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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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Verify that the given function(explicit or implicit)is a solution of the corresponding differential equation $y=e^x+1\qquad:\;y''-y'=0$

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Toolbox:
  • Differentiation of $e^x = e^x$
Step 1:
Given $y = e^x + 1$
Differentiating on both sides we get,
$y' = e^x + 0.$--------(1)
Step 2:
Differentiating again we get,
$y'' = e^x.$------(2)
Now subtracting equ (2) from equ (1)
$y'' - y' =e^2 -e^2 = 0$
Hence it is solution for the given function.
answered Aug 19, 2013 by sreemathi.v
 

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