y=f(x)===> y.y'' - ( y' )^2 = 0 what is y ?

Question

Its differential math problem

Answer 1

y.y'' - ( y' )^2 = 0 . This Eq. can be written as :
yy'' = (y')^2 or y''/y' = y'/y.
Integrating the two sides, we get:
ln(y') =ln(y) + ln(c), where c or ln (c) is a constant.
This leads to : y' =c y. Now we rewrite it as:
y'/y = c. integrating it again, we get:
ln(y) = cx + d, where d is a constant.
This leads to y= A exp(cx), where A=exp(d).
Answer: y = A exp( cx). Should we check the answer?
y'= Ac exp(cx) and y'' = A c^2 exp(cx),
Then y.y'' - ( y' )^2
=A^2 c^2 exp(2cx) -A^2 c^2 exp(2cx) = 0.
So the given Eq. is satisfied.Good

Answer 2

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One solution is y = cosh x. I let you find the others =) edit differentiate you get y'y'' + yy''' - 2y'y'' = 0 so yy''' = y'y'' or y'/y = y'''/y''. This gives ln y'' = ln y + C or y'' = e^C * y. You know solutions to such equations are of the form y = a cosh ut + b sinh u t with u^2 = e^C. Just plug into your equation and see what it gives for a,b,u. Can't be hard.

Answer 3

Yy Y 2 0

Answer 4

it looks like y'/y is constant right? So ln y = Cx + C' or y= K e^{ax}

Answer 5

ugg!

You can tell all the kids just got out of school! They're asking us all to do their homework!!!

Answer 6

I believe the answer is (_)_)////////////////////////D