Question

Solve the differential equation: (dy/dx)=(2y/x)

Answer 1

\[\frac{1}{2y}dy = \frac{1}{x}dx \]

Separable differential equation.

Answer 2

and then the antiderivative of (1/2y)dy
= the antiderivative of (1/x)dx

Answer 3

yes, you will get something logarithmic on both sides, usually they want you to term it as a function of y again.

Answer 4

okay, i got to: (1/2)lny= lnx+ C

Answer 5

\[ \frac{1}{2}\ln y = \ln x +C = \ln \sqrt{y} \]

Answer 6

Not sure if they want you to do that though, but you could use exponential properties now to get rid of the natural log terms and then square the entire solution.

Answer 7

so that leaves me with e(lnx+c) = e(ln sqrt y)
x+ Ce = sqrt y

Answer 8

?

Answer 9

\[\Large e^{\ln x} = x\]

Answer 10

\[\large \sqrt{y}=x+e^c \]

Answer 11

Note the e^C is usually just denoted as another constant, so you can write C again.

Answer 12

\[ \large y=x^2+2e^{c}x+e^{2c} \] I would keep it like that, or you can substitute arbitrary constants, usually they accept it like that.

Answer 13

ah okay. so then I get y=x^2 + C

Answer 14

i'm not familiar with how wolfram alpha denotes C
http://www.wolframalpha.com/input/?i=solve+differential+equation+%28dy%2Fdx%29%3D%282y%2Fx%29

Answer 15

let me check.

Answer 16

is x^2 + c equilavent to their answer?

Answer 17

yes I see that, well I guess wolfram ignores the constants we have derived above, and just takes y=cx^2 as the most general solution, it's not wrong when you write it as we did it above though.

Answer 18

yes it is.

Answer 19

okay that helps, thanks a lot

Answer 20

you're welcome.