15/05/2020
004
Families of DIFFERENTIAL EQUATIONS
EXACT ORDINARY DIFFERENTIAL EQUATION
a total derivative of a function is given by
u=f(x,y)
δu=(δu/δx)δx+(δu/δy)δy
where all δ assume any value except for δx,δy term
let
M = δu/δx and N= δu/δy
δM/δy=δ²u/δxδy
δN/δx=δ²u/δxδy
thus for an exact differential equation
the exactness test is verified by
δM/δy=δN/δx............(1)
For a non-exact case
δM/δy≠δN/δx
so we employ the use of an Integration factor
F(y)= 1/M(δN/δx-δM/δy)..............(2)
IF = e ^ ∫F(y)dy...........(3)
or
F(y) = 1/N(δM/δy-δN/δx)...............(4)
IF = e ^ ∫F(x)dx............(5)
thus the general form of an exact differential equation is given by
M(x,y)dx+N(x,y)=c
(c=0 in most cases)
lets disassemble some examples
dy-y²sinxdx=0
(this can be solved also by separation of variable )
inspecting we have
M=-y²sinx
N=1
dM/dy = -2ysinx
dN/dx=0
thus the theorem 1 is not verified(not exact)
to obtain an IF
trying theorem 2 we have
F(x)=1/(1)(0+2ysinx)=2ysinx
we can not use F(x) because an IF should be an Explicit function
(there are many cases of having IF as implicit function, but for simplicity we proceed)
trying theorem 4
F(y)= 1/(-y²sinx)(0+2ysinx)
= -2/y
thus we are good to go
from theorem 3
IF=e^∫F(y)dy=e^∫(-2/y)dy
=1/y²
dy-y²sinx=0
multiply throughout by IF
dy/y²-sinxdx=0
now
M=-sinx
N=1/y²
dM/dy=dN/dx=0
thus it is now exact
f(x)=∫Mdx+k(y)
(constant due to the other function)
=∫-sinxdx
=cosx+k(y)-----------------**
f(y)=∫1/y²dy
=-1/y+k(x)------------------- # #
comparing ** and # #
(by comparison,if both function have same and different terms,the same terms are left out and the different ones are used to compute k(x) and k(y))
k(y)=-1/y
and
k(x)=cosx
then the general solution is
u=cosx-1/y=C
another
(1+2x)cosydx+dy/cosy=0
M= (1+2x)cosy
dM/dy=-(1+2x)siny
N=1/cosy
dN/dx=0
so our DE is not exact
dM/dy≠dN/dx
F(y)=1/(1+2x)cosy[0+(1+2x)siny]
=siny/cosy
IF=1/cosy
multiply throughout we have
