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

Makanyaga · Dec 13, 2013

Qn. 11
(d[SUP]3[/SUP]x/dt[SUP]3[/SUP])-3(d[SUP]2[/SUP]x/dt[SUP]2[/SUP])+3(dx/dt)-x=0
Solution:
Let x= e[SUP]at [/SUP]as usual so that
dx/dt= ae[SUP]at[/SUP]

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]-3a[SUP]2[/SUP]+3a-1)=e[SUP]at[/SUP] (a-3)[SUP]3[/SUP]=0 ⟹a=3 since e[SUP]at[/SUP] ≠0
Therefore x= Ae[SUP]3t[/SUP]+Bte[SUP]3t[/SUP]+Ct[SUP]2[/SUP]e[SUP]3t[/SUP] where A, B, and C are constants, is a general solution to the above differential equation

Qn. 17
d[SUP]2[/SUP]y/dx[SUP]2[/SUP]+(dy/dx)[SUP]2[/SUP]+1=0

Let dy/dx= m; so that
d[SUP]2[/SUP]y/dx[SUP]2[/SUP]=dm/dx

Therefore dm/dx +m[SUP]2[/SUP]+1=0⟹ dm/dx=-m[SUP]2[/SUP]-1 or
dm/(=-m[SUP]2[/SUP]-1)=dx⟹- dm/(=m[SUP]2[/SUP]+1)

Again, if we let m=tanα
Then dm=sec[SUP]2[/SUP]αdα so that x= ⨜(sec[SUP]2[/SUP]αd&#945😉/(sec[SUP]2[/SUP]&#945😉 since
1+ m[SUP]2[/SUP]=1+tan[SUP]2[/SUP]α=sec[SUP]2[/SUP]α

Therefore, x=α=-tan[SUP]-1 [/SUP](m)
x=-tan[SUP]-1 [/SUP](m)⟹tan(x)=m= dy/dx
⟹dy= -tan(x)dx

Therefore y=⨜-tan(x)dx=lncos(x)+C where C is a constant of integration (Note that ⨜+tanxdx=-lncosx=lnsecx)

Therefore the solution to the above differential equation is given by y= lncos(x)+C

maroon7 · Dec 13, 2013

Mbona mengi tayari yanamajibu mkuu...tena mengine umeandika majibu yake ni zero...ila hilo la mwisho jibu lake ni 2

Makanyaga · Dec 13, 2013

Patrickn said:
Mbona mengi tayari yanamajibu mkuu...tena mengine umeandika majibu yake ni zero...ila hilo la mwisho jibu lake ni 2

Look here for exemple: Find the values of x in the expression (x-2)(x-3)=0. Kwa hiyo kinachotakiwa hapa ni kutafuta uhusiano uliopo kati ya "y" plane na "x" plane given their derivatives dy, dx. In other words, we are trying to look at the relationship between x and y or simply y as a function of x

maroon7 · Dec 13, 2013

Makanyaga said:
Look here for exemple: Find the values of x in the expression (x-2)(x-3)=0. Kwa hiyo kinachotakiwa hapa ni kutafuta uhusiano uliopo kati ya "y" plane na "x" plane given their derivatives dy, dx. In other words, we are trying to look at the relationship between x and y or simply y as a function of x

Naelewa mkuu nimezisoma enzi hizo kwenye BAM nilikua nanogesha jamvi tu...halafu nilijuaga ningeenda PCM so nilizipigia sana Tuition kabla matokeo hayajatoka maana tulitishwa sana na hizi mambo

mnogapavanu · Dec 13, 2013

Jaman mambo hayo ni matamu sema tu ukibukua sana hiyo PCM hapa nchini bado hakuna ataye kujali kivile.

Mentor · Dec 13, 2013

kabanga said:
wewe umekimbia BAM....

ahahahahahahahahahahahahahahahaha una utani na mwenyekiti lolest!!!!

kabanga · Dec 13, 2013

Mentor said:
ahahahahahahahahahahahahahahahaha una utani na mwenyekiti lolest!!!!

wewe...! ndio ukweli huo!

Makanyaga · Dec 14, 2013

Patrickn said:
Naelewa mkuu nimezisoma enzi hizo kwenye BAM nilikua nanogesha jamvi tu...halafu nilijuaga ningeenda PCM so nilizipigia sana Tuition kabla matokeo hayajatoka maana tulitishwa sana na hizi mambo

Hizi alizoleta huyu jamaa ziko advanced kidogo ukilinganisha na za A level. Hizi sana sana utakutana nazo kwenye university syllabi hasa hasa mwaka wa pili aidha kwenye course ya Advanced Calculus au kwenye Defferential Equations-II, maana yake ni baada ya kuwa umefanya tuseme Defferential Equations-I, labda ukiwa First Year university

Makanyaga · Dec 14, 2013

donlucchese said:
View attachment 124731

Qn. 13
(d[SUP]4[/SUP]y/dx[SUP]4[/SUP])+2(d[SUP]2[/SUP]y/dx[SUP]2[/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]4[/SUP]y/dx[SUP]4[/SUP]=a[SUP]4[/SUP]e[SUP]ax[/SUP] and
d[SUP]2[/SUP]y/dx[SUP]2[/SUP]=a[SUP]2[/SUP]e[SUP]ax[/SUP]

On substituting these values of y, d[SUP]2[/SUP]y/dx[SUP]2[/SUP] and d[SUP]4[/SUP]y/dx[SUP]4 [/SUP]in the original expression we get
e[SUP]ax[/SUP] (a[SUP]4[/SUP]+2a[SUP]2[/SUP]+1)=0
Since e[SUP]ax[/SUP] ≠0 then we must have (a[SUP]4[/SUP]+2a[SUP]2[/SUP]+1)=0 which has complex roots. Therefore a[SUP]2[/SUP]=i or a[SUP]2[/SUP]=-i

Case 1: a[SUP]2[/SUP]=i
a=√i or a=-√i so that y= e[SUP] ([/SUP][SUP]√[/SUP][SUP]i )(x)[/SUP] or y= e[SUP](-[/SUP][SUP]√[/SUP][SUP]i)(-x)[/SUP] which gives a general solution of the form y= c[SUB]1[/SUB] e[SUP] ([/SUP][SUP]√[/SUP][SUP]i )x[/SUP] +c[SUB]2[/SUB] e[SUP] (-[/SUP][SUP]√[/SUP][SUP]i )x[/SUP] where c[SUB]1[/SUB]and c[SUB]2[/SUB] are constants.

Therefore y=c[SUB]1[/SUB] e[SUP] ([/SUP][SUP]√[/SUP][SUP]i )x[/SUP] +c[SUB]2[/SUB] e[SUP] (-[/SUP][SUP]√[/SUP][SUP]i )x[/SUP] ,is a solution to the above differential equation in Question No. 13

Case 2: a[SUP]2[/SUP]=-i
The roots of the quadratic equation a[SUP]2[/SUP]=-i is a=√-i or a=-√-i
so that y= e[SUP] ([/SUP][SUP]√-[/SUP][SUP]i )(x)[/SUP] or y= e[SUP](-[/SUP][SUP]√-[/SUP][SUP]i)(x)[/SUP] which gives a general solution of the form
y= c[SUB]3[/SUB]e[SUP] ([/SUP][SUP]√-[/SUP][SUP]i )(x)[/SUP] + c[SUB]4[/SUB]e[SUP](-[/SUP][SUP]√-[/SUP][SUP]i)(x)[/SUP] where c[SUB]3[/SUB]and c[SUB]4[/SUB] are constants,

Therefore y=c[SUB]3[/SUB]e[SUP] ([/SUP][SUP]√-[/SUP][SUP]i )(x)[/SUP] + c[SUB]4[/SUB]e[SUP](-[/SUP][SUP]√-[/SUP][SUP]i)(x)[/SUP] , is also, another general solution of the same differential equation in Question No. 13

donlucchese · Dec 16, 2013

Makanyaga said:
 Qn. 13
(d[SUP]4[/SUP]y/dx[SUP]4[/SUP])+2(d[SUP]2[/SUP]y/dx[SUP]2[/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]4[/SUP]y/dx[SUP]4[/SUP]=a[SUP]4[/SUP]e[SUP]ax[/SUP] and
d[SUP]2[/SUP]y/dx[SUP]2[/SUP]=a[SUP]2[/SUP]e[SUP]ax[/SUP]

On substituting these values of y, d[SUP]2[/SUP]y/dx[SUP]2[/SUP] and d[SUP]4[/SUP]y/dx[SUP]4 [/SUP]in the original expression we get
e[SUP]ax[/SUP] (a[SUP]4[/SUP]+2a[SUP]2[/SUP]+1)=0
Since e[SUP]ax[/SUP] ≠0 then we must have (a[SUP]4[/SUP]+2a[SUP]2[/SUP]+1)=0 which has complex roots. Therefore a[SUP]2[/SUP]=i or a[SUP]2[/SUP]=-i

Case 1: a[SUP]2[/SUP]=i
a=√ior a=-√i so that y= e[SUP] ([/SUP][SUP]√[/SUP][SUP]i )(x)[/SUP] or y= e[SUP](-[/SUP][SUP]√[/SUP][SUP]i)(-x)[/SUP] which gives a general solution of the form y= c[SUB]1[/SUB] e[SUP] ([/SUP][SUP]√[/SUP][SUP]i )x[/SUP] +c[SUB]2[/SUB] e[SUP] (-[/SUP][SUP]√[/SUP][SUP]i )x[/SUP] where c[SUB]1[/SUB]and c[SUB]2[/SUB] are constants.

Therefore y= y= c[SUB]1[/SUB] e[SUP] ([/SUP][SUP]√[/SUP][SUP]i )x[/SUP] +c[SUB]2[/SUB] e[SUP] (-[/SUP][SUP]√[/SUP][SUP]i )x[/SUP] ,is a solution to the above differential equation in Question No. 13

Case 2: a[SUP]2[/SUP]=-i
The roots of the quadratic equation a[SUP]2[/SUP]=-i is a=√-i or a=-√-i
so that y= e[SUP] ([/SUP][SUP]√-[/SUP][SUP]i )(x)[/SUP] or y= e[SUP](-[/SUP][SUP]√-[/SUP][SUP]i)(x)[/SUP] which gives a general solution of the form
y= c[SUB]3[/SUB]e[SUP] ([/SUP][SUP]√-[/SUP][SUP]i )(x)[/SUP] + c[SUB]4[/SUB]e[SUP](-[/SUP][SUP]√-[/SUP][SUP]i)(x)[/SUP] where c[SUB]3[/SUB]and c[SUB]4[/SUB] are constants,

Therefore y= y= c[SUB]3[/SUB]e[SUP] ([/SUP][SUP]√-[/SUP][SUP]i )(x)[/SUP] + c[SUB]4[/SUB]e[SUP](-[/SUP][SUP]√-[/SUP][SUP]i)(x)[/SUP] , is also, another general solution of the same differential equation in Question No. 13

nimekubali we mkabe kaka!

Makanyaga · Dec 18, 2013

donlucchese said:
View attachment 124731

Qn. 18

d[SUP]4[/SUP]y/dx[SUP]4[/SUP]=d[SUP]2[/SUP]y/dx[SUP]2[/SUP]

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]4[/SUP]y/dx[SUP]4[/SUP]=a[SUP]4[/SUP]e[SUP]ax[/SUP]

On substituting these values of d[SUP]4[/SUP]y/dx[SUP]4[/SUP] and d[SUP]2[/SUP]y/dx[SUP]2 [/SUP]in the original expression we get

a[SUP]4[/SUP]e[SUP]ax[/SUP]= a[SUP]2[/SUP]e[SUP]ax[/SUP]

Since e[SUP]ax[/SUP]≠0 and a≠0 as well, then we must have (a[SUP]4[/SUP]-a[SUP]2[/SUP])=0

⟹a[SUP]2[/SUP](a[SUP]2[/SUP]-1)=0

⟹(a+1)(a-1)=0⟹a=a[SUB]1[/SUB]=1 or a=a[SUB]2[/SUB]=-1 which gives a general solution of the form 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 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 general solution of the above differential equation in the QN number 18

Note: The above problem suggests that differentiating the solution y=c[SUB]1[/SUB] e[SUP]x[/SUP] +c[SUB]2[/SUB]e-[SUP]x[/SUP] four (4) times gives similar results as differentiating the same solution y=c[SUB]1[/SUB] e[SUP]x[/SUP] +c[SUB]2[/SUB]e-[SUP]x[/SUP] two (2) times only.

Qn. 21
x(d[SUP]2[/SUP]x/dt[SUP]2[/SUP])=2(dx/dt)[SUP]2[/SUP]

Let dx/dt= m; so that dt= dx/m and x=mt upon integration

Moreover, d[SUP]2[/SUP]x/dt[SUP]2[/SUP]=dm/dt upon substitution (m=dx/dt).

Therefore x(m)(dm/dx) =2(m[SUP]2[/SUP])

⟹xdm/dx=2m

⟹dm/m=2(dx/x)

⟹ln(m)=2ln(x) ⟹m= x[SUP]2[/SUP]

Therefore, m= x[SUP]2[/SUP]

But m= dx/dt;

Therefore dx/dt= x[SUP]2[/SUP]⟹x=-(1/t)+C where C is a constant of integration

Hence, the general solution to the ebove differential equation is given by x=-(1/t)+C where C is a constant of integration

ALTERNATIVELY
Let x=e[SUP]at[/SUP]

dx/dt=ae[SUP]at[/SUP] so that

d[SUP]2[/SUP]x/dt[SUP]2[/SUP]=a[SUP]2[/SUP]e[SUP]at[/SUP]

On substituting these values of x, dx/dt and d[SUP]2[/SUP]x/dt[SUP]2 [/SUP]in the original expression we get

(e[SUP]at[/SUP])(a[SUP]2[/SUP])(e[SUP]at[/SUP])=2(e[SUP]at[/SUP])(e[SUP]at[/SUP]); since x= e[SUP]at[/SUP]

Upon cancelling like-terms on both sides of the expression, we arrive at a simplified mathematical expression a[SUP]2[/SUP]=2 from which we

obtain a solution

x= c[SUB]3[/SUB]e[SUP] (√2 )t[/SUP] + c[SUB]4[/SUB]e[SUP](-√2)t[/SUP] (c[SUB]3 [/SUB]and c[SUB]4[/SUB] are constants)

as a general solution of the above differential equation in Question number 21

Lalita · Dec 18, 2013

ugonjwa wangu ulionichafulia cheti changu lol!!!!

chipanga · Dec 18, 2013

donlucchese said:
View attachment 124731

Mbona mchekea tu!

Makanyaga · Dec 21, 2013

donlucchese said:
View attachment 124731

Qn. 19
(1+x[SUP]2[/SUP])(d[SUP]2[/SUP]y/dx[SUP]2[/SUP])+2x(dy/dx)=0

Solution

Let dy/dx=m

Therefore (1+x[SUP]2[/SUP])(dm/dx)+2xm=0

Divide both sides by 1+x[SUP]2[/SUP] and re-arrange to get

(dm/m)=-2xdx/(1+x[SUP]2[/SUP])

Therefore ln(m)= -ln(1+x[SUP]2[/SUP]) →m=1/(1+x[SUP]2[/SUP])

But m= dy/dx.=1/(1+x[SUP]2[/SUP])

→dy=dx/(1+x[SUP]2[/SUP])

Now, by letting let'say, x=tanβ and upon integrating we get

y= tan[SUP]-1[/SUP](x)+C where C is a constant of integration

Therefore y= tan[SUP]-1[/SUP](x)+C is a general solution of the above differential equation in Qn number 19

Qn.9 sinxcosydx+sinycosxdy=0

Solution

sinxcosydx+sinycosxdy=0 → sinycosxdy=-sinxcosydx

Divide throughout by the term cosxcosy to get tanydy=cotxdx

Therefore –ln(cosy)=ln(sinx) →ln(secy)=ln(sinx)

→secy=sinx→y=sec[SUP]-1[/SUP](sinx)+D where D is a constant of integration

Therefore y=sec[SUP]-1[/SUP](sinx)+D,is a general solution to the above differential equation in Qn numbe 9

Bavaria · Dec 21, 2013

wale wa BAM tunyooshe mikono juu? japo tulifanikiwa kusoma pure economics

Makanyaga · Dec 24, 2013

donlucchese said:
View attachment 124731

Qn. 22
(1-t[SUP]2[/SUP])(d[SUP]2[/SUP]s/dt[SUP]2[/SUP])-t(ds/dt)=2

Let ds/dt= m; so that d[SUP]2[/SUP]s/dt[SUP]2[/SUP]=dm/dt

Now, we can re-write the above differential as (a function of m and t)

(1-t[SUP]2[/SUP])(dm/dt)-t(m)=2

(dm/dt)-t(m)/(1-t[SUP]2[/SUP])=2/(1-t[SUP]2[/SUP]) .......................................................................................................(1)

Therefore equation (1) above is now reduced to an ODE of the form m'+pm=q with

p= -t/(1-t[SUP]2[/SUP]) and

q=2/(1-t[SUP]2[/SUP])

So we can use an integrating factor e[SUP]∫pdt[/SUP] to multiply throughout the equation to make the LHS of the equation exact and

thereafter integrate both sides. Moere over, you will find that the integrating factor is given by the expression

e[SUP]∫pdt[/SUP] =(1-t[SUP]2[/SUP])[SUP]1/2[/SUP]

On multiplying this factor throughout equation (1) above, we get

=>(1-t[SUP]2[/SUP])[SUP]1/2[/SUP](dm/dt)-t(1-t[SUP]2[/SUP])[SUP]1/2[/SUP] (m)/(1-t[SUP]2[/SUP])=2(1-t[SUP]2[/SUP])[SUP]1/2[/SUP]/(1-t[SUP]2[/SUP]) ........................................................................(2)

On integrating (bearing in mind the chain rule of differentiation and the Fundamental Theorem of Calculus), equation (2)

above can be re-written as

(1-t[SUP]2[/SUP])[SUP]1/2[/SUP](m)=2∫1/(1-t[SUP]2[/SUP])[SUP]1/2[/SUP]dt

=>(1-t[SUP]2[/SUP])[SUP]1/2[/SUP](m)=2∫1/(1-t[SUP]2[/SUP])[SUP]1/2[/SUP]dt upon simplification

Hence (1-t[SUP]2[/SUP])[SUP]1/2[/SUP](m)=2sin[SUP]-1[/SUP]t (note that ∫1/(1-t[SUP]2[/SUP])[SUP]1/2[/SUP]dt=sin[SUP]-1[/SUP]t)

But m=ds/dt

Therefore (1-t[SUP]2[/SUP])[SUP]1/2[/SUP](ds/dt)=2sin[SUP]-1[/SUP]t

=> (ds/dt)= 2(sin[SUP]-1[/SUP]t)/(1-t[SUP]2[/SUP])[SUP]1/2[/SUP]

=> ds= 2(sin[SUP]-1[/SUP]t)/(1-t[SUP]2[/SUP])[SUP]1/2[/SUP]dt

=>s=2∫(sin[SUP]-1[/SUP]t)/(1-t[SUP]2[/SUP])[SUP]1/2[/SUP]dt............................................................................................. (2a)

Using the techniques of integration by parts (using u and v as variables), we can let

du=1/(1-t[SUP]2[/SUP])[SUP]1/2[/SUP]dt and .....................................................................................................(2b)

v=sin[SUP]-1[/SUP]t so that ..............................................................................................................(2c)

u=∫1/(1-t[SUP]2[/SUP])[SUP]1/2[/SUP]dt= sin[SUP]-1[/SUP]t => u=sin[SUP]-1[/SUP]t ............................................................................... (2d)

dv=1/(1-t[SUP]2[/SUP])[SUP]1/2[/SUP] (i.e. d/dt)(sin[SUP]-1[/SUP]t)=1/(1-t[SUP]2[/SUP])[SUP]1/2[/SUP]) .................................................................(2e)

=>s=2∫(sin[SUP]-1[/SUP]t)/(1-t[SUP]2[/SUP])[SUP]1/2[/SUP]dt=∫vdu=uv-∫udv .......................................................................(3)

s=2(uv-∫udv)

From the above equations therefore

s=2[(sin[SUP]-1[/SUP]t)(sin[SUP]-1[/SUP]t)-∫(sin[SUP]-1[/SUP]t)/(1-t[SUP]2[/SUP])[SUP]1/2[/SUP]dt]

=>s=2(sin[SUP]-1[/SUP]t)[SUP]2[/SUP]-2∫(sin[SUP]-1[/SUP]t)/(1-t[SUP]2[/SUP])[SUP]1/2[/SUP]dt ..............................................................................(3a)

But from (2a) above, s=2∫(sin[SUP]-1[/SUP]t)/(1-t[SUP]2[/SUP])[SUP]1/2[/SUP]dt so that we can re-write equation (3a) above as

s=2(sin[SUP]-1[/SUP]t)[SUP]2[/SUP]-s

Therefore 2s=2(sin[SUP]-1[/SUP]t)[SUP]2[/SUP]

So that

s=(sin[SUP]-1[/SUP]t)[SUP]2[/SUP] is a solution of the above differential equation in Question number 22

Task:
Show that s=(sin[SUP]-1[/SUP]t)[SUP]2[/SUP] satisfies the DE (1-t[SUP]2[/SUP])(d[SUP]2[/SUP]s/dt[SUP]2[/SUP])-t(ds/dt)=2

Preta · Dec 24, 2013

Nikimaliza kuhesabu pesa......huku natafuta nini.....

Makanyaga · Dec 24, 2013

donlucchese said:
View attachment 124731

Task:
Show that s=(sin[SUP]-1[/SUP]t)[SUP]2[/SUP] satisfies the DE (1-t[SUP]2[/SUP])(d[SUP]2[/SUP]s/dt[SUP]2[/SUP])-t(ds/dt)=2

Note: Until now, questions which are not solved yet are question numbers 1 through 8, and question number 10 as well

donlucchese · Dec 25, 2013

Makanyaga said:
 Task:
Show that s=(sin[SUP]-1[/SUP]t)[SUP]2[/SUP] satisfies the DE (1-t[SUP]2[/SUP])(d[SUP]2[/SUP]s/dt[SUP]2[/SUP])-t(ds/dt)=2

Note: Until now, questions which are not solved yet are question numbers 1 through 8, and question number 10 as well

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

kadoda11 · Dec 25, 2013

math?NO......money counting?YES.(⊙o&#8857😉

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

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