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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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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} $

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Toolbox:
  • 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
 

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