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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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The general solution of the differential equation $(e^x+1)ydy=(y+1)e^xdx$ is

\[(A)\;(y+1)=k(e^x+1) \quad (B)\;y+1=e^x+1+k \quad (C)\;y=log\{k(y+1)(e^x+1)\} \quad (D)\;ylog{\frac{e^x+1}{y+1}}+k\]

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1 Answer

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Toolbox:
  • A linear differential equation is of the form $\large\frac{dy}{dx}$$=f(x)$, then it can be solved by seperating the variables.
Step 1:
Given $(e^x+1)y\;dy=(y+1)e^xdx$
Seperating the variables we get,
$\large\frac{y\;dy}{y+1}=\large\frac{e^x}{e^x+1}$$dx$
On integrtaing we get,
$\int \large\frac{y\;dy}{y+1}=\int \large\frac{e^x}{e^x+1}$$dx$
Consider the L.H.S
Add and subtract 1 to the numerator
$\int \large\frac{(y+1)-1}{y+1}$$dy$
On seperating the terms we get,
$\int dy-\int \large\frac{1}{y+1}$$dy$
$=> y-\log(y+1)$
Step 2:
Now Consider R.H.S
$\int \large \frac{e^x}{e^x+1}$$dx$
Put $ e^x+1=t$ On differentiating w.r.t x
$e^xdx=dt$ on substituting this
Hence $\int \large \frac{dt}{t} $$=\log t$
$=\log |(e^x+1)|$
Step 3:
Now combining the terms we get,
$y=\log (y+1)=\log(e^x+1)+\log c$
$y=\log [k(y+1)(e^x+1)]$
Hence the correct option is $C$
answered May 21, 2013 by meena.p
 

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