Question

The solution of differential equation $\cos x.\sin ydx+\sin x.\cos ydy=0$ is

• A. $\frac{\sin x}{\sin y}=c$
• B. $\sin x.\sin y=c$
• C. $\sin x+\sin y=c$
• D. $\cos x.\cos y=c$

Answer 1

Answer: (B)

$\sin x.\sin y=c$

Given,

$(\cos x.\sin y)dx+(\sin x.\cos y)dy=0$

$\Rightarrow (\cos x.\sin y)dx=-\sin x.\cos ydy$

$\Rightarrow \frac{\cos x}{\sin x}dx=\frac{-\cos x}{\sin y}dy$

( All we have done is separated the variables )

$\Rightarrow \cot xdx=-\cot ydy$

( Now integrating both the sides we get )

$\Rightarrow \int \cot xdx=-\int \cot ydy$

$\Rightarrow \int \cot ydy=-\int \cot xdx$

$ln\sin y=-ln\sin x+lnC$ ( $ln$ C is interpretation const. )

$ln\sin y=ln\frac{1}{\sin x}+lnC$

$ln\sin y=ln\frac{C}{\sin x}$ $(\because lna+lnb=lnab)$

$\sin y=\frac{C}{\sin x}$

$\Rightarrow \sin x.\sin y=C$

Hence, the answer is $\sin x.\sin y=C.$

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