+91-8076753736 [email protected]

-->

Differential Equations

Question

Wired Faculty App

Solve the following differential equation:
$dy over dx plus 1 space equals space straight e to the power of straight x plus straight y end exponent$

Solution

The given differential equation is

dy over dx plus 1 space equals straight e to the power of straight x plus straight y end exponent

...(1)
Put

straight x plus straight y space equals straight t comma space space space space space space space space space space space therefore space space space space 1 plus dy over dx equals space dt over dx space space or space space space dy over dx space equals dt over dx minus 1

therefore space space space from space left parenthesis 1 right parenthesis comma space space dt over dx minus 1 plus 1 space equals space straight e to the power of straight t space space space space space or space space space dt over dx space equals straight e to the power of straight t rightwards double arrow space space space space 1 over straight e to the power of straight t dt space equals space dx space space space rightwards double arrow space space space integral straight e to the power of negative straight t end exponent space dt space equals space integral 1. space dx therefore space space space space space space space space space space fraction numerator straight e to the power of negative straight t end exponent over denominator negative 1 end fraction space equals space straight x plus straight c space space space space rightwards double arrow space space space minus straight e to the power of negative left parenthesis straight x plus straight y right parenthesis end exponent space equals space straight x space plus straight c space space which space is space required space solution. space space space

Some More Questions From Differential Equations Chapter

Mock Test Series

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation