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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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Find the general solution $e^x\tan y\;dx+(1-e^x)\sec ^2y\;dy=0$

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Toolbox:
  • Method of substitution can be followed if the function inthe numerator is a derivative of the function of the denominator.
Step 1:
Given $e^x\tan xdx + (1 - e^x)\sec^2ydy = 0$
On rearranging we get,
$(1-e^x)\sec^2ydy = - e^x \tan y dx$
$\large\frac{\sec^2ydy}{\tan y} = \frac{- e^xdx}{1-e^x}$
Step 2:
Using the information in the tool box, we understand that differentiation of $\tan y$ is $\sec^2y$ on the LHS and differentiation of $1-e^x$ is $-e^x$ on the RHS.
Hence now taking tany as t, then $dt = \sec^2y$ and take $u$ as $(1 - e^x)$, then $du = -e^x$
Hence the function can be written as $\large\frac{dt}{t} = \frac{du}{u}$
Step 3:
Now integrating on both sides we get,
$\log t = \log u + \log C $
$\log \tan y = \log(1-e^x)C$
$\tan y = C(1-e^x)$
This is the required solution.
answered Aug 14, 2013 by sreemathi.v
 
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