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$Let y (x) be the solution of the differential equation (xlogx) \frac{dy}{dx} + y = 2 xlogx, (x \geq 1) then y (e) is$

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Correct option is C

dydx+1x ln⁡xy=2 I.F. =e∫1x ln⁡xdx=eln(ln⁡x)=ln x Sol of the D.E is y(ln⁡x)=∫2 ln⁡xdx=2ln x∫1 dx-∫1x∫1dxdxy(ln⁡x)=2xln x-∫1dxy(ln⁡x)=2x(ln⁡x−1)+c put x=10=−2+c→c=2y(ln⁡x)=2x(ln⁡x−1)+2 When x=ey=2e(1−1)+2y=2

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Let y(x) be the solution of the differential equation (xlogx)dydx+y=2xlogx,(x≥1) then y (e) is

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Ready to Test Your Skills?

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$Let y (x) be the solution of the differential equation (xlogx) \frac{dy}{dx} + y = 2 xlogx, (x \geq 1) then y (e) is$