 7 lakh students visiting us monthly. We want to keep our services free and improve with prompt help and advanced solutions by adding more teachers and infrastructure.

A small donation from you will help us reach that goal faster. Talk to your parents, teachers and school and spread the word about clay6. You can pay online or send a cheque.

Thanks for your support.
Continue
Please choose your payment mode to continue
Home Ask Homework Questions
Your payment for is successful.
Continue
Clay6 tutors use Telegram* chat app to help students with their questions and doubts.
Do you have the Telegram chat app installed?
Already installed Install now
*Telegram is a chat app like WhatsApp / Facebook Messenger / Skype.
...