Question

state and prove the necessary and sufficient condition for the equation
Mdx + Ndy =0 to be exact
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Answer 1

Answer:

if [∂M/∂y - ∂N/∂x]/N is a function of x alone, say f(x), then I.F. = e∫f(x)dx

if [∂N/∂x - ∂M/∂y]/M is a function of y alone, say f(y), then I.F. = e∫f(y)dy

if M is of the form M = yf1(xy) and N is of the form N = xf2(xy) then I.F. = 1/(Mx-Ny)

For this type of xmyn(aydx + bxdy) + xm'yn'(a'ydx + b'xdy) = 0, the integrating factor is xhyk.

I.F. = xh.yk

where (m + h + 1)/a = (n + k + 1)/b and (m' + h + 1)/a' = (n' + k + 1)/b'

Homogeneous equation

If Mdx + Ndy = 0 be a homogeneous equation in x and y, then

I.F. = 1/(Mx + Ny)

Step-by-step explanation: