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P3 review exercise Q (edexcel) watch

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    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
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    (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
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    (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
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    (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
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    (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
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    (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.
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    (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
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    (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)
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    (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
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    (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
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    actually your answer must be right because it differentiates to get sec^2 x tan x
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    If I got that question in an exam Id use u = tanx, I think everyone would?
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    (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
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    (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.
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    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.
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    (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!!
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    (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
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    (Original post by serious stuff)
    yeah i noticed that
    Lamer!
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    So is the integration:

    1/(2(sec2x)^2)?
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    (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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