JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

1 Easy

1 of 1

Application +4 | -1

The equation of the curve whose slope is

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

and which passes through the point (1,0) is

View answer and explanation

Correct option: A

xy + x + y - 1 = 0

Option A (

xy + x + y - 1 = 0

) is correct because it is the equation of the curve obtained by solving the differential equation and using the given point (1,0).

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

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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)