Find the solution of x-ydydxy-xdydx-xcos 2 x2-y2 -y3

The correct option is B tan ( x^2+y^2 )=x^2/y^2+c The given equation can be written as x dx+y dy/ ( y dx x dy )/y^2=y^2.x/y^3cos ^2 ( x^2+y^2 nbsp ...

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Standard XII

Mathematics

Variable Separable Method

Find the solu...

Question

Find the solution of $\frac{x + y \frac{d y}{d x}}{y - x \frac{d y}{d x}} = \frac{x {cos}^{2} (x^{2} + y^{2})}{y^{3}}$

tan (x^{2} + y^{2}) = \frac{x^{2}}{y^{2}} + c

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tan (x^{2} - y^{2}) = \frac{x^{2}}{y^{2}} + c

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tan (x^{2} + y^{2}) = \frac{y^{2}}{x^{2}} + c

No worries! We‘ve got your back. Try BYJU‘S free classes today!

tan (x^{2} - y^{2}) = \frac{y^{2}}{x^{2}} + c

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Solution

The correct option is B $tan (x^{2} + y^{2}) = \frac{x^{2}}{y^{2}} + c$
The given equation can be written as $\frac{x d x + y d y}{(y d x - x d y) / y^{2}} = y^{2} . \frac{x}{y^{3}} {cos}^{2} (x^{2} + y^{2})$
or ${sec}^{2} (x^{2} + y^{2}) \frac{1}{2} d (x^{2} + y^{2}) = \frac{x}{y} d (\frac{x}{y})$
integrating we get,
$\frac{1}{2} tan (x^{2} + y^{2}) = \frac{1}{2} {(\frac{x}{y})}^{2} + \frac{c}{2}$
or $tan (x^{2} + y^{2}) = \frac{x^{2}}{y^{2}} + c$

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Similar questions

Q. Assertion :The solution of

\frac{x + \frac{d y}{d x}}{y - x \frac{d y}{d x}} = \frac{x {cos}^{2} (x^{2} + y^{2})}{y^{3}}

tan (x^{2} + y^{2}) = \frac{x^{2}}{y^{2}} + c

Reason:

x d x + y d y = \frac{1}{2} d (x^{2} + y^{2})

\frac{y d x - x d y}{y^{2}} = d (\frac{x}{y})

Q. Find the solution of

\frac{x d x + y d y}{x d y - d x} = \sqrt{(\frac{a^{2} - x^{2} - y^{2}}{x^{2} + y^{2}})} .

The solution of differential equation $\frac{x + y \frac{d y}{d x}}{y - x \frac{d y}{d x}} = \frac{x {cos}^{2} (x^{2} + y^{2})}{y^{3}}$ is

Q. Solution of

\frac{x d x + y d y}{x d y - y d x} = \frac{\sqrt{1 - (x^{2} + y^{2})}}{\sqrt{(x^{2} + y^{2})}}

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Q. The solution of

\frac{x d x + y d y}{x d y - y d x} = \sqrt{\frac{1 - x^{2} - y^{2}}{x^{2} + y^{2}}}

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Find the solution of x+ydydxy−xdydx=xcos2(x2+y2)y3

The correct option is B tan(x2+y2)=x2y2+cThe given equation can be written as xdx+ydy(ydx−xdy)/y2=y2.xy3cos2(x2+y2) or sec2(x2+y2)12d(x2+y2)=xyd(xy) integrating we get,12tan(x2+y2)=12(xy)2+c2 or tan(x2+y2)=x2y2+c

Find the solution of $\frac{x + y \frac{d y}{d x}}{y - x \frac{d y}{d x}} = \frac{x {cos}^{2} (x^{2} + y^{2})}{y^{3}}$