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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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The solution of the differential equation $\cos x\sin y\;dx+\sin x\cos y\;dy=0$ is:

\[(A)\;\frac{\sin x}{\sin y}=c \quad (B)\;\sin x\sin y=c\quad(C)\;\sin x+\sin y=c \quad (D)\;\cos x\cos y=c\]

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  • If a linear differential equation is of the form $\large\frac{dy}{dx}$$=f(x)$ then it can be solved by seperating the variables and integrating it
Given $\cos x \sin ydx+\sin x \cos y dy=0$
This can be written as
$-\cos x \sin y dx=\sin x \cos y dy$
$=>\sin x \cos y dy =-\cos x \sin y dx$
Now seperating the variables we get
$\large\frac{\cos y dy}{\sin y}=\frac{-\cos x dx}{\sin x}$
Now integrating on both sides,
$\large\int \frac{\cos y dy}{\sin y}=\int \frac{-\cos x dx}{\sin x}$
Put $\sin y=t$ and differentiate it w.r.t y we get $\cos y dy=dt$
Similarly put $\sin x=u$ and differentiate it w.r.t x we get $\cos x dx=du$
Now substituing this we get
Hence $ \int \large\frac{dt}{t}=-\int \frac{du}{u}$
$=\log |t|=\log |u|+\log c$
Substituting for t and u
$\log |\sin y|=-\log |\sin x|+\log c$
$=> \log |\sin y|+\log |\sin x|=\log c$
or $\sin y.\sin x =c$
Hence $B$ is the correct option
answered May 21, 2013 by meena.p

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