Ask Questions, Get Answers

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

Find the general solution of the differential equation $(x+y)\large\frac{dy}{dx}$$=1$

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

Can you answer this question?

1 Answer

0 votes
  • The first order linear equation of the form $\large\frac{dx}{dy}$$ + Px = Q$ can be solved as follows:
  • (i) Write the given equation in the form of $\large\frac{dx}{dy}$$ + Px = Q$
  • (ii) Find the integrating factor (I.F) = $e^{\int Pdy}$.
  • (iii) Write the solution as $y(I.F) = \int Q(I.F) dy + C$
Step 1:
Using the given information in the tool box let us rewrite the given equation.
Divide throughout by $(x+y)$
$\large\frac{dy}{dx} = \frac{1}{(x+y)}$
Recirprocate on both sides we get,
$\large\frac{dx}{dy}$$ = x+y$
or $\large\frac{dx}{dy}$$ - x = y$
Here $P = -1$ and $Q = y$
To find the I.F integrate $P$
$\int Pdy = \int-dy = -y$
Step 2:
The required solution is $x.e^{-y} = \int y.e^{-y} dy + C$
$\int y.e^{-y}$ can be done by parts,
Let $u = y;du = dy$ and $dv = e^{-y}dy$ , hence $v = -e^{-y}$
$\int y.e^{-y}dy = (-ye^{-y}) - \int[- e^{-y}.dy]$
$\qquad\;\;\;\;\;\;\;= (-ye^{-y}) +\int e^ydy$
$\quad\;\;\;\;\;\;\;\;\;\;\;= -ye^{-y }+ (-e^{-y}) + C$
Hence the required solution is $x.e^y = (-ye^{-y}) -e^{-y} + C$
$x.e^y = -e^{-y}[y + 1] + C$
dividing throughout by $e^{-y}$ we get
$x = -y -1 + Ce^{-y}$
$x+y+1 = Ce^{-y}$
This is the required solution.
answered Jul 31, 2013 by sreemathi.v

Related questions

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