JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

1 Easy

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Application +4 | -1

\frac{d^2 x}{d y^2}=\ldots \ldots

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Correct option: B

-\left(\frac{d^2 y}{d x^2}\right)\left(\frac{d y}{d x}\right)^{-3}

Option B is correct because it shows the correct application of the chain rule and inverse function differentiation to find the second derivative.

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

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y=y(t)

\frac{d y}{d t}+\alpha y=\gamma e^{-\beta t}

\alpha>0, \beta>0

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\lim \limits_{{x \rightarrow \infty}} y(t)