Solve x d x-y d y-x d y-y d x-x2-y2

The correct option is D 1/2(x^2+y^2)=tan^ 1 ( y/x )+c1/2d(x^2+y^2)=d ( tan^ 1 ( y/x ) ) Integrating both sides w.r.t. x, we get 1/2(x^2+y^2)=tan^ 1 ( y/x )+ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Standard Formulae - 1

Solve x d x+y...

Question

Solve $x d x + y d y = \frac{x d y - y d x}{x^{2} + y^{2}}$

\frac{1}{2} (x^{2} + y^{2}) = {tan}^{- 1} (\frac{y}{x}) + c

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{1}{2} (x^{2} + y^{2}) + {tan}^{- 1} (\frac{y}{x}) + c = 0

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

\frac{1}{2} (x^{2} - y^{2}) = {tan}^{- 1} (\frac{y}{x}) + c

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

(x^{2} + y^{2}) = {tan}^{- 1} (\frac{y}{x}) + c

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

Open in App

Solution

The correct option is A $\frac{1}{2} (x^{2} + y^{2}) = {tan}^{- 1} (\frac{y}{x}) + c$
$\frac{1}{2} d (x^{2} + y^{2}) = d ({tan}^{- 1} (\frac{y}{x}))$

Integrating both sides w.r.t. x, we get

$\frac{1}{2} (x^{2} + y^{2}) = {tan}^{- 1} (\frac{y}{x}) + c$

Suggest Corrections

Similar questions

Q. Solve :

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

Solve:

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

Q. Solve:

x d x + y d y + \frac{x d y - y d x}{x^{2} + y^{2}} = 0

Q. Solve:

x d y + y d x + \frac{x d x + y d y}{x^{2} + y^{2}} = 0

Q. Solve the equation:-

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

Join BYJU'S Learning Program

Solve x dx+y dy=x dy−y dxx2+y2

The correct option is A 12(x2+y2)=tan−1(yx)+c12d(x2+y2)=d(tan−1(yx)) Integrating both sides w.r.t. x, we get 12(x2+y2)=tan−1(yx)+c

Solve $x d x + y d y = \frac{x d y - y d x}{x^{2} + y^{2}}$

The correct option is A $\frac{1}{2} (x^{2} + y^{2}) = {tan}^{- 1} (\frac{y}{x}) + c$
$\frac{1}{2} d (x^{2} + y^{2}) = d ({tan}^{- 1} (\frac{y}{x}))$

Integrating both sides w.r.t. x, we get

$\frac{1}{2} (x^{2} + y^{2}) = {tan}^{- 1} (\frac{y}{x}) + c$