Question

An integrating factor of the ODE (e^x*sec(y)-48*tan(y))dx+dy=0 is (e^(-a*x))*cosy. Find a.

Answer 1

What do you know about integrating factors?

Answer 2

In the question \[e^{-ax}\cos(y)\] is the integrating factor of the ODE

Answer 3

Yes, what properties does the integrating factor have?

Answer 4

Integrating factor can make ODE (Ordinary Differential Equations) as an Exact differential form like as M(x,y)dx + N(x,y)dy = 0 and partial derivatives of M(x,y)|y = N(x,y)|x

Answer 5

I tried to find integrating factor of above ODE using this method :
\[(M_{y}-N_{x})/M =\frac{ e^xcsc(y)\tan(y)-48\sec^2(y) }{ e^xsec(y)-48\tan(y) }\]
However I can't make \[e^{-ax}\cos(y)\] form.

Answer 6

Im not able to get it to come out either, but two corrections. First, should be e^xsec(y)tan(y) in the numerator. Second, you have it reversed for the formula. Its just
\[\frac{ N _{x}-M _{y} }{ M }\]
Other than that, Im pretty much in this chapter in my class, so not sure I would know much more than you x_x