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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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Solution of the differential equation $\tan y\sec^2x dx+\tan x\sec^2y dy=0$ is \[(A)\;\tan x+\tan y=c \quad (B)\;\tan x-\tan y=c \quad(C)\;\frac{\tan x}{\tan y}=c \quad (D\;\tan x.tan y=c\]

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1 Answer

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Toolbox:
  • A linear differential equation of the form $\large\frac{dy}{dx}$$=f(x)$ can be solved by seperating the variables and then integrating it.
  • $\int \large\frac{dx}{x}$$=\log |x|+c$
$\tan y \sec ^2 xdx+\tan x\sec ^2 y dy=0$
This can be written as: $\tan y \sec ^2 x dx=-\tan x \sec ^2 y dy$
Seperate the variables
$\large\frac{\sec ^2 y dy}{\tan y}=-\large\frac{\sec ^2 x dx}{\tan x}$
Put $\tan y=t $ and $\tan x=u$ on differentiating with respect to x we get,
$\sec ^2 ydy=dt$ and $\sec^2 xdx=du$
Substituting this we get,
$\large\frac{dt}{t}=-\frac{du}{u}$
Integrate on both sides,
$\int \large\frac{dt}{t}=-\int \frac{du}{u}$
$=\log |t|=-\log |u|+\log |c|$
Substitute for t and u,
$\log |\tan y|=-\log |tan x|+\log c$
$\log (\tan y).(\tan x)=\log c$
=>$ \tan x.\tan y=c$
Hence $D$ is the correct answer
answered May 20, 2013 by meena.p
 

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