Question

xy'-y= e^(y/x)* (x^3/y)

Answer 1

you want y?

Answer 2

Any help with x+y-2+(1-x)y'=0

Answer 3

Yes

Answer 4

just to be sure, plz write out xy'-y= e^(y/x)* (x^3/y) using the equation tool here.

as it is written now, it is a non-linear differential equation n the solution will not be simple.

Answer 5

@ganeshie8 @Directrix

Answer 6

\[xy'-y=e ^{y/x}*x^3/y\]

Answer 7

and if anyone can also help with this one x+y-2+(1-x)y'=0

Answer 8

not the exact prob but similar one here:

http://webcache.googleusercontent.com/search?q=cache:8FaD8dBjYcUJ:www.enotes.com/homework-help/xy-y-xe-y-x-372335+&cd=2&hl=en&ct=clnk&gl=us

let u=y/x
so u' = y'/x - y/x^2
substitute back into the original eqn...

Answer 9

\[x+y+2+(1-x)y' = 0\]\[y'(1-x) + y = -x -2\]\[y'+ \frac{y}{1-x} = \frac{-x-2}{1-x}\]\[let: \ P(x) = \frac{1}{1-x} \ and \ Q(x) = \frac{-x-2}{1-x}\] what we do here is trying to get y in terms of p(x) and q(x)\[y\times e^{\int P(x)d(x)} = \int P(x)\times Q(x)d(x)\]finally we get \[y = \frac{\int P(x)\times Q(x)d(x)}{e^{\int P(x)d(x)}}\]

Answer 10

references http://www.intmath.com/differential-equations/4-linear-des-order-1.php