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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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The solution of the equation $(2y-1)dx-(2x+3)dy=0$ is\[(A)\;\frac{2x-1}{2y+3}=k \quad (B)\;\frac{2y+1}{2x-3}=k \quad (C)\;\frac{2x+3}{2y-1}=k \quad (D)\;\frac{2x-1}{2y+1}=k \]

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Toolbox:
  • If the linear differential eqation is of the form $\large\frac{dy}{dx}$$=f(x)$, then it can be solved by seperaring the variable and integrating.
  • $\int \large\frac{dx}{x+1}$$= \log |x+1|+c$
Given $(2y-1)dx-(2x+3)dy=0$
Seperating the variables we get,
$\large\frac{dy}{(2y-1)}=\frac{dx}{(2x+3)}$
On integrating we get
$\int \large\frac{dy}{(2y-1)}=\int \frac{dx}{(2x+3)}$
$=>\large\frac{1}{2} $$\log(2y-1)=\large\frac{1}{2} $$\log (2x+3)+\log c$
$=>\log(2y-1)=\log 2c (2x+3)$
$ \large \frac{2y-1}{2x+3}$$=2c$
$ \large \frac{2x+3}{2y-1}=\frac{1}{2c}$$=k\qquad (where \;k=\large\frac{1}{2c})$
Hence $ \large \frac{2x+3}{2y-1}$$=k$
The correct option is $C$
answered May 20, 2013 by meena.p
 

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