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$$(6xy)dx = (4y+9x^2)dy$$ to find out if its exact $$M= 6xy, N =4y+9x^2 $$ $$ \frac {dM}{dy} = 6x, \frac {dN}{dx} = 18x$$ Hence its not exact. Please correct me if i did something wrong and help me to make it exact by multiplying integrating factor.

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You should write the equation as $M(x,y)dx+N(x,y)dy=0$ so $M_y=6x$ and $N_x=-(18x)$ – Babak S. Nov 15 '12 at 17:05
Thanks! is it right now? $$ (6xy)dx - (4y+9x^2)dy = 0 $$ – TPSstar Nov 15 '12 at 17:09
See this. – glebovg Nov 15 '12 at 17:17
@TPSstar: See the link glebovg attached and see the Case 2 in it. Then follow the rest of the material. – Babak S. Nov 15 '12 at 17:21
@glebovg yep i've got it but my question was written in that form that i posted first. is it appropriate to change it to your mentioned form or if equation is not in the said form then it would obviously be non-exact? – TPSstar Nov 15 '12 at 17:22
up vote 0 down vote accepted

You can multiply the original equation by anything and equality will always hold. However, multiplying it by an appropriate integration factor (see Case 2) makes it exact, hence easier to solve. If you are wondering where these integration factors come from, consider $$M(x,y)dx + N(x,y)dy = 0$$ and multiply by an integration factor $u(x,y)$ to make it exact. Note that we must have $$\frac{{\partial (uM)}}{{\partial y}} = \frac{{\partial (uN)}}{{\partial x}} \Leftrightarrow {u_y}M + u{M_y} = {u_x}N + u{N_x}$$ for exactness. If $u = u(x)$, then $$\frac{{du}}{{dx}} = u\frac{1}{N}\left( {{M_y} - {N_x}} \right)$$ which you know how to solve if the righthand side is a function of $x$ only. Similarly, if $u = u(y)$, then $$\frac{{du}}{{dy}} = u\frac{1}{M}\left( {{N_x} - {M_y}} \right).$$ Again, you know how to solve this if the righthand side is a function of $y$ only.

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This is what I meant. +1 – Babak S. Nov 15 '12 at 17:28
Who downvoted this? This answers the question perfectly. – glebovg Nov 26 '12 at 2:39

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