JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

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Application +4 | -1

The curve satisfying the differential equation

\frac{d y}{d x}=\frac{y\left(x+y^3\right)}{x\left(y^3-x\right)}

and passing through

(4,-2)

View answer and explanation

Correct option: B

y=-2 x

Option C,

y^3=-2 x

, is the correct curve obtained by solving the given differential equation and applying the given condition (4,-2).

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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

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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

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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)