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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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Find the equation of the curve passing through the point\(\bigg(0,\large\frac{\pi}{4}\bigg)\) whose differential equation is $\sin x\cos y\;dx+\cos x\sin y\;dy\;=\;0$

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  • A linear differential equation of the form dy/dx = F(x,y) where F(x,y) is of the form g(x).h(y) where g(x) is a fuunction of x and h(y) is a function of y, then this type of equation is called as variables seperable and the equation can be solved by seperating the variables.
Step 1:
Given :$\sin x\cos ydx+\cos x\sin ydy=0$
We can write above equation as
$\large\frac{dy}{dx} = -\large\frac{ (\sin x\cos y)}{(\cos x\sin y)}$
Using the information in the tool box, we understand that the given equation is variable seperable and it can be solved by seperating the variables.
Seperating the variables we get,
$\large\frac{\sin ydy}{\cos y }= -\large\frac{ \sin xdx}{\cos x}$
$\tan y dy=-\tan xdx$
Integrating on both sides we get,
$\int\tan y=-\int \tan x$
$\log\mid\sec y\mid=-\log\mid\sec x\mid+\log C$
$\log C=\log\mid\sec y\mid+\log\mid \sec x\mid$
$\log C=\log\mid\sec y.\sec x\mid$
$C=\sec y.\sec x$
Step 3:
Substituting the given values of x = 0 and $y = \pi/4$ to evaluate the value of C
$C=\sec\large\frac{\pi}{4}$$.\sec 0$
$C=1.\sqrt 2$
$C=\sqrt 2$
$\sqrt 2=\sec y.\sec x$
$\sec y=\large\frac{1}{\cos y}$
$\sqrt 2=\large\frac{\sec x}{\cos y}$
$\cos y=\large\frac{\sec x}{\sqrt 2}$
This is the required solution.
answered Jul 30, 2013 by sreemathi.v

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