- #1
daveed
- 138
- 0
Hey...
So the question is as stated:
Show that
[tex] \frac{1} {M_x + N_y} [/tex], where [tex] M_x+N_y [/tex] is not identically zero, is an integrating factor of the homogeneous equation [tex]M(x, y)dx+N(x, y)dy=0[/tex] of degree n.
So I am not too sure where to go with this. I suppose what it's saying is, that I'm supposed to show that with the integrating factor, it's an exact equation, so differentiating the [tex] \frac{M(x, y)} {M_x + N_y}[/tex] term with respect to y should equal the value from differentiating [tex] \frac{N(x, y)} {M_x + N_y}[/tex] with respect to x...
but that doesn't work,
and I'm not sure what else will.
I'm looking at my book, which says that only sometimes will an integrating factor make an equation like this exact; however, it does say that [tex]M(x, y)dx+N(x, y)dy=0[/tex] has degree n. Does that imply that they are polynomial equations, and if so, how would this help me? Does anyone have any suggestions?
So the question is as stated:
Show that
[tex] \frac{1} {M_x + N_y} [/tex], where [tex] M_x+N_y [/tex] is not identically zero, is an integrating factor of the homogeneous equation [tex]M(x, y)dx+N(x, y)dy=0[/tex] of degree n.
So I am not too sure where to go with this. I suppose what it's saying is, that I'm supposed to show that with the integrating factor, it's an exact equation, so differentiating the [tex] \frac{M(x, y)} {M_x + N_y}[/tex] term with respect to y should equal the value from differentiating [tex] \frac{N(x, y)} {M_x + N_y}[/tex] with respect to x...
but that doesn't work,
and I'm not sure what else will.
I'm looking at my book, which says that only sometimes will an integrating factor make an equation like this exact; however, it does say that [tex]M(x, y)dx+N(x, y)dy=0[/tex] has degree n. Does that imply that they are polynomial equations, and if so, how would this help me? Does anyone have any suggestions?
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