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

donlucchese · Dec 6, 2013

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

mkuu umenipa raha sana, saana! unakumbuka kwenye trig kuderive formula zile za double angles eg. kuderive cos2x= cos[SUP]2[/SUP]X-Sin[SUP]2[/SUP]X au Sin2x= 2SinXCosX? Raha sana

donlucchese · Dec 6, 2013

LexAid said:
Hazinipi shida Mkuu, Kama kumsukuma mlevi vile!

nakukubali sana mkuu

donlucchese · Dec 6, 2013

Threesixteen Himself said:
!
!
sifa tu lione kwanza.

hahaha kwanini tena mkuu?

donlucchese · Dec 6, 2013

sasa hapo kwenye D.E's kuna kitu inaitwa [FONT=Arial, sans-serif]Nonlinear differential equations and bifurcation theory[/FONT]

MjuviKitambo · Dec 6, 2013

Mna bwebwe Watu wa mahesabu!!!!hamuachi tuuu. Form five week one Pugu pale tuition ya physics ( mechanics) nikakutana na nyoka nyoka sijawai kuziona sielewi, namuuliza teacher anasema kalili tuuu dogo. Hatariiii

donlucchese · Dec 6, 2013

MjuviKitambo said:
Mna bwebwe Watu wa mahesabu!!!!hamuachi tuuu. Form five week one Pugu pale tuition ya physics ( mechanics) nikakutana na nyoka nyoka sijawai kuziona sielewi, namuuliza teacher anasema kalili tuuu dogo. Hatariiii

hahahaha

Omnibus · Dec 6, 2013

Tyta said:
i love mathematics

Shusha basi hizo nondo kama kweli unazipenda hisabati

Tyta · Dec 7, 2013

omnibus said:
shusha basi hizo nondo kama kweli unazipenda hisabati

sasa mkuu,hivi maswali yote hayo unataka nifanye......????

Ninaweza · Dec 7, 2013

Baba V said:
Nilipofika hapa ndo nikazikimbia hesabu...

dyscalculia?

cataliya · Dec 7, 2013

Honey Faith said:
kiukweli hizo hesabu sijawahi kuziona

A level hukusoma advance math!!!
Hated notes sana!!

DAWA YA SIKIO · Dec 7, 2013

Hivi haya makorokocho yana msaada gani katika maisha ya mtz?! Nimesoma BAM lakini huu mwaka wa 18 toka nitoke shule, sijaona faida yoyote.... Ni ukoloni tu.

Makanyaga · Dec 7, 2013

donlucchese said:
View attachment 124731

Nitakujibu maswali yote kwa vipindi tofauti tofauti ila sitaonyesha njia.
Tukianza na swali namba 16

Makanyaga · Dec 7, 2013

donlucchese said:
View attachment 124731

Qn No. 16
Let m=dy/dx

Makanyaga · Dec 7, 2013

donlucchese said:
View attachment 124731

Qn No. 16
Let m=dy/dx

Therefore (a)(dm/dx)=m

then dm/m=(1/a)(dx)
ln(m)=(x/a)+C where C is constant of integration and "ln" stands for a natural logarithm of a number i.e. the logarithm of a (variable) number ("m" in this case), to base "e".

m=e^[(x/a)+C]. But m=dy/dx

Therefore dy/dx=e^[(x/a)+C]

dy=(e^[(x/a)+C])dx

Therefore y=(a)(e^[(x/a)+C])+D where D is another constant of integration

simplemind · Dec 7, 2013

DAWA YA SIKIO said:
Hivi haya makorokocho yana msaada gani katika maisha ya mtz?! Nimesoma BAM lakini huu mwaka wa 18 toka nitoke shule, sijaona faida yoyote.... Ni ukoloni tu.

wanafunzi wa sayansi na uhandisi wanafahamu umuhimu wa "makorokocho" haya. Maxwell's Equations

MKATA KIU · Dec 7, 2013

Haya makorokocho ni moja ya sababu ya taifa hili kuwa maskini mpaka leo..

Makanyaga · Dec 11, 2013

MKATA KIU said:
Haya makorokocho ni moja ya sababu ya taifa hili kuwa maskini mpaka leo..

Tuachie wenyewe tunaoyapenda, bora kufa maskini kuliko kuyaacha. By the way, bila haya hata ndege isingeweza kutembea angani, kwa hiyo hata Ulaya au Afrika tusingekuwa tunajua kama ipo!

Makanyaga · Dec 11, 2013

donlucchese said:
View attachment 124731

Qn. 15
(d[SUP]8[/SUP]y/dx[SUP]8[/SUP])-2(d[SUP]4[/SUP]y/dx[SUP]4[/SUP])+y=0
Solution
Assume a solution (involving the variables x and y) of the form y=e[SUP]ax [/SUP] where a is a constant.
Then dy/dx=ae[SUP]ax [/SUP] , so that
d[SUP]8[/SUP]y/dx[SUP]8[/SUP]=a[SUP]8[/SUP]e[SUP]ax[/SUP] and
d[SUP]4[/SUP]y/dx[SUP]4[/SUP]=a[SUP]4[/SUP]e[SUP]ax[/SUP]

On substituting these values of y, d[SUP]8[/SUP]y/dx[SUP]8[/SUP] and d[SUP]4[/SUP]y/dx[SUP]4 [/SUP]in the original expression we get
e[SUP]ax[/SUP] (a[SUP]8[/SUP]-2a[SUP]4[/SUP]+1)= e[SUP]ax[/SUP] (a[SUP]4[/SUP]-1)(a[SUP]4[/SUP]-1)=0
Since e[SUP]ax[/SUP] ≠0 then we must have (a[SUP]4[/SUP]-1)(a[SUP]4[/SUP]-1)=0⟹[(a[SUP]2[/SUP]-1)(a[SUP]2[/SUP]-1)][SUP]2[/SUP]=0
Therefore a[SUP]2[/SUP]=1 or a[SUP]2[/SUP]=-1
Case 1: a[SUP]2[/SUP]=1
a=1 or a=-1 ⟹ y= e[SUP]x[/SUP] or y= e-[SUP]x[/SUP] which gives a general solution of the form y= c[SUB]1[/SUB] e[SUP]x[/SUP] +c[SUB]2[/SUB]e-[SUP]x[/SUP]where c[SUB]1[/SUB]and c[SUB]2[/SUB] are constants Therefore y= c[SUB]1[/SUB] e[SUP]x[/SUP] +c[SUB]2[/SUB]e-[SUP]x [/SUP] is a solution to the above differential equation in Question No. 15
Case 2: a[SUP]2[/SUP]=-1
The roots of the quadratic equation a[SUP]2[/SUP]=-1 are complex and hence a=i or a=-i ⟹ y=e[SUP]ix [/SUP]or y= e-[SUP]ix[/SUP] which gives a general solution of the form y= c[SUB]3[/SUB]e[SUP]ix [/SUP]+ c[SUB]4[/SUB]e-[SUP]ix[/SUP] where c[SUB]3[/SUB]and c[SUB]4[/SUB] are constants Therefore y= c[SUB]3[/SUB]e[SUP]ix [/SUP]+ c[SUB]4[/SUB]e-[SUP]ix[/SUP] is also, another solution to the same differential equation in Question No. 15

Note: Same/similar procedure solves the Differential Equations, Question numbers 11 to 14. In this case therefore, only the solutions and NOT the procedures of arriving at such solutions, will be shown.

COMING SOON TO QUESTIONS 11 TO 14

Makanyaga · Dec 12, 2013

donlucchese said:
View attachment 124731

Qn. 14
(d[SUP]3[/SUP]y/dx[SUP]3[/SUP])-(d[SUP]2[/SUP]y/dx[SUP]2[/SUP])+(dy/dx)=0
Solution
Let dy/dx=e[SUP]ax [/SUP] , so that
d[SUP]2[/SUP]y/dx[SUP]2[/SUP]=a[SUP]2[/SUP]e[SUP]ax[/SUP] and
d[SUP]3[/SUP]y/dx[SUP]3[/SUP]=a[SUP]3[/SUP]e[SUP]ax[/SUP]

On substituting these values of dy/dx, d[SUP]3[/SUP]y/dx[SUP]3[/SUP] and d[SUP]2[/SUP]y/dx[SUP]2 [/SUP]in the original expression we get
e[SUP]ax[/SUP] (a[SUP]3[/SUP]-a[SUP]2[/SUP]+a)= ae[SUP]ax[/SUP] (a[SUP]2[/SUP]-a+1)=0
Since ae[SUP]ax[/SUP] ≠0 then we must have (a[SUP]2[/SUP]-a+1)=0 which has complex roots.
Therefore
a=a[SUB]1[/SUB]=[-1+(3)[SUP]1/2[/SUP] i]/2 or
a= a[SUB]2[/SUB]=[-1-(3)[SUP]1/2[/SUP] i]/2

Therefore, the solution to the above differential equation is given by

y= c[SUB]1[/SUB] e[SUP]a[/SUP][SUB]1[/SUB][SUP]x[/SUP] +c[SUB]2[/SUB]e[SUP]a[/SUP][SUB]2[/SUB][SUP]x[/SUP]
where a[SUB]1=[/SUB][-1+(3)[SUP]1/2[/SUP] i]/2 and a[SUB]2[/SUB]=[-1-(3)[SUP]1/2[/SUP] i]/2

 Qn. 12
(d[SUP]3[/SUP]x/dt[SUP]3[/SUP])-6(d[SUP]2[/SUP]x/dt[SUP]2[/SUP])+9(dx/dt)=0
Solution:

Let dx/dt=e[SUP]at [/SUP]as usual so that
d[SUP]2[/SUP]x/dt[SUP]2[/SUP]=a[SUP]2[/SUP]e[SUP]at[/SUP] and
d[SUP]3[/SUP]x/dt[SUP]3[/SUP]=a[SUP]3[/SUP]e[SUP]at[/SUP]

On substituting these values of dx/dt, d[SUP]3[/SUP]x/dt[SUP]3[/SUP] and d[SUP]2[/SUP]x/dt[SUP]2 [/SUP]in the original expression we get
e[SUP]at[/SUP] (a[SUP]3[/SUP]-6a[SUP]2[/SUP]+9a)= ae[SUP]at[/SUP] (a[SUP]2[/SUP]-6a+9)=0
Since ae[SUP]at[/SUP] ≠0 then we must have (a[SUP]2[/SUP]-6a+9)=0 which has repeated real roots a=3
Now, since a=3. is a repeated root of the above quadratic equation, then the general solution to the above differential equation is given by x= c[SUB]1[/SUB] e[SUP]3t[/SUP] +c[SUB]2[/SUB] te[SUP]3t[/SUP] where c[SUB]1[/SUB] and c[SUB]2[/SUB] are constants.

PREPARATIONS ARE UNDERWAY FOR QUESTION NUMBERS 11 AND 13 WHOSE APPROACH OF FINDING THEIR ROOTS IS SLIGHTLY DIFFERENT FROM ALL QUESTIONS WHICH HAVE ALREADY BEEN SOLVED, NEED FOR REVISING THE APPROACH. Nina miaka karibia 20 tangu nimeachana na haya makorokocho BUT I LOVE THEM ANYWAY!

mij · Dec 12, 2013

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

Ha ha ha ha ha ha ha ha ha ha ha ha !!!!!! Mkuu umenivunja mbavu sana.

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

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