Richiez: Doubledx Ride on, you're a star
@Ositadima1 common show dis chic what you got!

lol, h i p for the hip, hippo hippo for the hip hippopotamus...

DoubleDx relax man osita was just being goofy na! U taking it too personal!

@My humble General Osita, You know i know what you got upstairs

Hope No problems?

Odunnu: Hope No problems?

Nope we're just catching some fun....u know all work without play makes jack a dullard

Odunnu: Hope No problems?

hehehe....we are cool ma/sir...

Ok guys, am waiting for the solution to question(b)...pleaseeeeee!!! @Osita and @Doubledx you both gurus and am glad we got pple like you guys on Nairaland, btw @Op thanks for opening this thread! I really wana become a Mathematician someday, and i guess this thread would really help me in achieving that dream, although i hate Maths with passion cause its just kinda too scary for me

Good. Carry on. Do not hesitate to call when I am needed

Do u kno y am sti on dis trend even when am nt on subscriptn?, z nt dat am nt gud n mth or realy 2 cum paste quest/soln......bt d way u give credit,ur presentatn of words amaziz me.4 a guru wt such atribute i av no choice dan 2 jus Learn @ DOUBLE DX......u rem my last word 2 u aft dat soln? i sti maintain dat...

SALUUUUTTEEEE.......GBAM!!!!.....hn,u guys cn badt gan oo....men,i'd like to knw u guys' qualifications,.....cos d way u guys r dismantling maths here is making me feel like ma problems are jes too creche-like........dont blame me though,i'm jes a young fresher too eager to scatter......is dere any awoite on dis thread,i go like jam gurus like una for skul oooo.....

Thanksss guys, have worked (b) my self, not that difficult...thanks alotttttttttttttt!!!!

SpicyMimi: Thanksss guys, have worked (b) my self, not that difficult...thanks alotttttttttttttt!!!!

I admire your courage keep it up

thats what id like to see
not just eating fish but fishin urself!

okay guys let's warm up a bit. who can attempt this;

what is the complete solution of;
y = 1/(D^2 + D)e^x

where D is differential operator
e^x = exponential x

^^^ chilling n sipping coke
this one pass me eh

biolabee: ^^^ chilling n sipping coke
this one pass me eh

it's easy oh just make research on the following topics
1. differential operators
2. shift theorem
3. complementary functions and particular integrals

oga no be so.. my brain don dry na.. that was years ago..

ODEs, PDEs and stuff

SpicyMimi:
Thank you!!! Thank You!!! Thank you!!! Thank You!!! Thank You!!! Thank You!!! Thank You!!! Thank You!!! Thank You!!! Thank You!!! ThaNk You!!! Thank You!!! And Thank You!!! God am sooooo happpyyyy!!!! And sooooo Grateful!!!!!!

You are welcome dear!

SpicyMimi: Thanksss guys, have worked (b) my self, not that difficult...thanks alotttttttttttttt!!!!

Oh okay, that's great!

Richiez: okay guys let's warm up a bit. who can attempt this;

what is the complete solution of;
y = 1/(D^2 + D)e^x

where D is differential operator
e^x = exponential x

Good idea, I think I'll post some questions today or tomorrow too. Where is our five-star General?

I'm a three-star General oh

2nioshine: Do u kno y am sti on dis trend even when am nt on subscriptn?, z nt dat am nt gud n mth or realy 2 cum paste quest/soln......bt d way u give credit,ur presentatn of words amaziz me.4 a guru wt such atribute i av no choice dan 2 jus Learn @ DOUBLE DX......u rem my last word 2 u aft dat soln? i sti maintain dat...

Thank you bruv.

There are a lot of math gurus on Nairaland, heck even on this thread but they don't have the time and/or are not willing to help. It's not easy to take out time to solve a question, type it on a device and then post it online, which explains why some folks ignore questions on threads like this, it's not as if we are the best!!

doubleDx:

Good idea, I think I'll post some questions today or tomorrow too. Where is our five-star General?

I'm a three-star General oh

You are modest, if anyone should be given five stars it's you. What happened on Sunday was a joke dat went bad, I salute my general DoubleDx...

ositadima1: You are modest, if anyone should be given five stars it's you. What happened on Sunday was a joke dat went bad, I salute my general DoubleDx...

dats maturity playing it's role...thanks guys but no one has answered my questions yet, or do i put a price tag

Richiez: okay guys let's warm up a bit. who can attempt this;

what is the complete solution of;
y = 1/(D^2 + D)e^x

where D is differential operator
e^x = exponential x

^^^

That baby is damn sexy She sure deserves a prize tag

I salute my generals!

Richiez: okay guys let's warm up a bit. who can attempt this;

what is the complete solution of;
y = 1/(D^2 + D)e^x

where D is differential operator
e^x = exponential x

Lemme give it a try General richiez.

Here it goes=>

y = 1/(D^2 +D)e^x
y(D^2 + D) = e^x
Diving through by e^x yields=>
y(D^2 + D)e^(-x) = 1
Applying shift theorem
Df(x)e^(ax) = f(x)( D + a)e^(ax), since a = -1 in this case =>

y[(D - 1)^2 + (D - 1)]e^(-x) = 1
y(D^2 - 2D +1 + D - 1)e^(-x) = 1
y(D^2 - D)e^(-x) = 1

Now, let's put ye^(-x) = u
:. u(D^2 - D) = 1
uD^2 - uD = 1
Integrating both sides yields=>
∫uD^2 - ∫uD = ∫1dx
Du - u = x + C1
Integrating again yields =>
∫Du - ∫u = ∫xdx + ∫C1dx
u = x^2/2 + C1x + C2

Remember that ye^(-x) = u
:.y = ue^x, substituting the value of u yields =>
:. The complete solution of the function is =>

y = [x^2/2 + C1x + C2]e^x or x^2e^x/2 + C1xe^x + C2e^x

Where C1 and C2 are constants.

We can verify our answer by computing the second derivative of the function y = (x^2/2 + C1x + C2)e^x or x^2e^x/2 + C1xe^x + C2e^x wrt x!

ositadima1:

You are modest, if anyone should be given five stars it's you. What happened on Sunday was a joke dat went bad, I salute my general DoubleDx...

No problemo bredda! I knew it was a joke and never took it seriously either. Greetings *Shakes General Osita*

masperano: DoubleDx relax man osita was just being goofy na! U taking it too personal!

All is cool bruv. I get your point!

Let's warm up people. Solve the differential equation=>

∂^3y(x)/∂x^3 - 6∂^2y(x)/∂x^2 +12∂y(x)/∂x - 8y(x) = 0.

i'm representing my class in a science quiz tomorrow base on the following subjects:maths, physics, chemistry and biology.
I'm feeling nervous right now i'm representing my class in a science quiz tomorrow base on the following subjects:maths, physics, chemistry and biology.
I'm feeling nervous right now

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https://www.nairaland.com/1171660/government-crs-online-tutorial-4jamb

doubleDx: Lemme give it a try General richiez. Here it goes=> y = 1/(D^2 +D)e^x y(D^2 + D) = e^x Diving through by e^x yields=> y(D^2 + D)e^(-x) = 1 Applying shift theorem Df(x)e^(ax) = f(x)( D + a)e^(ax), since a = -1 in this case => y[(D - 1)^2 + (D - 1)]e^(-x) = 1 y(D^2 - 2D +1 + D - 1)e^(-x) = 1 y(D^2 - D)e^(-x) = 1 Now, let's put ye^(-x) = u :. u(D^2 - D) = 1 uD^2 - uD = 1 Integrating both sides yields=> ∫uD^2 - ∫uD = ∫1dx Du - u = x + C1 Integrating again yields => ∫Du - ∫u = ∫xdx + ∫C1dx u = x^2/2 + C1x + C2 Remember that ye^(-x) = u :.y = ue^x, substituting the value of u yields => :. The complete solution of the function is => y = [x^2/2 + C1x + C2]e^x or x^2e^x/2 + C1xe^x + C2e^x Where C1 and C2 are constants. We can verify our answer by computing the second derivative of the function y = (x^2/2 + C1x + C2)e^x or x^2e^x/2 + C1xe^x + C2e^x wrt x!

your have the true mark of a General, u fearlessly attempted the question

^lolz!

johnpaul1101: i'm representing my class in a science quiz tomorrow base on the following subjects:maths, physics, chemistry and biology.
I'm feeling nervous right now i'm representing my class in a science quiz tomorrow base on the following subjects:maths, physics, chemistry and biology.
I'm feeling nervous right now

You needn't feel nervous.Just pray,plan and revise what you've learnt so far and leave the rest to GOD.Wishing you the best.

Richiez: okay guys let's warm up a bit. who can attempt this; what is the complete solution of; y = 1/(D^2 + D)e^x where D is differential operator e^x = exponential x

Now, here's the long awaited solution;
Y = 1/(D^2 + D)*e^x
(D^2 + D)Y = e^x
to obtain the complementary function, we need an auxillary equation from the L.H.S
D^2 + D = 0 ......auxillary eqn D= 0 or -1
let these roots be m1=0 and m2= -1
C.F = C1e^m1x + C2e^m2x
C.F = C1e^0x + C2e^-x
C.F = C1 + C2e^-x
the 2nd part of the solution is the particular integral, and it's gotten from the R.H.S of the eqn.
1/(D^2 + D)*e^x
from shift theorem;
1/f(D)*e^ax = 1/f(a)*e^ax
this simply means replacing 'D' with 'a' 1/(D^2 + D)*e^x = 1/(1^2 + 1)*e^x
P.I = 1/2(e^x)
the complete solution is; Y = C.F + P.I
Y = C1 + C2(e^-x) + 1/2(e^x)
to check; recall that (D^2 + D)Y = e^x
(D^2 + D)[C1 + C2(e^-x) + 1/2(e^x) = e^x where;
D^2 = d^2y/dx^2
D = dy/dx
C1 and C2 are constants.