Question

Solve the differential equation y dx+(2xy-e^(-2y))dy=0. I know the integrating factor is e^2y/y but what do I do next?

Answer 1

\[\frac{ e^{2y} }{ y }y dx+\frac{ e^{2y} }{ y }(2xy-e^{-2y})dy=0\]
scale equation and simplify

Answer 2

can you do that?

Answer 3

Wait a minute.

Answer 4

ok :)

Answer 5

I got e^2y dx+2x(e^2y)-1/y dy=0, now what?

Answer 6

Do I integrate now?

Answer 7

oh i'm sorry for the late reply

Answer 8

:\[M_1(x,y)=e ^{2y} and N(x,y)=2xe ^{2y}-1/y\]

Answer 9

I already know that. From e^2y dx+2x(e^2y)-1/y dy=0, do I need to integrate? The answer is xe^2y-ln abs(y)=c, y=0. How do I get that?

Answer 10

\[M_1(x,y)=\int\limits_{}^{} M (x,y) dx=\int\limits_{}^{}e^{2y} dx= xe ^{2y}\]

Answer 11

I don't get what you wrote.

Answer 12

i just confirmed

Answer 13

but looks like,constant of integration can't be added on what i wrote above

Answer 14

Why is y=0 part of the answer?

Answer 15

this is not the answer?

Answer 16

The answer is xe^2y-ln abs(y)=c, y=0. But I don't know how to get y=0 as the answer.

Answer 17

X is not dependent

Answer 18

How so?

Answer 19

\[M_1(x,y) + \int\limits_{}^{}[ N(x,y)-\frac{ ∂ }{ ∂y } M_1(x,y)] dy=xe^{2y}-\int\limits_{}^{}\frac{ 1 }{ y }dy=xe ^{2y}-In|y|+C=0\]

Answer 20

btw, C is an arbitrary constant

Answer 21

what, it didn't show fully

Answer 22

\[\frac{ 1 }{ y }dy=xe ^{2y}-In|y|+C=0\]

Answer 23

I still don't know how to get y=0.

Answer 24

Basically, the original equation in the form dy/dx = f(x, y)

Answer 25

e^integral(2dy) = e^(2y)
(x*e^(2y))' = e^(2y)*e^(-2y)/y = 1/y
Integrate:x*e^(2y) = ln(y) + c
then you'll get for the solution: x(y) = (ln(y) + c) * e^(-2y)

Answer 26

M=e^2y and N=2xe^2y-1/y,
that means My=Nx=2e^2y