If $x d y-y d x+x \cos \left(\log _e^x\right) d x=0, y(1)=1$, then $y(e)=$

JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

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Application +4 | -1

x d y-y d x+x \cos \left(\log _e^x\right) d x=0, y(1)=1

, then

y(e)=

View answer and explanation

Correct option: B

e(1-\sin 1)

Option B (e(1-sin 1)) is correct because solving the differential equation and applying the initial condition y(1)=1 leads to y(e) = e(1-sin 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

y(1)=1

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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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y=y_{2}(x)

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y_{1}(0)=0

y_{2}(0)=1

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)