The solution of the differential equation $x{\rm{ }}dy - y{\rm{ }}dx = \sqrt {{x^3} + {y^2}} dx$, is

$x + \sqrt {{x^2} + {y^2}} = C{x^2}$

$\frac{{{x^2} - 1}}{{{x^2} + 1}}$

JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

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The solution of the differential equation

x\mathrm{ }dy - y\mathrm{ }dx = \sqrt {{x^3} + {y^2}} dx

, is

View answer and explanation

Correct option: D

None of these

Option B (

x

) is correct because it is the integrating factor of the given differential equation after rearranging it into the standard form.

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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

y(1)=1

\left|(y(2))^{3}-12 y(2)\right|

x

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

\left( {y' = \frac{{dy}}{{dx}},y'' = \frac{{{d^2}y}}{{d{x^2}}}} \right)

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

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

y=y_{1}(x)

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)