- #1

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^{2}x+y)dy + (x

^{2}y+2x)dy = 0 the only thing i know is that the integrating factor should be a function of xy

Can some one pleas explain how to do this? Thank you.

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- Thread starter Happydog
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- #1

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Can some one pleas explain how to do this? Thank you.

- #2

Defennder

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You appear to have mistyped your problem. There 2 "dy"s here but no "dx".

- #3

HallsofIvy

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Assuming that you mean [itex](y^2x+ y)dx+ (x^2y+ 2x)dy= 0[/itex], if you know that the integrating factor is a function of xy, the obvious thing to do is to put multiply that equation by u(xy) and see what happens. In order that [itex]u(xy)(y^2x+ y)dx+ u(xy)(x^2y+ 2x)dy= 0[/itex] be exact, we must have [itex](u(y^2x+ y))_y[/itex] and [itex](u(x^2y+ 2x)_x)[/itex] equal. That is, [itex]u_x(y^2x+ y)+ u(2xy+1)= u_x(x^2y+ 2x)+ u(2xy+ 2)[/itex]. Since u is a function of xy specifically, [itex]u_x= yu'[/itex] and [itex]u_y= xu'[/itex]. That is, u must satisfy [itex]xu(y^2x+ y)+ u(2xy+1)= yu'(x^2y+ 2x)+ u(2xy+ 2)[/itex]. That gives you a differential equation for u which may or may not be solvable, depending upon whether there^{2}x+y)dy + (x^{2}y+2x)dy = 0 the only thing i know is that the integrating factor should be a function of xy

Can some one pleas explain how to do this? Thank you.

- #4

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so can any one also give me some advice on how to solve this problem i am having. Thank you.

- #5

Defennder

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- #6

HallsofIvy

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You told us, originally, that the integrating factor^{2}x+y)dx + (x^{2}y+2x)dy = 0, i have taken your sugestion but it did not work, i tryed a more general form of x^{n}y^{m}and multiplied it through to see if i could solve for the n and m and equate the two of them but the two are not equal i end up getting (m+2)=(n+2) and (m+1)=2(n+1).

so can any one also give me some advice on how to solve this problem i am having. Thank you.

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