Question

Solve the differential equation by using substitution for homogeneous equations. see attachment.

Answer 1

I swear I'm sooo close to getting x = (y-x)ln(y-x)+c

Answer 2

just needs rearranging... maybe shifting the 1/y-x to the right...?

Answer 3

@wio

Answer 4

You have this differential equation?:
\[2\frac{dy}{dx}=\frac{2y-x}{y}\]

Answer 5

uumm let me recheck

Answer 6

it's y ' = 2y-x
------------
y

Answer 7

sorry lagging on the real whiteboard on here

Answer 8

\[y'=\frac{2y-x}{y}\]
???

Answer 9

yessssss

Answer 10

well...I can get:
\[y'=2-\frac{x}{y}\]
\[y'=2-v\]
\[v=\frac{x}{y}\rightarrow y=\frac{x}{v}\rightarrow y'=\frac{1}{v}+v'x\]
\[\frac{1}{v}+\frac{dv}{dx}x=2-v\]
\[\frac{1}{vx}+\frac{dv}{dx}=\frac{2-v}{x}\]
\[\frac{dv}{dx}=\frac{2}{x}-\frac{v}{x}-\frac{1}{vx}\]
\[\frac{dv}{dx}=\frac{1}{x}\left(2-v-\frac{1}{v}\right)\]
\[\frac{dv}{\left(2-v-\frac{1}{v}\right)}=\frac{dx}{x}\]
\[\int{\frac{1}{2-v-\frac{1}{v}}dv=\int{\frac{1}{x}}dx}\]

Answer 11

\[\int{\frac{1}{\frac{2v-v^2-1}{v}}}=ln(x)+c_1\]

Answer 12

wait a sec...I did that earlier, but I got a partial fraction that didn't go anywhere

Answer 13

yeah I remember doing it that way...a sign must have messed it up which was why my A's and B's canceled

Answer 14

\[\int{\frac{v}{-v^2+2v-1}}dv=ln(x)+c_1\]
\[\int{\frac{v}{-(v-1)^2}}dv=ln(x)+c_1\]
Just take your time integrating that
\[-\left(\frac{1}{1-v}+ln(v-1)\right)=ln(x)+c_1\]
\[-\left(\frac{1}{1-\frac{x}{y}}+ln\left(\frac{x}{y}-1\right)\right)=ln(x)+c_1\]

Answer 15

\[\frac{-y}{y-x}-[ln(x-y)-ln(y)]=ln(x)+c_1\]
...I think

Answer 16

damn manipulation and this half a century old textbook....ughhhhhhhh it's making me look dumb!!!!!

Answer 17

hmmm split the fraction up instead of really using partial fractions....interesting....

Answer 18

To finish:
\[\eqalign{
&\frac{-y}{y-x}-[ln(x-y)-ln(y)]=ln(x)+c_1 \\
&\frac{-y}{y-x}+ln(y)-ln(x-y)=ln(x)+c_1 \\
&\ln\left(\frac{e^{\frac{-y}{y-x}}\times y}{x(x-y)}\right)=c_1 \\
&\frac{e^{\frac{-y}{y-x}}\times y}{x(x-y)}=e^{c_1}
}\]