Question

Solve (x-y)dy=ydx help!

Answer 1

solve for differential equation

Answer 2

\(\dfrac{dy}{dx} = \dfrac{y}{x-y}\)

Answer 3

\(y' = \dfrac{y}{x-y}\)

Answer 4

sibstitute \(y = vx \implies y' = v + v'x \)

Answer 5

the equation becomes :

\(v + v'x = \dfrac{v}{1-v}\)

Answer 6

hmm so I plug in y=u(x)*x
got du/dx = u^2/ x(1-u)
how to switch it back?

Answer 7

separate variables

Answer 8

separate variables, integrate and solve \(v\) first

Answer 9

in the end substitute back the \(y\) using : \(y=vx\)

Answer 10

seperate variable and got
u= (u^2 / 1-u) lnx
still not seeing how you got the solution

Answer 11

@Rinru, you made a mistake somewhere:
\[y'=\frac{y}{x-y}~~\Rightarrow~~v+xv'=\frac{vx}{x-vx}\\
~~~~~~~~~~~~~~~~~~~~~~~~~~~~v+xv'=\frac{v}{1-v}\\
~~~~~~~~~~~~~~~~~~~~~~~~~~~~xv'=\frac{v}{1-v}-\frac{1-v}{1-v}\\
~~~~~~~~~~~~~~~~~~~~~~~~~~~~xv'=\frac{2v-1}{1-v}\\
~~~~~~~~~~~~~~~~~~~~~~~~~~~~\frac{1-v}{2v-1}dv=\frac{1}{x}~dx
\]

Answer 12

should get

\(\dfrac{1-v}{v^2} dv = \dfrac{1}{x} dx\)

right ? @SithsAndGiggles

Answer 13

\(\dfrac{1-v}{v^2} dv = \dfrac{1}{x} dx\)

\(\int \dfrac{1-v}{v^2} dv = \int \dfrac{1}{x} dx\)


\(\dfrac{-1}{v} - \ln|v|= \ln |x| + c\)

Answer 14

replace \(v\) by \(\dfrac{y}{x}\), and you're done.

Answer 15

\[
y'=\frac{y}{x-y}~~\Rightarrow~~v+xv'=\frac{vx}{x-vx}\\
~~~~~~~~~~~~~~~~~~~~~~~~~~~~v+xv'=\frac{v}{1-v}\\
~~~~~~~~~~~~~~~~~~~~~~~~~~~~xv'=\frac{v}{1-v}-v\\
~~~~~~~~~~~~~~~~~~~~~~~~~~~~xv'=\frac{v}{1-v}-\color{red}{\frac{v(1-v)}{1-v}}\\
~~~~~~~~~~~~~~~~~~~~~~~~~~~~xv'=\frac{v^2}{1-v}\\

\]

Answer 16

Oh right, sorry!

Answer 17

so the final solution is....
-lny-x/y = C
x=-(c+lny) it seem

Answer 18

\( \ln y +\dfrac{x}{y} = C\) looks good to me

Answer 19

anyway to check it?

Answer 20

there's too many sign change with the constant variable...

Answer 21

checking is easy :

take ur final solution,
\(\ln y +\dfrac{x}{y} = C \)

differentiate and find out \(\dfrac{dy}{dx}\)

Answer 22

you should get back ur differential equation