JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

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The equation of family of a curve is

{y^2} = 4a\left( {x + a} \right)

then differential equation of the family is

View answer and explanation

Correct option: C

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

Option C (

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

) is the correct differential equation obtained by differentiating the given equation and eliminating the arbitrary constant 'a'.

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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

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y=y_{1}(x)

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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)