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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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Solve the differential equation $\Large\frac{dy}{dx}\normalsize+1=e^{x+y}$

$\begin{array}{1 1}(A)\;(x+c)e^{(\large x-y)}+1=0 \\ (B)\;(x+c)e^{(\large x+y)}-1=0 \\(C)\;(x+c)e^{(\large x+y)}+1=0 \\(D)\;(x-c)e^{(\large x+y)}-1=0\end{array} $

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Toolbox:
  • A linear differential equation is of the form $\large\frac{dy}{dx}$$=f(x+y),$ then $(x+y) $ can be substituted as z and $\large\frac{dy}{dx}=\frac{dz}{dx}-1$.Then the variable can be seperated and integrated, to get the solution.
  • $\int e^{-x}dx=-e^{-x}+c$
Given $\large\frac{dy}{dx}$$+1=e^{x+y}$-----(1)
Put $x+y=z$ on differentiating w.r.t x we get,
$1+\large\frac{dy}{dx}=\frac{dz}{dx}$$=>\large\frac{dy}{dx}=\frac{dz}{dx}$$-1$
Now substituting in equ(1) we get
$\large\frac{dz}{dx}$$-1+1=e^z$
=>$ \large\frac{dz}{dx}$$=e^z$
Now seperating the variables we get
$\Large\frac{dz}{e^z}$$=dx$
=>$e^{-z}dz=dx$
On integrating we get: $ \int e^{-z}dz=\int dx$
$-e^{-z}=x+c$
Substituting for z we get
$-e^{-(x+y)}=x+c$
Dividing throughout by $e^{-(x+y)}$ we get
$-1=e^{(x+y)}(x+c)$
Therefore $(x+c)e^{(\large x+y)}$$+1=0$
answered May 8, 2013 by meena.p
 
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