Ask your own question, for FREE!
Mathematics 12 Online
OpenStudy (wade123):

@phi

OpenStudy (phi):

yes. y is inversely proportional to x means y = k/x in this case dy/dx is inversely proportional to y^4 means dy/dx = k/y^4

OpenStudy (phi):

so solve multiply both sides by y^4 multiply both sides by dx

OpenStudy (wade123):

yay(: thanks for rechecking my work(:

OpenStudy (phi):

you should get (1/5) y^5 = k x + C

OpenStudy (wade123):

oh wait for my final answer i got dy/dx=k/y^4

OpenStudy (wade123):

@phi

OpenStudy (phi):

the question says Write and then solve

OpenStudy (wade123):

oh so that wasnt my final answer..

OpenStudy (wade123):

ohh! youre correct , sorry

OpenStudy (phi):

you did the "write" part now you have to do the solve

OpenStudy (phi):

but it's the same process as the previous post (and I posted the steps up above)

OpenStudy (wade123):

yeah i got that, so it would be (1/5)y^5=kx+C?

OpenStudy (phi):

can you post the steps?

OpenStudy (wade123):

just multipy both sides by dx and y^4

OpenStudy (phi):

and what do you get ?

OpenStudy (wade123):

y^4dy=kdx

OpenStudy (phi):

now what is the next step ?

OpenStudy (phi):

if you see "infinitesimals" like dy or dx I would think about integrating

OpenStudy (wade123):

integral?

OpenStudy (wade123):

okay ill fin the integral(:

OpenStudy (phi):

yes, because that is how you "get rid of dx" for example

OpenStudy (wade123):

(1/5)y^5=kx+C(:

OpenStudy (phi):

unless they give you a couple of values for (x,y), we cannot solve for the C or k

OpenStudy (wade123):

thank you so much, you explain it very well, would you mind helping me on two more questions? later on? or will you be leaving soon?

OpenStudy (phi):

I can go another 15 minutes now, otherwise I'm done for today.

OpenStudy (wade123):

okay(:

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!