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tchiwinnie:
"Idiopathic" means cause is unknown, not cure

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englishmart:
same thing I did in Miami 3weeks ago

Two stvpid robbers came to my flat to rob. When they knocked, I hid behind the door and asked my brother to open the door for them. As soon as they were inside, one of them pointed his gun at him.
I removed my double edged axe to divide his head.
The other one ran away leaving his bag that contains about $75,000

No be today na.

winwinx:
what is the website for this College.at Umunze

Scrypt:
Can I backwards march? . Who will be held accountable for your posts on here then?

Laplacian:
Mr. A can complete a farm work in t₁ hours. Mr. B can finish the same work in t₂ hours. How would it take them if they worked together?

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?

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?

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

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?

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

Laplacian:
And u think initial values will not settle that?

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

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.

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

tohero:
YES! All differential equations has a degree! Some might not be seen easily until you rationalise them.
In your case, the degree is one. Because one (1) is the power of the highest order(y'') in your differential equation.
So I would define a DEGREE as the power of the highest order of a differential equation after it has been rationalised.

tohero:
Comparing it to (x-h)²=4p(y-k)
So that
Vertex=(h,k)
Focus=(h,k+p)
Directrix is (y=k-p)
Lenght of lactus rectum=(4p)

The question is equivalent to
(x-2)²=4*8(y-(-1))

Hence,
Vertex=(2,-1)
Focus=(2,7)
Directrix is at y=-9
Lenght... =32

Japkelly:
Your true life story is in deed a sad one but all hope is not lost To God be the glory u are alive. this might be a blessing to someone, somewhere. This is not the first time this department is deploying or mobilizing corp members, but a day begins a story. You never can tell the end of your one yr story. Mr Bright you are going about this the wrong way. You have better ways of achieving this and not necessarily the media. Some of the affected students met the VC sometime last month and efforts have been made, to which am sure about. How much powers do u have to fight the Federal Government ( MOUAU)? Pls take it easy. You might end up winning a lost battle.

henryobinna:
when I read that guy's post I git confused.

was talking or RAM or graphics card.

and its laptop not desktop