Question

Differential Equations:
For what domain does y = x^2 satisfy the differential equation xy' = 2y

Book says answer, for all x except x = 0, but why is this if y is define for all x including 0??? book error?

Answer 1

When you separate the variables, you put x and y into the denominator; clearly they can't be zero in that position.

Answer 2

i dont understand what you mean by seperate the variables... is that some kind of method? because if so I didnt get to that yet so I dont see why the book would ask a question on something is didnt cover yet o_O

what i got is
\[y=x^2\]
\[y' = 2x\]

substituing in the DE:
\[x(2x) = 2(x^2)\]
\[2x^2 = 2x^2\]

Answer 3

it being a solution is not the issue; the domain for the solution is the issue

Answer 4

but y and y' are define for all x so whats the problem, how does x get in the denominator?

Answer 5

sorry i literally just started DE's today

Answer 6

then itll make more sense to you later :)

how do we find y=2x to be a solution?

Answer 7

and by that i meant y = x^2 :)

Answer 8

xy' = 2y

y' - 2y = 1/x

e^-2x y' - 2e^-2x y = e^-2x 1/x

integrate: e^-2x y' - 2e^-2x y = e^-2x 1/x

e^-2x y = integrate (e^-2x)/x dx

y = integrate (e^-2x)/x dx
-------------------
e^-2x

Answer 9

at any rate; you see that we get an x in a denominator position

Answer 10

not sure why you integrating the DE but I guess you're right and it will make sense later. So far the only definition the book gave for it being a solutions is this:
http://i.imgur.com/dGxWh.png
and there is no mention of integration but I guess ill move on and come back to this later when I get better understanding

Answer 11

An alternate method.....\[xy' = 2y \implies x \frac{dy}{dx}=2y \implies \frac{dy}{y}=2\frac{dx}{x}\implies \ln(y)=2\ln(x)+C=\ln(x^2)+C\]\[\implies y=C_1x^2\]Clearly if the constant C1 is one, we get the desired solution.