y''+y'/x-y/x^2=0 y(1)=1, y(R)=n

y''+y'/x-y/x^2=0 y(1)=1, y(R)=n
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2 Answers

Best answer

Let p=y/x, then p'=dp/dx=y'/x-y/x^2.

y"+y'/x-y/x^2=0=y"+p' so y"=-p'.

Integrating: y'=-y/x.

So dy/dx=-y/x; dy/y=-dx/x and ln(y)=-ln(x)+C, ln(y)+ln(x)=C or xy=A, a constant (A=e^C).

y(1)=1 so A=1 and xy=1. y(r)=1/r=n.

CHECK

xy=1 so y=1/x=x^-1; y'=-x^-2=-1/x^2; y"=2x^-3.

So y"+y'/x-y/x^2=(2x^-3)-(x^-3)-(x^-3)=0. OK.

by Top Rated User (1.2m points)
OH, your to late! Anonymous probably went to http://brainly.com, so huh...
by Level 10 User (57.4k points)
HEY ROD! YOU SELECTED YOURSELF DIDN'T YOU!? TELL THE TRUTH!
by Level 10 User (57.4k points)
OK. I'll tell you the truth. I didn't select my answer. Someone else did.
by Top Rated User (1.2m points)

I just found out who selected Best Answer. It's in my email notifications and it was at 6:20am my time yesterday (Wednesday 21 December). It's still Wednesday for you so that makes it 1.20am your time. It was anonymous so it was probably the person who posted the question. (Thank you whoever you are.)

by Top Rated User (1.2m points)

And, no, my intention wasn't to steal anything from you; someone needed help and in this case I was able to provide it.

by Top Rated User (1.2m points)
Well I got nothing from returne I tried to help but ugh, you always get the best answers...
by Level 10 User (57.4k points)

if you want help faster, go to http://brainly.com and you'll able to get your question answered.

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by Level 10 User (57.4k points)

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