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The general solution of the differential equation\(\frac{\large dy}{\large dx}=e^{x+y}\)is

\[ \begin{array}{1 1}(A)\;e^x+e^{-y}=C\qquad(B)\;e^x+e^y=C\\(C)\;e^{-x}+e^y=C\qquad(D)\;e^{-x}+e^{-y}=C\end{array}\]

1 Answer

  • $e^{(x+y)}= e^x.e^y$
Step 1:
Given $\large\frac{dy}{dx}=$$e^{(x+y)}$
using the information in the tool box we get,
$\large\frac{dy}{dx} $$= e^x.e^y$
Now seperating the variables,
$\large\frac{dy}{e^y }=$$e^x.dx$
Step 2:
Now integrating on both sides we get,
$\int e^{-y}dy =\int e^x$
$-ye^{-y }= e^x + C$
$e^x + e^{-y} = C$
Hence the correct answer is A
answered Aug 13, 2013 by sreemathi.v

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