JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

1 Easy

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Application +4 | -1

The solution curve of the differential equation,

(1 + {e^{ - x}})(1 + {y^2})\frac{{dy}}{{dx}} = {y^2}

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

View answer and explanation

Correct option: C

{y^2} = 1 + y{\log _e}\left[ {\frac{{1 + {e^x}}}{2}} \right]

Option C (

{y^2} = 1 + y{\log _e}\left[ {\frac{{1 + {e^x}}}{2}} \right]

) is correct because it represents the solution to the given differential equation that passes through the point (0, 1).

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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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\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)