Miscellaneous:


why is it nonlinear? because of the "sin" ?

Yes sire!

doubleDx:

It's a 2nd order nonlinear ODE with degree 1 . Happy new year to y'all!

2nd order, yes. Nonlinear, yes.

But I doubt if it is of degree 1. To see this, recall that
sin(y'')=y''- (y'')³/3!+ (y'')⁵/5!-(y'')⁷/7!+. . .

So, inherently, we see that the degree is infinite (or does not exist?)


that's what I feel ooooh. Please, criticize.


Happy New Year, Sire.
Happy New Year, NMCites

jackpot:
2nd order, yes. Nonlinear, yes.

But I doubt if it is of degree 1. To see this, recall that
sin(y'')=y''- (y'')³/3!+ (y'')⁵/5!-(y'')⁷/7!+. . .

So, inherently, we see that the degree is infinite (or does not exist?)


that's what I feel ooooh. Please, criticize.


Happy New Year, Sire.
Happy New Year, NMCites

You are actually right! The degree doesn't exist... I have rechecked! Thanks for pointing that out...

Happy new year! It's been a while....howdy?

jackpot:
2nd order, yes. Nonlinear, yes.

But I doubt if it is of degree 1. To see this, recall that
sin(y'')=y''- (y'')³/3!+ (y'')⁵/5!-(y'')⁷/7!+. . .

So, inherently, we see that the degree is infinite (or does not exist?)


that's what I feel ooooh. Please, criticize.


Happy New Year, Sire.
Happy New Year, NMCites

It Is a NON-LINEAR, SECOND DEGREE, THIRD ORDER differential equation

PROOF;

given; sin(y'')+y'=0 or

sin(y'') = -y'............(1.)

first we differentiate the above

using function of a function to get;

d[sin(y" )]/dy" x dy"/dt + d(y')/dt=0 where we take t as the independent variable; we obtain

cos(y" ) x y'" + y"=0 or

cos(y" ) = -y"/y'" ..............(2.)

square (1.) and (2.) and add;

sin²(y'' ) + cos²(y'' )= (y')² + (y"/y'" )²=1 or

[1-(y')²] x (y'" )²-(y" )²=0

Q.E.D

if u doubt that the two differential equations are inequivalent, solve the first one (since it is easier) and show that it satisfies the second.

A RULE ON MULTIPLICATION
suppose we are to multiply two numbers with 5 as unit digit each.
We proceed as follows;
let the numbers be n5 & m5 then
n5 x m5 = (10n+5) x (10m+5) = 100nm+50(n+m)+25=100nm+100(n+m)/2+25=100[nm+(n+m)/2]+25

Hence; n5 x m5 = [nm+(n+m)/2]25

Theorem;
i.) to multiply two numbers ending with 5, take the product of their non-unit digit, take the average of their non unit digit, sum them and add 25 @ the end. Provided the two non-unit digits are either both odd or both even.
E.g 135 x 15= (13+7)25=2025 since 13 and 1 are both odd, 13 x 1 =13 and (13+1)/2=7 so 13+7=20

again, 25 x 65= (12+4)25=1625 since 2 and 6 are both even, 6 x 2 =12 and (6+2)/2=4 so 12+4=16

ii.) if one of the non-unit digit is odd and the other is even, take their product, add the non-units to 1 and divide by 2, sum the result with the product, put 75 @ the end.
E.g 125 x 35= (36+75=4475
since 12 and 3 have opposite parity, 12x3=36, and (12+3+1)/2=8

COLORALLY;
i.) if two or more numbers ending with 5 are multplied, the last two digits of the result must either be 25 or 75
ii.) to multiply a number by 5, if its non unit digit is even, divide it by 2 and add 25. E.g 65 x 5=(6/2)25=325.
It its non-unit digit is odd, add 1 to it and divide by 2 and put 75 behind the result. E.g 35x5=(3+1)/2=275.
All these follow from the above result; n5xm5=[nm+(n+m)/2]25 if m=0 then n5x05=[0+(n+0)/2]25=(n/2)25
iii.) to square a number that ends with 5, multply the non-unit digit by the nearest integer greater than it. From our formula;
n5xm5=[nm+(n+m)/2]25 put n=m then, m5xm5=(m5)²=[m²+m]25=m(m+1)25
e.g
65²=(6x7)25=4225
or
85²=(8x9)25=7225
or
15²=(1x2)25=225
or
5²=05²=(0x1)25=025=25
or
35²=(3x4)25=1225
or
e.t.c

Laplacian:


It Is a NON-LINEAR, SECOND DEGREE, THIRD ORDER differential equation

PROOF;

given; sin(y'')+y'=0 or

sin(y'') = -y'............(1.)

first we differentiate the above

using function of a function to get;

d[sin(y" )]/dy" x dy"/dt + d(y')/dt=0 where we take t as the independent variable; we obtain

cos(y" ) x y'" + y"=0 or

cos(y" ) = -y"/y'" ..............(2.)

square (1.) and (2.) and add;

sin²(y'' ) + cos²(y'' )= (y')² + (y"/y'" )²=1 or

[1-(y')²] x (y'" )²-(y" )²=0

Q.E.D

if u doubt that the two differential equations are inequivalent, solve the first one (since it is easier) and show that it satisfies the second.

hay man , the prof. himself. .

but seems its of a non-existence /infinite degree since even after every successive derivatives we obtain different D.E ......we employ your approach for P.S ( power series solutions) .

hope you get the gist ..?

you're great man !!

greetings .....


happy new year

please help out with this. integral of ln(sinx) dx = ?

Laplacian:
A RULE ON MULTIPLICATION
suppose we are to multiply two numbers with 5 as unit digit each.
We proceed as follows;
let the numbers be n5 & m5 then
n5 x m5 = (10n+5) x (10m+5) = 100nm+50(n+m)+25=100nm+100(n+m)/2+25=100[nm+(n+m)/2]+25

Hence; n5 x m5 = [nm+(n+m)/2]25



COLORALLY;
i.) if two or more numbers ending with 5 are multplied, the last two digits of the result must either be 25 or 75
ii.) to multiply a number by 5, if its non unit digit is even, divide it by 2 and add 25. E.g 65 x 5=(6/2)25=325.
It its non-unit digit is odd, add 1 to it and divide by 2 and put 75 behind the result. E.g 35x5=(3+1)/2=275.
All these follow from the above result; n5xm5=[nm+(n+m)/2]25 if m=0 then n5x05=[0+(n+0)/2]25=(n/2)25
iii.) to square a number that ends with 5, multply the non-unit digit by the nearest integer greater than it. From our formula;
n5xm5=[nm+(n+m)/2]25 put n=m then, m5xm5=(m5)²=[m²+m]25=m(m+1)25
e.g
65²=(6x7)25=4225
or
85²=(8x9)25=7225
or
15²=(1x2)25=225
or
5²=05²=(0x1)25=025=25
or
35²=(3x4)25=1225
or
e.t.c

Highly creative!

agentofchange1:
greetings .....


happy new year

please help out with this. integral of ln(sinx) dx = ?

hi ben!
Let I=§In(sinz)dz,

let y=In(sinz).........(1.), dy/dz=cosz/sinz, so ,

I=§ytanz.dy............(2.) from (1.)

e^y=sinz or e^-2y=cosec²z

so that (e^-2y-1)^1/2=cotz so that (2.) becomes;

I=§y/(e^-2y-1)^1/2dy or

I=§ ye^y/(1-e^2y)^1/2dy

if we integrate by part by keepin y constant first we get;

I=-y(1-e^2y)^1/2+§(1-e^2y)^1/2dy

now let
H=§(1-e^2y)^1/2dy

let u=e^y so that du/dy=u

Hence

H=§(1-u²)^1/2du/u

Or

H=§(u^-2-1)^1/2du

now let u=cosw then du/dw=-sinw, Hence,
H=§tanw.-sinwdw

or

H=§-sin²w/cosw dw

=§(cos²w-1)/cosw dw

=§(cosw - 1/cosw) dw

=sinw-§1/cosw dw

let G=§1/cosw dw

=§cosw/cos²w dw

let k=sinw then dk/dw=cosw
hence
G=§1/(1-k²) dk

using partial fraction and resolvin we get

G=1/2 In[(1+k)/(1-k)]

= 1/2 In[(1+sinw)/(1-sinw)]

therefore
H=sinw-1/2 In[(1+sinw)/(1-sinw)]
but
I=-y(1-e^2y)^1/2+H
hence

I=-y(1-e^2y)^1/2+sinw-1/2 In[(1+sinw)/(1-sinw)]+Constant

where w=cos^-1(e^y)
and y=In(sinz)

simplifying further if u wish,

I=[1-In(sinz)]cosz +1/2 In[(1+cosz)/(1-cosz)]+constant

Laplacian:


It Is a NON-LINEAR, SECOND DEGREE, THIRD ORDER differential equation

PROOF;

given; sin(y'')+y'=0 or

sin(y'') = -y'............(1.)

first we differentiate the above

using function of a function to get;

d[sin(y" )]/dy" x dy"/dt + d(y')/dt=0 where we take t as the independent variable; we obtain

cos(y" ) x y'" + y"=0 or

cos(y" ) = -y"/y'" ..............(2.)

square (1.) and (2.) and add;

sin²(y'' ) + cos²(y'' )= (y')² + (y"/y'" )²=1 or

[1-(y')²] x (y'" )²-(y" )²=0

Q.E.D

if u doubt that the two differential equations are inequivalent, solve the first one (since it is easier) and show that it satisfies the second.

hmmm. I can see the mathematics. But. . .

Let's take another look.

Given y'+y=0, differentiating both sides w.r.t. x gives a new differential equation
y''+y'=0.

It is really a heresy to say that both differential equations are equivalent since y'+y=0 has the solution y=C e^-x, whereas y''+y'=0 has the solution y=C₁+C₂e^-x.


Because y'+y=0 is easier to handle and it's solution satisfies y''+y'=0, I can not conclude that both differential equations has the same (general) solution.

So, I will say that the two differential equations you compared are not equivalent.

Or, what do you think?

jackpot:
hmmm. I can see the mathematics. But. . .

Let's take another look.

Given y'+y=0, differentiating both sides w.r.t. x gives a new differential equation
y''+y'=0.

It is really a heresy to say that both differential equations are equivalent since y'+y=0 has the solution y=C e^-x, whereas y''+y'=0 has the solution y=C₁+C₂e^-x.


Because y'+y=0 is easier to handle and it's solution satisfies y''+y'=0, I can not conclude that both differential equations has the same (general) solution.

So, I will say that the two differential equations you compared are not equivalent.

Or, what do you think?

jackpot:
hmmm. I can see the mathematics. But. . .

Let's take another look.

Given y'+y=0, differentiating both sides w.r.t. x gives a new differential equation
y''+y'=0.

It is really a heresy to say that both differential equations are equivalent since y'+y=0 has the solution y=C e^-x, whereas y''+y'=0 has the solution y=C₁+C₂e^-x.


Because y'+y=0 is easier to handle and it's solution satisfies y''+y'=0, I can not conclude that both differential equations has the same (general) solution.

So, I will say that the two differential equations you compared are not equivalent.

Or, what do you think?

Wait o, are you sure you're actually feminine?

jackpot:
hmmm. I can see the mathematics. But. . .

Let's take another look.

Given y'+y=0, differentiating both sides w.r.t. x gives a new differential equation
y''+y'=0.

It is really a heresy to say that both differential equations are equivalent since y'+y=0 has the solution y=C e^-x, whereas y''+y'=0 has the solution y=C₁+C₂e^-x.


Because y'+y=0 is easier to handle and it's solution satisfies y''+y'=0, I can not conclude that both differential equations has the same (general) solution.

So, I will say that the two differential equations you compared are not equivalent.

Or, what do you think?

y should they not be equivalent?
Integration as we know is d inverse of differentiation. Does it not imply that if one differentiates a function and integrates it he should still have the same general solution? Let's use ur example.
Given
y'+y=0 after differentiating, y"+y'=0, should we not have same function (in a more general form) if we integrate? §y"dt+§y'dt=§0xdt so that y'+y=C₁ and this is slightly more general and does no harm to the original function (in terms of generality). Let's solve the two equations; from dy/dt=C₁-y
we get
dy/(C₁-y)=dt so that
-In(C₁-y)=t+K
or
C₁-y=e^-t-K
or
y=C₁-e^-Kxe^-t
so that y=C₁+C₂e^-t. This tells u how C₁ came into the picture. And as i have explained above, the are one and the same. Questioning its validity amounts to doubtin if the following is true; given y=t²+3........(1.)
differentiate, y'=2t, now let's integrat d last to get y=t²+C............(2.)
and u say (1.) and (2.) are not the same? Leave it or take it INITIAL VALUE SETTLES EVERYTHING MY DEAR.
So on what ground do u accuse me of heresy, and what is the FOUNDATION of ur claim of inequivalence?

Laplacian:


And u think initial values will not settle that?

take it or leave it. Stick 2 what u think is best.

Both differential equations won't even have the same number of initial conditions. The one you differentiated will need extra one initial condition.

For example, y'+y=0 needs only one initial condition before you obtain the particular solution free from arbitrary constants, whereas y''+y'=0 needs two initial conditions. So, i am thinking it makes them even far from being equivalent?

jackpot:
Both differential equations won't even have the same number of initial conditions. The one you differentiated will need extra one initial condition.

For example, y'+y=0 needs only one initial condition before you obtain the particular solution free from arbitrary constants, whereas y''+y'=0 needs two initial conditions. So, i am thinking it makes them even far from being equivalent?

C₁ came as a result of an integration (see above). So u are saying in essence that, differentiating a function and integratin the result will not yield the initial function.
Because i differentiate; y'+y=0 to get y"+y'=0 and i integrate this to get y'+y=C₁

Laplacian:

C₁ came as a result of an integration (see above). So u are saying in essence that, differentiating a function and integratin the result will not yield the initial function.
Because i differentiate; y'+y=0 to get y"+y'=0 and i integrate this to get y'+y=C₁

maybe in the world of cartoons(where things happen the way we want), we may say that y'+y=C is equivalent to y'+y=0 for all values of the constant C.

jackpot:
maybe in the world of cartoons(where things happen the way we want), we may say that y'+y=C is equivalent to y'+y=0 for all values of the constant C.

everyone is entitled to his opinion

Laplacian:


y should they not be equivalent?
Integration as we know is d inverse of differentiation. Does it not imply that if one differentiates a function and integrates it he should still have the same general solution? Let's use ur example.
Given
y'+y=0 after differentiating, y"+y'=0, should we not have same function (in a more general form) if we integrate? §y"dt+§y'dt=§0xdt so that y'+y=C₁ and this is slightly more general and does no harm to the original function (in terms of generality). Let's solve the two equations; from dy/dt=C₁-y
we get
dy/(C₁-y)=dt so that
-In(C₁-y)=t+K
or
C₁-y=e^-t-K
or
y=C₁-e^-Kxe^-t
so that y=C₁+C₂e^-t. This tells u how C₁ came into the picture. And as i have explained above, the are one and the same. Questioning its validity amounts to doubtin if the following is true; given y=t²+3........(1.)
differentiate, y'=2t, now let's integrat d last to get y=t²+C............(2.)
and u say (1.) and (2.) are not the same? Leave it or take it INITIAL VALUE SETTLES EVERYTHING MY DEAR.
So on what ground do u accuse me of heresy, and what is the FOUNDATION of ur claim of inequivalence?

hmmm.

Suppose for contradiction that y'+y=0 is equivalent to y''+y'=0. By differentiating once more, we must have that y''+y'=0 is also equivalent to y'''+y''=0.

But these guys are an equivalence relation, so by transitivity,
y'+y=0 is equivalent to y''+y'=0 which is equivalent to y'''+y''=0 implies that y'+y=0 is equivalent to y'''+y''=0.
Now, y'+y=0 has the solution y=C^-x, but y'''+y''=0 has the solution y=C₁+C₂x+C₃e^-x.
Are these two solutions equivalent?

jackpot:
hmmm.

Suppose for contradiction that y'+y=0 is equivalent to y''+y'=0. By differentiating once more, we must have that y''+y'=0 is also equivalent to y'''+y''=0.

But these guys are an equivalence relation, so by transitivity,
y'+y=0 is equivalent to y''+y'=0 which is equivalent to y'''+y''=0 implies that y'+y=0 is equivalent to y'''+y''=0.
Now, y'+y=0 has the solution y=C^-x, but y'''+y''=0 has the solution y=C₁+C₂x+C₃e^-x.
Are these two solutions equivalent?

yes!
A simple substitution C₁=C₂=0 will pay the price. It stil boils down to initial value.

Laplacian:

yes!
A simple substitution C₁=C₂=0 will pay the price. It stil boils down to initial value.

using your logic,
y=Ce-^x is equivalent to y=C₁e^x+C₂e^-x which is (again, by your logic) equivalent to y=Ce^x.

Then, by transitivity of an equivalence relation, y=Ce^-x is equivalent to y=Ce^x.

Do you really believe that both solutions are equivalent?

jackpot:
using your logic,
y=Ce-^x is equivalent to y=C₁e^x+C₂e^-x which is (again, by your logic) equivalent to y=Ce^x.

Then, by transitivity of an equivalence relation, y=Ce^-x is equivalent to y=Ce^x.

Do you really believe that both solutions are equivalent?

what extension of the D.E for y=Ce-^x did u get the solution y=C₁e^x+C₂e^-x if u want to play by my logic?

hmmm. think am enjoying this. drama. as long it doesn't leads to something else .

hmm ride profs .


na wa ooo. Differential equations sha ..


^^hay guys those are. mathematical models that assume a real life phenomenon , why not use both model to solve a problem let's see what happens .?

just a suggestion sha ..


## just passing by


c.ya lera .


1luv

Laplacian:

what extension of the D.E for y=Ce-^x did u get the solution y=C₁e^x+C₂e^-x if u want to play by my logic?

since y'+y=0 is equivalent (you said so) to y''+y'=0, it is equivalent to y''+y'-(y'+y)=0, which is the same thing as y''-y=0.

The DE y''-y=0 now has the solution y=C₁e^x+C₂e^-x.

jackpot:
since y'+y=0 is equivalent (you said so) to y''+y'=0, it is equivalent to y''+y'-(y'+y)=0, which is the same thing as y''-y=0.

The DE y''-y=0 now has the solution y=C₁e^x+C₂e^-x.

but the solution of y'+y=0 always satisfies any of its extended D.E doesn't it? But the matter of the extended D.E's solution (y"-y=0) satisfying the original D.E (either y'+y=0 or y'-y=0) is decided by initial value. The values on e^x are not on e^-x, so with d initial value inputed in the solution of the extended D.E, either C₁ or C₂=0 as appropriate. Turning tables around, i ask u; if the D.E y'+sin(y" )=0 is of infinite degree, why should the D.E obtained from it have degree 2?

Laplacian:

but the solution of y'+y=0 always satisfies any of its extended D.E doesn't it? But the matter of the extended D.E's solution (y"-y=0) satisfying the original D.E (either y'+y=0 or y'-y=0) is decided by initial value. The values on e^x are not on e^-x, so with d initial value inputed in the solution of the extended D.E, either C₁ or C₂=0 as appropriate. Turning tables around, i ask u; if the D.E y'+sin(y" )=0 is of infinite degree, why should the D.E obtained from it have degree 2?

There's no theorem saying that if you differentiate a DE, the degree will increase or decrease. In other words, there is no theorem comparing the degree of a DE with that of the differentiated DE. So, you shouldn't compare both degrees.

Maybe you're trying to draw analogy from the fact that the difference between the order of a DE and with that of the differentiated DE is 1.

Catch my drift, Sir?

jackpot:
There's no theorem saying that if you differentiate a DE, the degree will increase or decrease. In other words, there is no theorem comparing the degree of a DE with that of the differentiated DE. So, you shouldn't compare both degrees.

Maybe you're trying to draw analogy from the fact that the difference between the order of a DE and with that of the differentiated DE is 1.

Catch my drift, Sir?

i catch ur drift ma. But, with due respect, just because u've not seen a theorem on that does not erase d fact that it does exist. And do i always need a theorem to confirm what i know?

Battle of the Titans!


They made me thought if I ever sat in a mathematics class.

Do you need any information or need to be enlightened on
OAU admission,visit the school social website
www.oaucampus.com to get all necessary information
from stalites to get ready for the admission race.

Evaluate

|5 1 0 0 0|
|1 4 0 1 2|
|3 2 1 4 0|
|0 1 2 1 0|
|1 2 0 1 2|

tohero:
Battle of the Titans!


They made me thought if I ever sat in a mathematics class.

no be small tin oo

tohero:
Battle of the Titans!


They made me thought if I ever sat in a mathematics class.

no be only you

Nastydroid:
no be only you

agentofchange1:


no be small tin oo

All is well. At least we sabi count 1, 2, 3... So we arnt getting freaked!