guys!!!further math help!first order differential

[liuweijian94]

how can i solve this question~~~i cannot carry on when i get there~~~

 

[OakMoon!]

As far as I know first order differentials are not in the syllabus of CIE Furthter Maths 9231.

 

[liuweijian94]

oh no~~~i am using the text book which is written by Braian and Mark Gaulter~~~Oxford press ~~~~am i using the wrong text book? :sorry:

 

[OakMoon!]

The textbooks for Further Maths are for both the Ed-excel and the CIE board so you'll not get one book covering all the CIE syllabus. I actually don't use a book, so I don't know much about that. You can check the syllabus of Further Maths and go about it.

 

[liuweijian94]

right ~~~thanks man ~~~

 

[OakMoon!]

You are welcome.

 

[usman]

oh no~~~i am using the text book which is written by Braian and Mark Gaulter~~~Oxford press ~~~~am i using the wrong text book? :sorry:

Assalam-o-alaikum

Hi there 'liuweijian94', according to my experience, the book "Further Pure Mathematics by Brian and Mark Gaulter, Oxford University press" is infact a very good book covering, in a very nice way, many topics of CIE Further Maths (and I use it frequently).

So don't worry, you may well use this book, along with "Further Pure Mathematics by L. Bostock, S. Chandler, C. Rourke; Stanley Thomes Publishers," which is also a similar, but a bit better choice (though it also contains a considerable amount of irrelevant material for CIE Further Maths, but covers many more topics than the book by B.M Gaulter as a whole) + Regularly go through the past-papers of each topic when you are done with it e.g. differential equations in this case (though there are questions relating only to Second Order Differential Equations in the CIE -papers, but practising First Order Differential Equations as well, will atleast give you a nice background for 2nd O.D.E).

As far as your problem is concerned, i worked out the following solution for it--->

Continuing on from your last step.... (Don’t lose your patience, cos it’s gonna get xtremely messy)
....y*sin(x) = int [ exp(2x)*sin(x) ] dx
Using integration by parts on int [ exp(2x)*sin(x) ] dx, setting:-
du/dx = exp(2x) while v = sin(x)
Then, u = 0.5*exp(2x) while dv/dx = cos(x)

---> y*sin(x) = 0.5*[ exp(2x)*sin(x) ] - 0.5*int [ exp(2x)*cos(x) ] dx

Let's use integration by parts again, on int [ exp(2x)*cos(x) ] dx, setting:-
dp/dx = exp(2x) while q = cos(x)
Then, p = 0.5*exp(2x) while dq/dx = -sin(x)
So, int [ exp(2x)*cos(x) ] dx = 0.5*exp(2x)*cos(x) + 0.5*int [ exp(2x)*sin(x) ] dx

We can observe that the integral we started with, i.e. int [ exp(2x)*sin(x) ] dx, has now appeared again as the last term in the expression for “int [ exp(2x)*cos(x) ] dx.” Bingo!

I know it’s gettin’ messy as hell…;p Bt anyways, let’s bear with it.

So, upto now, we’ve found that

int [ exp(2x)*sin(x) ] dx = 0.5*[ exp(2x)*sin(x) ] – 0.5*[0.5*exp(2x)*cos(x) + 0.5*int { exp(2x)*sin(x) } dx ]
int [ exp(2x)*sin(x) ] dx = 0.5*[ exp(2x)*sin(x) ] – 0.25*[ exp(2x)*cos(x) ] - 0.25*int { exp(2x)*sin(x) } dx ]
(1 + 0.25)* int [ exp(2x)*sin(x) ] dx = 0.5*exp(2x)*sin(x) – 0.25*[ exp(2x)*cos(x) ]
(5/4)* int [ exp(2x)*sin(x) ] dx = [0.25*exp(2x)] *[2sin(x) – cos(x)]
int [ exp(2x)*sin(x) ] dx = [0.25*(4/5)*exp(2x)] *[2sin(x) – cos(x)]
int [ exp(2x)*sin(x) ] dx = 0.2*exp(2x)*[2sin(x) – cos(x)]

--->int [ exp(2x)*sin(x) ] dx = (1/5)*exp(2x)*[2sin(x) – cos(x)] + constant of integration (let it be ‘C’)

Therefore, y*sin(x) = (1/5)*exp(2x)*[2sin(x) – cos(x)] + C

NB: The constant of integration is shown only in the last step, to avoid repetition of constants of integration, over and over again.


P.S.
The question you asked, as I know, is from Further Maths by B.M. Gaulter --> Exercise 4A --> Q.7, which is very similar to the Q.15 of Exercise 7a of Further Maths by L.Bostock [the difference being that there is an exp(x) in that question rather than an exp(2x) as in this case]…;p

Anyways, hope that helps (rather than confuses yew further….;p)….!

 

[anillatoo]

oh no~~~i am using the text book which is written by Braian and Mark Gaulter~~~Oxford press ~~~~am i using the wrong text book? :sorry:

Assalam-o-alaikum

Hi there 'liuweijian94', according to my experience, the book "Further Pure Mathematics by Brian and Mark Gaulter, Oxford University press" is infact a very good book covering, in a very nice way, many topics of CIE Further Maths (and I use it frequently).

So don't worry, you may well use this book, along with "Further Pure Mathematics by L. Bostock, S. Chandler, C. Rourke; Stanley Thomes Publishers," which is also a similar, but a bit better choice (though it also contains a considerable amount of irrelevant material for CIE Further Maths, but covers many more topics than the book by B.M Gaulter as a whole) + Regularly go through the past-papers of each topic when you are done with it e.g. differential equations in this case (though there are questions relating only to Second Order Differential Equations in the CIE -papers, but practising First Order Differential Equations as well, will atleast give you a nice background for 2nd O.D.E).

As far as your problem is concerned, i worked out the following solution for it--->

Continuing on from your last step.... (Don’t lose your patience, cos it’s gonna get xtremely messy)
....y*sin(x) = int [ exp(2x)*sin(x) ] dx
Using integration by parts on int [ exp(2x)*sin(x) ] dx, setting:-
du/dx = exp(2x) while v = sin(x)
Then, u = 0.5*exp(2x) while dv/dx = cos(x)

---> y*sin(x) = 0.5*[ exp(2x)*sin(x) ] - 0.5*int [ exp(2x)*cos(x) ] dx

Let's use integration by parts again, on int [ exp(2x)*cos(x) ] dx, setting:-
dp/dx = exp(2x) while q = cos(x)
Then, p = 0.5*exp(2x) while dq/dx = -sin(x)
So, int [ exp(2x)*cos(x) ] dx = 0.5*exp(2x)*cos(x) + 0.5*int [ exp(2x)*sin(x) ] dx

We can observe that the integral we started with, i.e. int [ exp(2x)*sin(x) ] dx, has now appeared again as the last term in the expression for “int [ exp(2x)*cos(x) ] dx.” Bingo!

I know it’s gettin’ messy as hell…;p Bt anyways, let’s bear with it.

So, upto now, we’ve found that

int [ exp(2x)*sin(x) ] dx = 0.5*[ exp(2x)*sin(x) ] – 0.5*[0.5*exp(2x)*cos(x) + 0.5*int { exp(2x)*sin(x) } dx ]
int [ exp(2x)*sin(x) ] dx = 0.5*[ exp(2x)*sin(x) ] – 0.25*[ exp(2x)*cos(x) ] - 0.25*int { exp(2x)*sin(x) } dx ]
(1 + 0.25)* int [ exp(2x)*sin(x) ] dx = 0.5*exp(2x)*sin(x) – 0.25*[ exp(2x)*cos(x) ]
(5/4)* int [ exp(2x)*sin(x) ] dx = [0.25*exp(2x)] *[2sin(x) – cos(x)]
int [ exp(2x)*sin(x) ] dx = [0.25*(4/5)*exp(2x)] *[2sin(x) – cos(x)]
int [ exp(2x)*sin(x) ] dx = 0.2*exp(2x)*[2sin(x) – cos(x)]

--->int [ exp(2x)*sin(x) ] dx = (1/5)*exp(2x)*[2sin(x) – cos(x)] + constant of integration (let it be ‘C’)

Therefore, y*sin(x) = (1/5)*exp(2x)*[2sin(x) – cos(x)] + C

NB: The constant of integration is shown only in the last step, to avoid repetition of constants of integration, over and over again.


P.S.
The question you asked, as I know, is from Further Maths by B.M. Gaulter --> Exercise 4A --> Q.7, which is very similar to the Q.15 of Exercise 7a of Further Maths by L.Bostock [the difference being that there is an exp(x) in that question rather than an exp(2x) as in this case]…;p

Anyways, hope that helps (rather than confuses yew further….;p)….!

Lol, that's is some serious stuff for a first order differential!
@liuweijian, second order differentials are the easiest questions in paper 1, so don't worry! U just have to remember some formulae...