JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

1 Easy

1 of 1

Application +4 | -1

x^3 \frac{d y}{d x}=y^3+y^2 \sqrt{y^2-x^2}

and

y=1

when

x=1

, then

y=2

when

5 x=\ldots (x \neq 0)

View answer and explanation

Correct option: A

Option A (8) is correct because solving the differential equation and using the given condition, we find that 5x = 8 when y = 2.

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y=y(x)

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x

{x^2} + {y^2} = 1

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x = A\cos 4t + B\sin 4t,

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

y_{1}(0)=0

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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)