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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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Find the general solution $(e^x+e^{-x})\;dy-(e^x-e^{-x})\;dx=0$

$\begin{array}{1 1} y = \log(e^x + e^{-x}) + C \\ y = \log(e^x - e^{-x}) + C \\ y = \large\frac{x}{\log(e^x )} + C\\y = 1 - \log(e^x + e^{-x}) + C \end{array} $

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1 Answer

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Toolbox:
  • If the numerator of a given function is the derivative of the denominator we can follow the substitution method.
Step 1:
Given:$ (e^x + e^{-x})dy - (e^x - e^{-x}) = 0$
$(e^x + e^{-x})dy = (e^x - e^{-x})dx$
seperating the variables we get
$dy =\large\frac{ (e^x - e^{-x})dx}{(e^x + e^{-x})}$
Step 2:
Using the information from the tool box
Let us follow the substitution method
Let $t = e^x + e^{-x}$ then $e^x - e^{-x} dx= dt$
$dy = \large\frac{dt}{t}$
Step 3:
Integrating on both sides
$y = \log t + C$
Substituting for t
$y = \log(e^x + e^{-x}) + C$
This is the required general equation.
answered Aug 15, 2013 by sreemathi.v
 
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