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Miscellaneous Exercise 6(II) · Q138

Q.Find the particular solution: y(1+log⁡x)=(log⁡xx)dydxy(1+\log x)=(\log x^x)\dfrac{dy}{dx}, when y(e)=e2y(e)=e^2.

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y(1+log⁡x)=(log⁡xx)dydxy(1+\log x)=(\log x^x)\dfrac{dy}{dx}, and log⁡xx=xlog⁡x\log x^x=x\log x, so dydx=y(1+log⁡x)xlog⁡x\dfrac{dy}{dx}=\dfrac{y(1+\log x)}{x\log x} — exactly the reciprocal of Exercise 6.3 Q3(iii)'s equation, with the same initial condition. So (as solved there) $y=e\ …

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