JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

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Application +4 | -1

The degree of the differential equation satisfying

\sqrt {1 - {x^2}} + \sqrt {1 - {y^2}} = a\left( {x - y} \right)

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Correct option: A

Option A (1) is correct because after simplification and differentiation, the highest order derivative is of order 1.

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y=y(x)

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

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