AliaaAliHassan
  • AliaaAliHassan
xy'-y= e^(y/x)* (x^3/y)
Differential Equations
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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lochana
  • lochana
you want y?
AliaaAliHassan
  • AliaaAliHassan
Any help with x+y-2+(1-x)y'=0
AliaaAliHassan
  • AliaaAliHassan
Yes

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superdavesuper
  • superdavesuper
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.
lochana
  • lochana
@ganeshie8 @Directrix
AliaaAliHassan
  • AliaaAliHassan
\[xy'-y=e ^{y/x}*x^3/y\]
AliaaAliHassan
  • AliaaAliHassan
and if anyone can also help with this one x+y-2+(1-x)y'=0
superdavesuper
  • superdavesuper
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...
lochana
  • lochana
\[x+y+2+(1-x)y' = 0\]\[y'(1-x) + y = -x -2\]\[y'+ \frac{y}{1-x} = \frac{-x-2}{1-x}\]\[let: \ P(x) = \frac{1}{1-x} \ and \ Q(x) = \frac{-x-2}{1-x}\] 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 \[y = \frac{\int P(x)\times Q(x)d(x)}{e^{\int P(x)d(x)}}\]
lochana
  • lochana
references http://www.intmath.com/differential-equations/4-linear-des-order-1.php

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