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[Chapa Nalo Jr]

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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, vipi ndio yule ninaemjua, original Machame

 

[Gwangambo]

JF Porojo tu, no mathematitians, all ngwinis

 

[Rua]

Zzzzzz! Mi ni mtaalam wa 2+2=4 mambo ya let x = y, given x + y=3,nawaachia manguli.

 

[SaaMbovu]

Ukisoma Engineering hasa Electrical ndio utajua zinakua applied vipi hasa kwenye Magnetic kama ilivyo Complex na Matrices kwenye Power System. Ndio maana Engineers hasa wa umeme wanajua sana hesabu.

 

[SaaMbovu]

JF Porojo tu, no mathematitians, all ngwinis

Wewe acha hizo. Mm mbona sio ngwini.

 

[kisumapai]

Hapo mi simo kwanxa hesabu yenyewe nilikuwa nalala mbele

 

[mamanalia]

Differential equations zinatumika sana kwenye modelling, huo mziki wa modelling usikie tu, ninaweza kuufananisha na turbo machinery au energy n materials balance

 

[Gwangambo]

Wewe acha hizo. Mm mbona sio ngwini.

kOKOTOA MOJA BASI

 

[Makanyaga]

Umezitoa wapi? Trunter,Shayo au Mzumbe Project?

Nadhani itakuwa ni kwenye Schaum's Outline Series

 

[Makanyaga]

View attachment 124731

<!--[if !mso]> <style> v\:* {behavior:url(#default#VML);} o\:* {behavior:url(#default#VML);} w\:* {behavior:url(#default#VML);} .shape {behavior:url(#default#VML);} </style> <![endif]-->Qn1 (1+y[SUP]2[/SUP])dx-[y-(1+y)[SUP]1/2[/SUP]](1+x)[SUP]3/2[/SUP]dy

Rearanging the above equations, colleting similar variables together (separating thevariables) and thereafter simplifying, gives the following expressions

(1+x)-[SUP]3/2[/SUP]dx=[y-(1+y)[SUP]1/2[/SUP]]/(1+y[SUP]2[/SUP])dy

&#10233;(1+x)-[SUP]3/2[/SUP]dx=[y/(1+y[SUP]2[/SUP])]dy&#8211;[(1+y)[SUP]1/2[/SUP]/(1+y[SUP]2[/SUP])] dy

The LHS integrates to

-2(1+x)-[SUP]1/2[/SUP] .............................................................(1) whereas;

the first term [y/(1+y[SUP]2[/SUP])]dyon the RHS integrates to
[SUP]
1[/SUP]/[SUB]2[/SUB]ln(1+y[SUP]2[/SUP])................................................................(2)

Now it remains to find theintegral of [(1+y)[SUP]1/2[/SUP]/(1+y[SUP]2[/SUP])] dy.

Let I=&#10780;[(1+y)[SUP]1/2[/SUP]/(1+y[SUP]2[/SUP])]dy.............................(3)

Again, if we let u=(1+y)[SUP]1/2[/SUP]

Then u[SUP]2[/SUP]=1+y

&#10233;2udu=dy

Therefore I=&#10780;[(1+y)[SUP]1/2[/SUP]/(1+y[SUP]2[/SUP])]dy=&#10780;[2u[SUP]2[/SUP]/(u[SUP]4[/SUP]-2u[SUP]2[/SUP]+2)]du.

Therefore, the change ofvariable from y to u transforms the integral to

I=&#10780;[2u[SUP]2[/SUP]/(u[SUP]4[/SUP]-2u[SUP]2[/SUP]+2)]du,

which can be split intopartial fractions whose denominators must be the roots of the polynomialequation u[SUP]4[/SUP]-2u[SUP]2[/SUP]+2=0



Solving for the roots of the polynomial u[SUP]4[/SUP]-2u[SUP]2[/SUP]+2

By considering the polynomialu[SUP]4[/SUP]-2u[SUP]2[/SUP]+2 as a quadratic equation in u[SUP]2[/SUP] i.e. u[SUP]4[/SUP]-2u[SUP]2[/SUP]+2=(u[SUP]2[/SUP])[SUP]2[/SUP]-2(u[SUP]2[/SUP])+2,then we can solve for u[SUP]2 [/SUP]in the quadratic equation (u[SUP]2[/SUP])[SUP]
2[/SUP]-2(u[SUP]2[/SUP])+2=0

Therefore u[SUP]2[/SUP]<v:shapetype id="_x0000_t75" stroked="f" filled="f" path="m@4@5l@4@11@9@11@9@5xe" o😛referrelative="t" o:spt="75" coordsize="21600,21600"> <v:stroke joinstyle="miter"> <v:f eqn="if lineDrawn pixelLineWidth 0"> <v:f eqn="sum @0 1 0"> <v:f eqn="sum 0 0 @1"> <v:f eqn="prod @2 1 2"> <v:f eqn="prod @3 21600 pixelWidth"> <v:f eqn="prod @3 21600 pixelHeight"> <v:f eqn="sum @0 0 1"> <v:f eqn="prod @6 1 2"> <v:f eqn="prod @7 21600 pixelWidth"> <v:f eqn="sum @8 21600 0"> <v:f eqn="prod @7 21600 pixelHeight"> <v:f eqn="sum @10 21600 0"> </v:f> <v😛ath o:connecttype="rect" gradientshapeok="t" o:extrusionok="f"> <o:lock aspectratio="t" v:ext="edit"></o:lock><v:shape id="_x0000_i1025" style="width: 90pt; height: 34.5pt;" type="#_x0000_t75"> <v:imagedata o:title="" chromakey="white" src="file:///C:\Users\machibya\AppData\Local\Temp\msohtmlclip1\01\clip_image001.png"></v:imagedata></v:shape></v😛ath></v:f></v:f></v:f></v:f></v:f></v:f></v:f></v:f></v:f></v:f></v:f></v:stroke></v:shapetype> where <v:shape id="_x0000_i1025" style="width: 39pt; height: 21.75pt;" type="#_x0000_t75"> <v:imagedata o:title="" chromakey="white" src="file:///C:\Users\machibya\AppData\Local\Temp\msohtmlclip1\01\clip_image002.png"></v:imagedata></v:shape> , b=-2 and c=+2 which gives two conjugatecomplex roots:

u[SUP]2[/SUP]=1+i or

u[SUP]2[/SUP]=1-i

Therefore 2u[SUP]2[/SUP]/(u[SUP]4[/SUP]-2u[SUP]2[/SUP]+2can be written in the partial fractions form as

A/[u[SUP]2[/SUP]-(1+i)]+B/[u[SUP]2[/SUP]-(1-i)]=2u[SUP]2[/SUP]/(u[SUP]4[/SUP]-2u[SUP]2[/SUP]+2) where A and B are coefficientswhich need to be detemined. On solving however, we get

A=1-i and

B=1+i

&#10233;2u[SUP]2[/SUP]/(u[SUP]4[/SUP]-2u[SUP]2[/SUP]+2)can be written as

(1-i)/[u[SUP]2[/SUP]-(1+i)]+ (1+i)/[u[SUP]2[/SUP]-(1-i)]

Therefore

2u[SUP]2[/SUP]/(u[SUP]4[/SUP]-2u[SUP]2[/SUP]+2)=(1-i)/[u[SUP]2[/SUP]-(1+i)]+ (1+i)/[u[SUP]2[/SUP]-(1-i)] and hence

I=&#10780;[2u[SUP]2[/SUP]/(u[SUP]4[/SUP]-2u[SUP]2[/SUP]+2)]du=&#10780;{(1-i)/[u[SUP]2[/SUP]-(1+i)]+ (1+i)/[u[SUP]2[/SUP]-(1-i)]}du...................................................................................................................(4)

I=&#10780;{(1-i)/[u[SUP]2[/SUP]-(1+i)]}du+&#10780;{(1+i)/[u[SUP]2[/SUP]-(1-i)]} du

Now let &#945;[SUP]2[/SUP]=1+i and&#946;[SUP]2[/SUP]=1-i

Therefore I=&#10780;{(1-i)/[u[SUP]2[/SUP]-(1+i)]}du+&#10780;{(1+i)/[u[SUP]2[/SUP]-(1-i)]} du

Changes to the expressions interms of &#945; and &#946; as

I=&#10780;[&#946;[SUP]2[/SUP]/(u[SUP]2[/SUP]-&#945;[SUP]2[/SUP])]du+&#10780;[&#945;[SUP]2[/SUP]/(u[SUP]2[/SUP]-&#946;[SUP]2[/SUP])]du............................(5)

Let L=&#10780;[&#946;[SUP]2[/SUP]/(u[SUP]2[/SUP]-&#945;[SUP]2[/SUP])]du and

M=&#10780;[&#945;[SUP]2[/SUP]/(u[SUP]2[/SUP]-&#946;[SUP]2[/SUP])]du

So that I=L+M

Now consider the first partof the integral in equation (5)

L=&#10780;[&#946;[SUP]2[/SUP]/(u[SUP]2[/SUP]-&#945;[SUP]2[/SUP])]du

Now, if we let u/&#945;=sec&#952;

&#10233;u=&#945;sec&#952; &#10233;du= &#945;sec&#952;tan&#952;d&#952;

Therefore L=&#10780;[&#946;[SUP]2[/SUP]/(u[SUP]2[/SUP]-&#945;[SUP]2[/SUP])]du=&#946;[SUP]2[/SUP]/&#945;[SUP] 2[/SUP]&#10780;&#945;sec&#952;tan&#952;d&#952;/(sec[SUP]2[/SUP]&#952;-1)

L=&#946;[SUP]2[/SUP]/&#945;[SUP] 2[/SUP]&#10780;&#945;sec&#952;tan&#952;d&#952;/(sec[SUP]2[/SUP]&#952;-1)=&#946;[SUP]2[/SUP]/&#945;[SUP] 2[/SUP]&#10780;&#945;sec&#952;tan&#952;d&#952;/tan[SUP]2[/SUP]&#952;

&#10233;L= &#946;[SUP]2[/SUP]/&#945;&#10780;sec&#952;d&#952;/tan&#952;

=&#946;[SUP]2[/SUP]/&#945;&#10780;sec&#952;d&#952;/tan&#952;

=&#946;[SUP]2[/SUP]/&#945;&#10780;cosec&#952;d&#952;=(-&#946;[SUP]2[/SUP]/&#945😉ln(cosec&#952;+cot&#952😉

However, in any right angledtriangle where sec&#952;=u/&#945; (or cos &#952;=&#945;/u), one must be able to show that

cosec&#952;=u/(u[SUP]2[/SUP]-&#945;[SUP]2[/SUP])[SUP]1/2[/SUP]and cot&#952;= &#945;/(u[SUP]2[/SUP]-&#945;[SUP]2[/SUP])[SUP]1/2[/SUP] .Therefore

L=(-&#946;[SUP]2[/SUP]/&#945😉ln[u/(u[SUP]2[/SUP]-&#945;[SUP]2[/SUP])[SUP]1/2[/SUP]+&#945;/(u[SUP]2[/SUP]-&#945;[SUP]2[/SUP])[SUP]1/2[/SUP]]

&#10233; L=(-&#946;[SUP]2[/SUP]/&#945😉ln[(u+ &#945; )/(u[SUP]2[/SUP]-&#945;[SUP]2[/SUP])[SUP]1/2[/SUP]]

L=(-&#946;[SUP]2[/SUP]/&#945😉ln{(u+&#945; )/[(u+ &#945😉[SUP]1/2[/SUP] (u-&#945😉[SUP] 1/2[/SUP]]}

L=(-&#946;[SUP]2[/SUP]/&#945😉ln[(u+&#945😉[SUP]1/2[/SUP]/(u-&#945😉[SUP]1/2[/SUP]]

L=(&#946;[SUP]2[/SUP]/2&#945😉ln[(u-&#945😉/(u+&#945😉].............................................(7)

Logically, in a similarfashion

M=(&#945;[SUP]2[/SUP]/2&#946😉ln[(u-&#946😉/(u+&#946😉].............................................(8)

Therefore

I=L+M

I=&#10780;{(&#946;[SUP]2[/SUP]/2&#945😉ln[(u-&#945😉/(u+&#945😉}du+&#10780;{(&#945;[SUP]2[/SUP]/2&#946😉ln[(u-&#946😉/(u+&#946😉]}du

can now be written as

I=(&#946;[SUP]2[/SUP]/2&#945😉ln[(u-&#945😉/(u+ &#945😉]+ (&#945;[SUP]2[/SUP]/2&#946😉ln[(u-&#946😉/(u+&#946😉]..................................................................................................................................(9)

where

u=(1+y)[SUP]1/2[/SUP],

&#945;=[(1+i)[SUP]1/2[/SUP] ,-(1+i)[SUP]1/2[/SUP])]and

&#946;=[(1-i)[SUP]1/2[/SUP],-(1-i)[SUP]1/2[/SUP])]

Therefore relatingand/combining equations (1), (2) and (9) above gives

-2(1+x)-[SUP]1/2[/SUP]=[SUP]1[/SUP]/[SUB]2[/SUB]ln(1+y[SUP]2[/SUP])+(&#946;[SUP]2[/SUP]/2&#945😉ln[(u-&#945😉/(u+&#945😉]+ (&#945;[SUP]2[/SUP]/2&#946😉ln[(u-&#946😉/(u+&#946😉] .....................................................................................................(10).

If we put u=(1+y)[SUP]1/2[/SUP],&#945;=(1+i)[SUP]1/2[/SUP], and &#946;=(1-i)[SUP]1/2[/SUP] in equation (10) above we get

-2(1+x)-[SUP]1/2[/SUP]=[SUP]1[/SUP]/[SUB]2[/SUB]ln(1+y[SUP]2[/SUP])+[(1-i)/2(1+i)[SUP]1/2[/SUP]]ln{[(1+y)[SUP]1/2[/SUP]-(1+i)[SUP]1/2[/SUP]]/[(1+y)[SUP]1/2[/SUP]+(1+i)[SUP]1/2[/SUP]]}+[(1+i)/2(1-i)[SUP]1/2[/SUP] ]ln{[(1+y)[SUP]1/2[/SUP]-(1-i)[SUP]1/2[/SUP]]/[(1+y)[SUP]1/2[/SUP]+(1-i)[SUP]1/2[/SUP]]}

&#10233;f(x,y)= 2(1+x)-[SUP]1/2[/SUP]+[SUP]1[/SUP]/[SUB]2[/SUB]ln(1+y[SUP]2[/SUP])+[(1-i)/2(1+i)[SUP]1/2[/SUP]]ln{[(1+y)[SUP]1/2[/SUP]-(1+i)[SUP]1/2[/SUP]]/[(1+y)[SUP]1/2[/SUP]+(1+i)[SUP]1/2[/SUP]]}+[(1+i)/2(1-i)[SUP]1/2[/SUP] ]ln{[(1+y)[SUP]1/2[/SUP]-(1-i)[SUP]1/2[/SUP]]/[(1+y)[SUP]1/2[/SUP]+(1-i)[SUP]1/2[/SUP]]}=0

is one implicit solution tothe above DE in Question1

Again, if we put u=(1+y)[SUP]1/2[/SUP],&#945;=-(1+i)[SUP]1/2[/SUP], and &#946;=-(1-i)[SUP]1/2[/SUP] in equation (10) above weget

-2(1+x)-[SUP]1/2[/SUP]=[SUP]1[/SUP]/[SUB]2[/SUB]ln(1+y[SUP]2[/SUP])-[(1-i)/2(1+i)[SUP]1/2[/SUP]]ln{[(1+y)[SUP]1/2[/SUP]+(1+i)[SUP]1/2[/SUP]]/[(1+y)[SUP]1/2[/SUP]-(1+i)[SUP]1/2[/SUP]]}-[(1+i)/2(1-i)[SUP]1/2[/SUP] ]ln{[(1+y)[SUP]1/2[/SUP]+(1-i)[SUP]1/2[/SUP]]/[(1+y)[SUP]1/2[/SUP]-(1-i)[SUP]1/2[/SUP]]}

&#10233;f(x,y)= -2(1+x)-[SUP]1/2[/SUP]+[SUP]1[/SUP]/[SUB]2[/SUB]ln(1+y[SUP]2[/SUP])-[(1-i)/2(1+i)[SUP]1/2[/SUP]]ln{[(1+y)[SUP]1/2[/SUP]+(1+i)[SUP]1/2[/SUP]]/[(1+y)[SUP]1/2[/SUP]-(1+i)[SUP]1/2[/SUP]]}-[(1+i)/2(1-i)[SUP]1/2[/SUP] ]ln{[(1+y)[SUP]1/2[/SUP]+(1-i)[SUP]1/2[/SUP]]/[(1+y)[SUP]1/2[/SUP]-(1-i)[SUP]1/2[/SUP]]}=0

is another implicit solutionto the above DE in Question1

 

[Makanyaga]

View attachment 124731

Solutions zote kwa maswali haya zipo, zitatoka muda siyo mrefu ndani ya mwaka huu. Ubarikiwe na Bwana

 

[Mawembasa1979]

Heheh kama ulifika hadi huku basi ulijitahidi maana hii makitu ni ya A-Level au Chuo...

Wale waliosoma Additional O-Level waliinusa nusa tu kwa mbaliiiii

Kule kwenye Vitabu vya Backhouse 2 na C. J . Tranter hizi makitu zipo sana.

 

[Makanyaga]

View attachment 124731

Haya maswali uliyapa wapi? Ulipewa na mtu? Swali la 3 liko tofauti kabisa na maswali mengine, very advanced

 

[donlucchese]

Nilipewa na mkali wangu ZeGrebo

Haya maswali uliyapa wapi? Ulipewa na mtu? Swali la 3 liko tofauti kabisa na maswali mengine, very advanced

 

[Makanyaga]

Nilipewa na mkali wangu ZeGrebo

Safi kabisa. Naomba uniunganishe naye kama bado yupo humu jukwaani. Nahitaji kujadili naye kuhusu hilo swali la 3.

cc: @ZeGrebo unatafutwa huku kuna kitu!

 

[masopakyindi]

View attachment 124731

Du... niipokuwa A Level nilikuwa nazipiga kama mswaki.
Lakini sasa zimebaki hesabu za noti tu!!
Kweli dunia duara!!

 

[donlucchese]

Hatari mkuu hazina hata application

Du... niipokuwa A Level nilikuwa nazipiga kama mswaki.
Lakini sasa zimebaki hesabu za noti tu!!
Kweli dunia duara!!