Question

hi. can someone help me explain how to get a general soln for (x^3+(y/x)dx + (y^2+lnx)dy=0?

Answer 1

sorry cant use this.

Answer 2

\[\left(x^3 + \frac{y}{x}\right)\:dx + (y^2 + \ln x)\:dy = 0\]my first guess would be exact DE

Answer 3

Oh hey! welcome back!

Answer 4

hey, exraven, could you verify the general soln in the previous thread that we were both in? I want to make sure that's correct.

Answer 5

I have already written the general soln there

Answer 6

oh. I plugged in the value for z like you said. did i end up with the correct soln?

Answer 7

disregard that last comment. just saw that post just now.

Answer 8

im going to go ahead and try the exact method and ill check back with my answer. i hope you don't mind.

Answer 9

ok, go ahead

Answer 10

the exact method for the new problem*

Answer 11

ok, im back

Answer 12

should the general soln be ln|x|+ln|x|+c

Answer 13

are you sure?

Answer 14

i said dm/dy = 1/x =dn/dx

Answer 15

and then I said this proved that this is an exact eq and that there exists f(x,y) such that df/dx=M(x,y) and df/dy=N(x,y)

Answer 16

and then I proceeded to integrate

Answer 17

\[M = x^3 + \frac{y}{x}\]\[N = y^2 + \ln x\]\[\frac{\partial M}{\partial y} = \frac{1}{x}\]\[\frac{\partial N}{\partial x}=\frac{1}{x}\]the DE is exact

Answer 18

wait! i just noticed something. the integral of df/dx with respect to x of (1/x) is -1/x^2 not lnx|

Answer 19

yup. so far so good. I have what you wrote.

Answer 20

then there exist the solution\[f(x,y) = C\]such that\[\frac{\partial f}{\partial x} = M\]and\[\frac{\partial f}{\partial y} = N\]

Answer 21

\[\frac{\partial f}{\partial x} = x^3 + \frac{y}{x}\]integrate both sides w.r.t x\[f(x,y) = \frac{1}{4}x^4 + y \ln x + h(y)\]

Answer 22

what about df/dy? i integrated that w.r.t. y and i got y^3/3 +ylnx.

Also, how do we solve for h? would that eventually turn into C?

Answer 23

\[h(y) = \int\limits y^2 \:dy = \frac{1}{3}y^3\]\[\frac{\partial f}{\partial y} = \ln x + h'(y) = N\]\[\ln x + h'(y) = y^2 + \ln x\]\[h'(y) = y^2\]\[h(y) = \int\limits y^2 \: dy=\frac{1}{3}y^3\]we don't need to add the constanct because the solution already has the constant C\[f(x,y) = \frac{1}{4}x^4 + y \ln x + \frac{1}{3}y^3 \]therefore, the solution\[\frac{1}{4}x^4 + y \ln x + \frac{1}{3}y^3 = C\]

Answer 24

disregard the first line

Answer 25

the h(y)=...? that line?

Answer 26

yes

Answer 27

everything is almost crystal clear. had I left that constant (ylnx) in my f(x,y), would that be wrong?