JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

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Application +4 | -1

The solution of

(x \mathrm{~d} x+y \mathrm{~d} y)\left(x^2+y^2\right)+(x \mathrm{~d} y-y \mathrm{~d} x)

=0

View answer and explanation

Correct option: D

\frac{x^2+y^2}{2}+\tan ^{-1}\left(\frac{x}{y}\right)=C

Option D,

\frac{x^2+y^2}{2}+\tan ^{-1}\left(\frac{x}{y}\right)=C

, is the correct solution. It is obtained by recognizing exact differentials and integrating.

More Questions from DIFFERENTIAL EQUATIONS

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y=y(x)

\left(3 y^{2}-5 x^{2}\right) y d x+2 x\left(x^{2}-y^{2}\right) d y=0

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x

{x^2} + {y^2} = 1

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x = A\cos 4t + B\sin 4t,

\frac{{{d^2}x}}{{d{t^2}}}

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y=y_{2}(x)

\frac{d y}{d x}=y+7

y_{1}(0)=0

y_{2}(0)=1

y=y_{1}(x)

y=y_{2}(x)

y=y(t)

\frac{d y}{d t}+\alpha y=\gamma e^{-\beta t}

\alpha>0, \beta>0

\gamma>0

\lim \limits_{{x \rightarrow \infty}} y(t)