JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

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Application +4 | -1

\left(x^2+y^2\right) d y=x y d x

. If

y\left(x_0\right)=e, y(1)=1

, then value of

x_0=

View answer and explanation

Correct option: A

\sqrt{3} e

Option A (

\sqrt{3} e

) is the correct value of

x_0

, found by solving the differential equation and applying the given conditions.

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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)