Question

Find the particular solution of the differential equation (x^2/y^2-3)(dy/dx)=1/2y satisfying the initial condition y(1)=sqrt(4). solve for y. your answer should be a function of x.

Answer 1

\[\int\limits_{}^{} \frac{1}{x^{2}} dx = 2 \int\limits_{}^{} \frac{1}{y(y^2-3)} dy\]
are you familiar with partial fractions decomposition?

Answer 2

it looks like separation of variables can be done

Answer 3

not really. i know the formula for exponential decay

Answer 4

in order to separate variables, we need to have all y's or all x's right

Answer 5

no we need to be able to write f(y) dy =g(x) dx to use separation of variables

Answer 6

john already set it up gj john

Answer 7

ok. where do i go from there?

Answer 8

use partial fractions for right hand side

Answer 9

\[\int\limits_{}^{} \frac{2}{y(y^{2}-3}} = \frac{A}{y} + \frac{B}{y-\sqrt{3}} +\frac{C}{y+\sqrt{3}\]

Answer 10

bah

Answer 11

use A/y + B/y-sqrt(3) + C/y+sqrt(3)

Answer 12

2= A (y^2-3) + B y (y+sqrt3) + C y (y-sqrt3) and find A,B and C by plugging in y=0 sqrt(3) and - sqrt(3)

Answer 13

isn't the 2y in the numerator??

Answer 14

assuming 1/2y is really 1/(2y)

Answer 15

if it is, that makes life so much easier :)

Answer 16

Yes

Answer 17

lol

Answer 18

\[-\frac{1}{x}+C = \ln(y^2-3) \]

Answer 19

solve for C?

Answer 20

\[y= \sqrt{e^{-1/x+c} +3}\]

Answer 21

@bbaker plug in x=1 and y=2 and solve for C

Answer 22

4=e^-1 * e^c+3...............0=-1+C and C=1

Answer 23

use the equation before the second one john wrote
i think it is much easier to see what C is in that equation if x=1 and y=2

Answer 24

-1/x+c=ln(y^2-3)
-1/1+c=ln(2^2-3)
-1+c=ln(1)
-1+c=0
c=1

Answer 25

if c=1 then y= 3+e^-1/x+1?

Answer 26

if c=1, we can write
-1/x+1=ln(y^2-3)

Answer 27

we should put absolute value around that y^2-3 by the way

Answer 28

Though typically people use the exponential form.
I think you should add some parens to make your answer more clear.
and put a square root around it

Answer 29

from there solve for y?

Answer 30

e^(-1/x+1)=y^2-3
3+e^(-1/x+1)=y^2

take square root of both sides
don't forget to write plus or minus

Answer 31

so as a function of x y=sqrt(3+e^(-1/x+1)) or -sqrt(3+e^(-1/x+1))