Here's the problem:(adsbygoogle = window.adsbygoogle || []).push({});

[tex]y\left( 2x-y-1 \right) dx + x\left( 2y-x-1 \right) dy = 0[/tex]

So rewrite it as:

[tex]\left( 2xy-y^2-y \right) dx + \left( 2xy-x^2-x \right) dy = 0[/tex]

Now it's in the form,…[itex]P(x,y) dx + Q(x,y) dy = 0[/itex]

[tex]\frac{\partial P(x,y)}{\partial y} = 2x-2y-1 [/tex]

[tex]\frac {\partial Q(x,y)}{\partial x} = 2y-2x-1 [/tex]

These are not equal so we need an integrating factor,…

[tex]h(u) =e^{\int{F(u) du}}[/tex]

Where: [itex]u=xy[/itex], and,...

[tex]F(u)=\frac {\frac {\partial}{\partial y} P(x,y)- \frac {\partial}{\partial x}Q(x,y)}{y Q(x,y)-x P(x,y)} [/tex]

[tex]F(u)=\frac{2x-2y-1-(2y-2x-1)}{y(2xy-x^2-x)-x(2xy-y^2-y)}[/tex]

[tex]F(u)=\frac{4x-4y}{2xy^2-x^2y-xy-2x^2y+xy^2+xy}[/tex]

[tex]F(u)=\frac{4(x-y)}{3xy^2-3x^2y}=\frac{4(x-y)}{3xy(y-x)} =-\frac{4(x-y)}{3xy(x-y)} =-\frac{4}{3xy} [/tex]

[tex]h(u) =e^{\int -\frac{4}{3}\frac {du}{u}}}=e^{ -\frac{4}{3}\ln{u}}}=e^{ \ln{(u^{-\frac{4}{3}}})}} =u^{-\frac{4}{3}}=(xy) ^{-\frac{4}{3}} [/tex]

Except this doesn't seem to work. It's also quite different from the book answer.

Book answer for the integrating factor is,..

[tex]x^{-1} y^{-1}\left( x+y+1\right)^{-1}[/tex]

So where did I go wrong?

Just for the record this problem is from the Dover BookOrdinary Differential Equationsby Tenenbaum and Pollard. It's problem # 14 on page 91 in exercise 10.

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# Need Help - Exact Integrating Factor?

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