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# P3 review exercise Q (edexcel) watch

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1. from green book page 187 Q35

find the general solution of the diff. equation

e^x(dy/dx)+ y^2 = x(y^2)

cheers

i been going over the whole integration chapter today and i still cant get my head around integration by substitution such as Q6 P114
2. (Original post by eddiedaboss)
from green book page 187 Q35

find the general solution of the diff. equation

e^x(dy/dx)+ y^2 = x(y^2)

cheers

i been going over the whole integration chapter today and i still cant get my head around integration by substitution such as Q6 P114
Q6 would be easier using u = tanx
3. (Original post by eddiedaboss)
from green book page 187 Q35

find the general solution of the diff. equation

e^x(dy/dx)+ y^2 = x(y^2)

cheers

i been going over the whole integration chapter today and i still cant get my head around integration by substitution such as Q6 P114
{ 1+ y^2/y^2 dy ={ x/e^x dx

-1/y + y = {x/e^x dx

u =x
du/dx = 1
dv/dx = e^x
v =e^x
{ x/e^x = xe^x - { e^x = xe^x -e^x + c

-1/y + y = e^x(x-1) + C
4. (Original post by imasillynarb)
Q6 would be easier using u = tanx
Using u = tanx dx = du/sec^2x

sec^2xtanx . du/sec^2x = { tanx du = {u du = u^2/2 = tan^2x/2 + C
5. (Original post by imasillynarb)
{ 1+ y^2/y^2 dy ={ x/e^x dx

-1/y + y = {x/e^x dx

u =x
du/dx = 1
dv/dx = e^x
v =e^x
{ x/e^x = xe^x - { e^x = xe^x -e^x + c

-1/y + y = e^x(x-1) + C

they get 1/y = xe^-x + C
6. (Original post by imasillynarb)
{ 1+ y^2/y^2 dy ={ x/e^x dx

-1/y + y = {x/e^x dx

u =x
du/dx = 1
dv/dx = e^x
v =e^x
{ x/e^x = xe^x - { e^x = xe^x -e^x + c

-1/y + y = e^x(x-1) + C
I don't get this answer. Firstly rearrange to obtain

(1/y^2)(dy/dx) = (x-1)e^(-x)

Then,

INT 1/y^2 dy = INT (x-1)e^(-x) dx

The RHS can be evaluated by parts which should eventually give the answer

1/y = xe^(-x) + C

If you need me to write out the details, please write back.
7. (Original post by mikesgt2)
I don't get this answer. Firstly rearrange to obtain

(1/y^2)(dy/dx) = (x-1)e^(-x)

Then,

INT 1/y^2 dy = INT (x-1)e^(-x) dx

The RHS can be evaluated by parts which should eventually give the answer

1/y = xe^(-x) + C

If you need me to write out the details, please write back.

how did you get from

e^x(dy/dx)+ y^2 = x(y^2)

TO

(1/y^2)(dy/dx) = (x-1)e^(-x)

cheers
8. (Original post by eddiedaboss)
how did you get from

e^x(dy/dx)+ y^2 = x(y^2)

TO

(1/y^2)(dy/dx) = (x-1)e^(-x)

cheers
Sorry for my other answer, it was rushed and retarded

Firstly divide by y^2 to get e^x/y^2.dy/dx + 1 = x

Then divide by e^x to get 1/y^2.dy/dx + 1/e^x = x/e^x

Rearrange to get 1/y^2 . dy/dx = x/e^x - 1/e^x

1/y^2.dy/dx = xe^-x - e^-x

Factorise to get

1/y^2.dy/dx = e^-x(x-1)
9. (Original post by imasillynarb)
Sorry for my other answer, it was rushed and retarded

Firstly divide by y^2 to get e^x/y^2.dy/dx + 1 = x

Then divide by e^x to get 1/y^2.dy/dx + 1/e^x = x/e^x

Rearrange to get 1/y^2 . dy/dx = x/e^x - 1/e^x

1/y^2.dy/dx = xe^-x - e^-x

Factorise to get

1/y^2.dy/dx = e^-x(x-1)

Thanks alot there mate, very helpful
10. (Original post by imasillynarb)
Using u = tanx dx = du/sec^2x

sec^2xtanx . du/sec^2x = { tanx du = {u du = u^2/2 = tan^2x/2 + C
well if you use sec x like it says, you get the answer to be 1/2 sec^2 x which is what it says in the back.. strange, your method seemed fine
11. actually your answer must be right because it differentiates to get sec^2 x tan x
12. If I got that question in an exam Id use u = tanx, I think everyone would?
13. (Original post by imasillynarb)
If I got that question in an exam Id use u = tanx, I think everyone would?

yeh but dont you get told what substitution to use
14. (Original post by imasillynarb)
If I got that question in an exam Id use u = tanx, I think everyone would?
I would too, it seems the most obvious substitution to use.
15. Can someone tell me the question, i'm feeling left out - i am in a mathematical mood and can't be bothered to get my review exercise book.
16. (Original post by kimoni)
well if you use sec x like it says, you get the answer to be 1/2 sec^2 x which is what it says in the back.. strange, your method seemed fine
1/2sec^2x = 1/2(1 + tan^2x)

= 1/2 + tan^2x/2 + c

c + 1/2 = C

= tan^2x/2 + C = my answer

You just get a different integration constant!!
17. (Original post by imasillynarb)
1/2sec^2x = 1/2(1 + tan^2x)

= 1/2 + tan^2x/2 + c

c + 1/2 = C

= tan^2x/2 + C = my answer

You just get a different integration constant!!
yeah i noticed that
18. (Original post by serious stuff)
yeah i noticed that
Lamer!
19. So is the integration:

1/(2(sec2x)^2)?
20. (Original post by imasillynarb)
If I got that question in an exam Id use u = tanx, I think everyone would?
i guess you could but if you use u = secx, then since d/dx(secx) = secxtanx, you get rid of more stuff, so youre left with just u - integrating that gives 1/2(sec^2x) as required

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