Nawaachia Homework wale wa mambo yetu yaleee back in days (still got it?).....!!!!!

Hornet · Dec 25, 2013

donlucchese said:
View attachment 124731

😱😱;>

King Kong III · Dec 25, 2013

Umezitoa wapi? Trunter,Shayo au Mzumbe Project?

jacjaz · Dec 25, 2013

Mambo ya KSS hapo,kitu ndani ya Library na uko na Pure 2.. Kuna mtu alikuwa anajiita baba wa differentials,yan swali kwa hii topic halikatizi kwake!

Janjaweed · Dec 25, 2013

The.... Haina posho

Makanyaga · Dec 25, 2013

donlucchese said:
kaka uko vizuri sana and i like the fact that you do it for fun!

I'm actually, not doing it for funny because it makes me revise for one, and it as well takes my memory of events to about 17 years back in time!

Makanyaga · Jan 2, 2014

donlucchese said:
View attachment 124731

Qn. 8

3y[SUP]2[/SUP]y'+y[SUP]3[/SUP]=x-1

⟹3y'+y=y-[SUP]2[/SUP](x-1). This is Bernolli's equation with n=-2..........(1)

Let v=y[SUP]1-(-2)[/SUP] so that

v=y[SUP]3[/SUP] ⟹v'=3y[SUP]2[/SUP]y'

⟹ y'=⅓v'y-[SUP]2[/SUP] and

y=vy-[SUP]2[/SUP]

Substituting these values of y' and y in equation (1) above, we get

3(⅓v'y-[SUP]2[/SUP])+ vy-[SUP]2[/SUP]=y-[SUP]2[/SUP](x-1).

⟹v'+v=(x-1)...................................................................................(2)

At this stage, we have an inexact first order linear differential equation which can be solved by

using the technique of an integrating factor. Until now, one must be able to show that the

integrating factor (i.e. a factor which makes the above inexact DE exact) is e[SUP]x[/SUP]

Hence;

e[SUP]x[/SUP](v'+v)=e[SUP]x[/SUP](x-1)

⟹d/dx(e[SUP]x[/SUP]v)= e[SUP]x[/SUP] (x-1)

⟹e[SUP]x[/SUP]v=⨜e[SUP]x[/SUP](x-1)dx

=⨜xe[SUP]x[/SUP]dx-⨜e[SUP]x[/SUP]dx=xe[SUP]x[/SUP]-e[SUP]x[/SUP]-e[SUP]x[/SUP]

Therefore e[SUP]x[/SUP]v=e[SUP]x[/SUP](x-2) ⟹v=x-2+C

But, v=y[SUP]3[/SUP]=x-2+C, where C is a constant of integration

⟹y=[(x-2)+C][SUP]1/3[/SUP] is a solution to the above differential equation in Qn No. 8

Qn. 6

yy'+[SUP]1[/SUP]/[SUB]2[/SUB]y[SUP]2[/SUP]=sinx

⟹y'+[SUP]1[/SUP]/[SUB]2[/SUB]y=y[SUP]-1[/SUP]sinx..........................................................................................(1)

Similary, this is a Bernolli's equation with n=-1 and hence we let v=y[SUP]2[/SUP] so that

v'=2yy'⟹y'=[SUP]1[/SUP]/[SUB]2[/SUB](v')y[SUP]-1[/SUP] and

y=vy[SUP]-1[/SUP]

Substituting these values of y' and y in equation (1) above, we get
[SUP]
1[/SUP]/[SUB]2[/SUB](v')y[SUP]-1[/SUP]+[SUP]1[/SUP]/[SUB]2[/SUB]vy[SUP]-1[/SUP]=sinx

⟹v'y[SUP]-1[/SUP]+vy[SUP]-1[/SUP]=2y[SUP]-1[/SUP]sinx

⟹v'+v=2sinx

Again, the integrating factor of the DE is e[SUP]x[/SUP]

Hence

e[SUP]x[/SUP](v'+v)=2e[SUP]x[/SUP]sinx

⟹d/dx(e[SUP]x[/SUP]v)=2e[SUP]x[/SUP]sinx

⟹e[SUP]x[/SUP]v=2⨜e[SUP]x[/SUP]sinxdx................................................................................ (2)

=2[e[SUP]x[/SUP]sinx-⨜e[SUP]x[/SUP]cosxdx]

=2[e[SUP]x[/SUP]sinx-(e[SUP]x[/SUP]cosx+⨜e[SUP]x[/SUP]sinxdx)]............................................(3)

But, from equation (2) above,⨜e[SUP]x[/SUP]sinxdx=[SUP]1[/SUP]/[SUB]2[/SUB]e[SUP]x[/SUP]v. Therefore, substituting this value of the integral

in equation (3) above

we get

e[SUP]x[/SUP]v=2[e[SUP]x[/SUP]sinx-(e[SUP]x[/SUP]cosx+[SUP]1[/SUP]/[SUB]2[/SUB]e[SUP]x[/SUP]v)]

⟹e[SUP]x[/SUP]v=2e[SUP]x[/SUP]sinx-2e[SUP]x[/SUP]cosx-e[SUP]x[/SUP]v

⟹2e[SUP]x[/SUP]v=2e[SUP]x[/SUP]sinx-2e[SUP]x[/SUP]cosx

⟹v=sinx-cosx+C, where C is a constant of integration

But v=y[SUP]2[/SUP]

Therefore y[SUP]2[/SUP]= sinx-cosx+C ⟹y=[SUP]+/-[/SUP](sinx-cosx+C)[SUP]1/2[/SUP]

Therefore y=[SUP]+/-[/SUP](sinx-cosx+C)[SUP]1/2[/SUP] is a solution to the above DE in Qn number 6

Note: The operator "+/-" here stands for the plus, "+" or minus, "-" signs

The Wolf · Jan 2, 2014

JF raha sana

Makanyaga · Jan 4, 2014

13 mega pixel said:
JF raha sana

Ni raha tupu kweli. Maana hata assignment kuna uwezekano ikasoviwa JF halafu wewe ukafanya ku-copy tu na ku-submit kwa lecturer

Threesixteen Himself · Jan 4, 2014

Makanyaga said:
Ni raha tupu kweli. Maana hata assignment kuna uwezekano ikasoviwa JF halafu wewe ukafanya ku-copy tu na ku-submit kwa lecturer

!
!
Jf sio rahatupu, rahatupu wanarahatupuisha

Makanyaga · Jan 4, 2014

Threesixteen Himself said:
!
!
Jf sio rahatupu, rahatupu wanarahatupuisha

Rahatupulization ya ku-copy na ku-paste sasa haitakiwi tena siku hizi

Threesixteen Himself · Jan 4, 2014

!
!
Kumbe ni mdau au ulikuwa mdau wa rahatupu. Kwa sababu inaonekana unaijua vizuri hiyo blog. Umejua zamani ilikuwa vipi na sasa ikoje hahahahahahahahahahahaha haya mkuu salute

Makanyaga said:
Rahatupulization ya ku-copy na ku-paste sasa haitakiwi tena siku hizi

MOORIINGEE · Jan 4, 2014

bysange said:
na mimi namshangaa kwani zile za magazijuto hazipo siku hizi...

vipi vigawe na vigawo? vipeo na vipeuo?

Kappa · Jan 4, 2014

Hesabu zilianza kuwa ngumu pale tu walipochanganya namba na herufi.

Chocs · Jan 4, 2014

Ugonjwaaa!!

MOORIINGEE · Jan 4, 2014

ilikuwa zamani siyo leo, pesa mbele unawea ukawa mchawi na mbwembwe za hesabu wenzio tuna make money

Lucas gerald · Jan 4, 2014

hilo halinihusu mim najuwa husabu za kichina 2

masopakyindi · Jan 4, 2014

kyanaKyoMuhaya said:
Question:
Show that

y=c1sin2x+3cos2x
is a general solution for the differential equation

d2y/dx2+4y=0

Solution:

We have a second order differential equation and we have been given the general solution. Our job is to show that the solution is correct.

We do this by substituting the answer into the original 2nd order differential equation.

We need to find the second derivative of y:

y = c1 sin 2x + 3 cos 2x

First derivative:

dy/dx=2c1 cos 2x−6 sin 2x
Second derivative:

d2y/dx2=−4c1 sin 2x−12 cos 2x
Now for the check step:

LHS=d2y/dx2+4y=[−4c1sin 2x−12 cos 2x]+4(c1sin 2x+3 cos 2x)=0=RHS

Wewe kyanaKyoMuhaya utanifaa sana kutafutia hela hayo madudu.
Ni PM tungeneze roketi ya kutengeneza pesa

kiboje · Jan 4, 2014

CALCULUS.

Differential equations, Integrations.........................!

Makanyaga · Jan 4, 2014

Threesixteen Himself said:
!
!
Kumbe ni mdau au ulikuwa mdau wa rahatupu. Kwa sababu inaonekana unaijua vizuri hiyo blog. Umejua zamani ilikuwa vipi na sasa ikoje hahahahahahahahahahahaha haya mkuu salute

Sijakusoma vizuri unaposema mimi mdau, mdau wa nini?

Swat · Jan 4, 2014

King Kong III said:
Umezitoa wapi? Trunter,Shayo au Mzumbe Project?

Kinunda ha ha ha!

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