Question

2xy' - y = x^3 -x

Answer 1

\[2xy' -y=x^3 -x\]

Answer 2

its not exact, but i made it exact

Answer 3

http://www.wolframalpha.com/input/?i=2xy%27-y%3Dx%5E3-x

Answer 4

you know how to do it, right?

Answer 5

kind of

Answer 6

the linear way doesn't make any sense

Answer 7

you have to make it exact first, right?
show me your stuff

Answer 8

okay

Answer 9

u(x)=x^(-3/2) or 1/x^3/2

Answer 10

\[-x^{\frac{ 3 }{ 2 }}+\frac{ 1 }{ \sqrt{x} }-\frac{ y }{ x ^{\frac{ 3 }{ 2 }} }+\frac{ 2y' }{ \sqrt{x} }\]

Answer 11

M(-x^3/2 + 1/x^1/2 - y/x^3/2)
N(2y'/x^1/2)

Answer 12

-1/x^3/2 = -1/x^3/2 so now its exact

Answer 13

divide by 2x
\[y'-\frac{ y }{2x }=\frac{ x ^{2} }{2 }-\frac{ 1 }{ 2 }\]
\[I.F=e ^{\int\limits -\frac{ 1 }{2x }dx}=e ^{-\frac{ 1 }{2 }\ln x}=e ^{\ln x ^{\frac{ -1 }{ 2 }}}=x ^{-\frac{ 1 }{ 2 }}\]
\[C.S.~ is~ y*x ^{\frac{ -1 }{2 }}=\frac{ 1 }{ 2 } \int\limits x ^{-\frac{ 1 }{ 2 }}\left( x ^{2}-1 \right)dx+c\]
\[=\frac{ 1 }{2 }\int\limits \left( x ^{\frac{ 3 }{2 }} -x ^{\frac{ -1 }{2 }}\right)dx+c\]
\[=\frac{ 1 }{2 }\left( \frac{ x ^{\frac{ 5 }{2 }} }{ \frac{ 5 }{ 2 } }-\frac{ x^ \frac{ 1 }{2 } }{\frac{ 1 }{2 } } \right)+c\]

Answer 14

you can solve further.

Answer 15

Oh that makes PERFECT sense!

Answer 16

okay i got it, THANK!

Answer 17

yw

Answer 18

i got
\[y=\frac{ 1 }{ 5}x^3 -x +C \sqrt{x}\]

Answer 19

correct