Question

exact differential equations

Answer 1

this looks like bernouli to me

Answer 2

Have no idea bout this. I just want to know if this is an exact DE. =)

Answer 3

\[\frac{xdy - ydx}{y^2} = x^3 dx\]

\[xdy - ydx = y^2x^3 dx\]

\[xdy -(y+y^2x^3)dx = 0 \]

Answer 4

Oh you just want to check whether it is exact or not ?

Answer 5

yes.

Answer 6

then its easy :
1) differentiate the coefficient of `dy` with respect to `x`
2) differentiate the coefficient of `dx` with respect to `y`

if you get same expression for both, then it is exact

Answer 7

I guess it isnt exact

Answer 8

you're right

Answer 9

i got 1 and 1 + 2y

Answer 10

\[\color{red}{x}dy \color{blue}{-(y+y^2x^3)}dx = 0 \]


\(\frac{\partial }{\partial x} (\color{red}{x}) = 1\)

\(\frac{\partial }{\partial y} (\color{red}{-y-y^2x^3}) = -1-2yx^3\)

Answer 11

since the partials are not equal the DE is not exact

Answer 12

But in my notes. This problem falls in exact diff eq

Answer 13

thats because ...

partial derivative of x/y^2 w.r.t x = 1/y^2

partial derivative of -1/y +x^3 w.r.t y = 1/y^2

its indeed exact

Answer 14

by multiplying with y^2, it became non-exact lol

Answer 15

*** -1/y -x^3

Answer 16

@hartnn how did u get -1/y - x^3 ?

Answer 17

combining dx terms

Answer 18

-y/y^2 dx - x^3 dx

Answer 19

Oh in the start itself ! nice i missed that completely

Answer 20

yeah, if multiplying DE with some variable can make a
non-exact equation as exact ...
reverse is also true :P

Answer 21

thats so true yeah :)

Answer 22

I got it. Thank you so much @ganeshie8 @hartnn

Answer 23

an other method
\[\frac{ x dy-ydx }{ y^2 }=x^3dx\]
\[x dy-ydx=x^3 y^2 dx\]
divide by dx
\[x \frac{ dy }{ dx }-y=x^3 y^2\]
divide by x y^2
\[y ^{-2}\frac{ dy }{ dx }-\frac{ 1 }{ x }y ^{-1}=x^2\]
\[put ~y ^{-1}=t,-1 ~y ^{-2}\frac{ dy }{ dx }=\frac{ dt }{ dx }\]
\[-\frac{ dt }{ dx }-\frac{ 1 }{ x }t=x^2\]
\[\frac{ dt }{ dx }+\frac{ 1 }{ x }t=-x^2\]
\[I.F.=e ^{\frac{ 1 }{ x }dx}=e ^{\ln x}=x\]
\[C.S.~is~t*x=\int\limits \left( -x^2 \right)*x ~dx+c\]
now you can complete.