Question

Why is
\[\frac{dy}{dx}=y\]an exact ODE?

When I do the math, I obtain this:
\[\frac{dy}{dx}=y,\]\[dy=ydx,\]\[-ydx+dy=0,\]This, according to its definition, is not an exact ODE because
\[\frac{∂}{∂y}[-y]=-1≠0=\frac{∂}{∂x}[1].\]My professor claims it is exact, but I do not understand why.

Answer 1

i don't see it as exact either

Answer 2

the test didn't work just like you said

Answer 3

oh wait maybe it is
depends on how you write it

Answer 4

\[\frac{1}{y} \frac{dy}{dx}=1\]

\[-1+\frac{1}{y} \frac{dy}{dx}=0\]

Answer 5

\[M=-1=>M_y=0\]
\[N=\frac{-1}{y}=> N_x=0\]

Answer 6

since \[M_y=N_x\], it is exact

Answer 7

I didnt get exact too. \[M_{y}=-1 \]
\[N_{x}=1/y\]
*im curious of my answer too.:P

Answer 8

but we do get it is exact arranging it like i did

Answer 9

we can write it as -1+y'/y=0 like i did above
try your test now

Answer 10

But isn't
\[\frac{\partial }{\partial x}\left [ \frac{1}{y} \right ]=\frac{\partial }{\partial x}\left [ y^{-1} \right ]=-y^{-2}y'\]because y is a function of x?

Answer 11

remember we treat y like a constant if we are taking the partial with respect to x

Answer 12

\[N=\frac{-1}{y} => N_x=0 \]

Answer 13

Last question, how can you do this:
\[\frac{dy}{dx}=y,\]\[\frac{1}{y}\frac{dy}{dx}=1?\]What if
\[y=0?\]

Answer 14

im assuming y does not equal zero