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 August 6th, 2012, 02:54 AM #1 Member   Joined: Jun 2012 From: Uzbekistan Posts: 59 Thanks: 0 differential equation Solve the equation: $(2xy^2-3y^3)dx+(7-3xy^2)dy=0$. I solved equations such $M(x,y)dx+N(x,y)dy=0$ when $M_y'=N_x#39;$. In my case $M_y'=4xy-9y^2$ and $N_x'=-3y^2$ i.e. $M_y'\neq N_x'$. I know in this case I should find $K(x,y)$ such that $(KM)_y'=(KN)_x#39;$, but I really no idea how to find $K(x,y)$. Maybe there is a method which allows to find $K(x,y)?$
 August 6th, 2012, 03:04 AM #2 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: differational equation Use an integrating factor, $\mu$
 August 6th, 2012, 03:10 AM #3 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: differational equation You want to find a special integrating factor. We may compute: $\frac{\frac{\delta N}{\delta x}-\frac{\delta M}{\delta y}}{M(x,y)}=-\frac{2}{y}$ Since this is a function of y alone, our integrating factor is found by: $\mu(y)=e^{-2\int\frac{1}{y}\,dy}=\frac{1}{y^2}$ Multiplying the given ODE by this factor, we obtain: $$$2x-3y$$dx+$$7y^{-2}-3x$$dy=0$ Now you have an exact ODE, and we note that in multiplying by the integrating factor, we have lost the solution $y\equiv0$.
 August 6th, 2012, 03:22 AM #4 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: differential equation Oops, I didn't read the whole post Sorry.
August 6th, 2012, 09:40 AM   #5
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Re: differational equation

Quote:
 Multiplying the given ODE by this factor, we obtain: $$$2x-3y$$dx+$$7y^{-2}-3x$$dy=0$ Now you have an exact ODE, and we note that in multiplying by the integrating factor, we have lost the solution $y\equiv0$.
thank you MarkFL... How I didn't note it before.

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