If a curve passes through the origin and the slope of the tangent to it at any point $\left( {x,\,y} \right)$ is $\frac{{\left( {{x^2} - 4x + y + 8} \right)}}{{x - 2}}$, then this curve also passes through the point

JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

1 Easy

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Application +4 | -1

If a curve passes through the origin and the slope of the tangent to it at any point

\left( {x,\,y} \right)

\frac{{\left( {{x^2} - 4x + y + 8} \right)}}{{x - 2}}

, then this curve also passes through the point

View answer and explanation

Correct option: D

\left( {5,\,5} \right)

Option D (5,5) is correct because the curve passes through this point after solving the differential equation.

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y=y(x)

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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

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)