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Home  >>  CBSE XII  >>  Math  >>  Differential Equations

Find the general solution $\large\frac{dy}{dx}$$+y=1(y\neq1)$

1 Answer

Toolbox:
  • $\int\large\frac{ 1}{(y - 1)} $$= \log(y - 1)$
Step 1:
Given : $\large\frac{dy}{dx}$$ + y = 1$
$\large\frac{dy}{dx} $$= 1-y$
seperating the variables we get
$dy(y - 1)= - dx$
Step 2:
Integrating on both sides
Using the information from the tool box
$\log(y-1) = -x - \log A$
$\log_eA(y-1) = -x$
Writing this in the exponential form
$Ae^{-x} = y - 1.$
This is the required solution.
answered Aug 15, 2013 by sreemathi.v
 
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