The Calculus Phoenix

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A nice example on volume of a solid revolution

Solution to the series question

prove that
if
eˣ=Σ(0,∞)xⁿ/n!

and

eʸ=Σ(0,∞)yⁿ/n!

then

eˣ⁺ʸ=eˣ.eʸ

continuation.....

(1+2x)dx+sec²ydy=0

M=1+2x

N=sec²y

f(x)=x+2x²+k(y)

f(y)=tany + k(x)

then

u=x+2x²+tany=c

try and upload

(x²+y²)dx-2xydy=0

section 2

sinxcosydx+cosxsinydy=0

M=sinxcosy

dM/dx=-sinxsiny

N=cosxsiny

dN/dx=-sinxsiny

dM/dy=dN/dx

so our DE is exact

f(x)=∫Mdx

=∫sinxcosydx

=-cosxcosy+k(y)

f(y)=∫Ndy

= ∫cosxsinydy

=-cosxcosy+k(x)

by comparison

k(x)=0,k(y)=0

u=-cosxcosy-cosxcosy=c

=-2cosxcosy=c

enjoy.........

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