Email
Chat with tutor
logo

Ask Questions, Get Answers

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

The solution of $\large\frac{dy}{dx}$$+y=e^{-x},y(0)=0$ is

\[(A)\;y=e^{-x}(x-1) \quad (B)\;y=xe^x \quad (C)\;y=xe^x+1 \quad (D)\;y=xe^{-x}\]

1 Answer

Comment
A)
Toolbox:
  • A linear differential equation of the form $\large\frac{dy}{dx}+Py=Q$, has the general solution as $ye^{\int pdx}=\int Q e^{\int pdx}.dx+c$
Given $\large\frac{dy}{dx}$$+y=e^{-x}$
Clearly this is a linear differential equation of the form $ \large\frac{dy}{dx}$$+Py=Q$ where $P$ and $Q$ are $ 1$ and $e^{-x}$ respectively
The solution to this equation is
$y.e^{\int pdx}=\int Q.e^{\int pdx}+c$
Where $ e^{\int pdx} $ is the integrating factor(I.F)
$\int pdx=\int dx$ on integrating we get
$=x$
Hence $I.F=e^{\int pdx}=e^x$
Now the required solution is
$ye^x=\int e^{-x}.e^xdx+c$
$=> ye^{x}=\int dx+c$
On integrating we get
$ye^x=x+c$
It is given $y(0)=0=>when \;x=0,y=0$
Now substituting the value of x and y we get,
$0.e^x=0+c$
Hence the solution is
$ye^x=x=>y=xe^{-x}$
Hence the correct option is $D$
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.
...