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Find the general solution of xy'+(ln x)y=0.
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@ganeshie8
looks separable ?
But how do I do this?
\[\large \begin{align} \\ xy'+(\ln x)y &=0 \\\; \\ x\frac{dy}{dx} &=-(\ln x)y \\\; \\ \frac{dy}{y} &=-\frac{\ln x}{x} dx \\ \; \\ \int \frac{dy}{y} &=-\int \frac{\ln x}{x} dx \\ \; \\ \end{align}\]
Let me try. Please don't leave.
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integrate ^^
ok good luck !
How do I integrate ln(x)/x?
yes! don't forget negative sign haha
Do by part haha
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u susbtitution may work :) u = ln(x)
actually u is easy
try and see!
Yes, I got the right answer. Thank you, ganeshie8.
Good!
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yw :)
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