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Mathematics 18 Online
OpenStudy (anonymous):

Find the general solution of xy'+(ln x)y=0.

OpenStudy (anonymous):

@ganeshie8

ganeshie8 (ganeshie8):

looks separable ?

OpenStudy (anonymous):

But how do I do this?

ganeshie8 (ganeshie8):

\[\large \begin{align} \\ xy'+(\ln x)y &=0 \\\; \\ x\frac{dy}{dx} &=-(\ln x)y \\\; \\ \frac{dy}{y} &=-\frac{\ln x}{x} dx \\ \; \\ \int \frac{dy}{y} &=-\int \frac{\ln x}{x} dx \\ \; \\ \end{align}\]

OpenStudy (anonymous):

Let me try. Please don't leave.

ganeshie8 (ganeshie8):

integrate ^^

ganeshie8 (ganeshie8):

ok good luck !

OpenStudy (anonymous):

How do I integrate ln(x)/x?

OpenStudy (xapproachesinfinity):

yes! don't forget negative sign haha

OpenStudy (xapproachesinfinity):

Do by part haha

ganeshie8 (ganeshie8):

u susbtitution may work :) u = ln(x)

OpenStudy (xapproachesinfinity):

actually u is easy

OpenStudy (xapproachesinfinity):

try and see!

OpenStudy (anonymous):

Yes, I got the right answer. Thank you, ganeshie8.

OpenStudy (xapproachesinfinity):

Good!

ganeshie8 (ganeshie8):

yw :)

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