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  1. #1
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    integrate

    Can anyone help to integrate this ? thanks

    \Rightarrow \frac {dy}{(y + 3)^2} = (x - 1)^2~dx
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by aircraft View Post
    Can anyone help to integrate this ? thanks

    \Rightarrow \frac {dy}{(y + 3)^2} = (x - 1)^2~dx
    isn't this related to that thread you posted on ODE's? if you have problems with what was done in that thread, or were not sure on how to proceed in that thread then you should have made that known in THAT thread

    for \int \frac {dy}{(y + 3)^2} use a substitution, u = y + 3

    for \int (x - 1)^2~dx expand (x - 1)^2 and use the power rule

    (and you need to wrap math tags around your LaTeX code to get it to work, that is, you must type [tex]\frac {dy}{(y + 3)^2} = (x - 1)^2~dx[/tex] to get \frac {dy}{(y + 3)^2} = (x - 1)^2~dx
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  3. #3
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    Quote Originally Posted by aircraft View Post
    Can anyone help to integrate this ? thanks

    \Rightarrow \frac {dy}{(y + 3)^2} = (x - 1)^2~dx
    It would have been tidier to post this question as part of this thread since it's a follow-up question ....
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  4. #4
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    Quote Originally Posted by Jhevon View Post
    for \int (x - 1)^2~dx expand (x - 1)^2 and use the power rule
    By makin' u=x+1 there's no problem ('cause du=dx), so we can use the power rule directly

    \int {(x - 1)^2 \,dx}  = \frac{{(x - 1)^3 }}<br />
{3} + k.

    ---

    Solving differential equations without knowing basic integration techniques is quite dangerous.

    Brush up some of integration aircraft, from there get in on this.
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Krizalid View Post
    By makin' u=x+1 there's no problem ('cause du=dx), so we can use the power rule directly

    \int {(x - 1)^2 \,dx}  = \frac{{(x - 1)^3 }}<br />
{3} + k.

    ---

    Solving differential equations without knowing basic integration techniques is quite dangerous.

    Brush up some of integration aircraft, from there get in on this.
    yes, i was aware that we could do that. however, it was evident to me that aircraft was shaky on basic integration, so i opted for the more conceptually easy solution
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  6. #6
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    Thanks alot for your help guys. =) very weak on Maths , need to work hard..
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by aircraft View Post
    need to work hard..
    we all do. good luck

    it seems you need to brush up on calculus, differential equations will be hard if you don't. what math class are you in?
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  8. #8
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    Quote Originally Posted by Jhevon View Post
    [snip]

    it seems you need to brush up on calculus, differential equations will be hard if you don't. [snip]
    Hmmmm ...... 'dangerous' even
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  9. #9
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    t's t 8 years since I've touched Maths.
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  10. #10
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    Quote Originally Posted by aircraft View Post
    t's t 8 years since I've touched Maths.
    Well you've certainly dived into the deep end. Good luck. Try to extensively review and revise the mathematics this subject will assume as pre-requisite knowledge ......
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  11. #11
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    Hi Guys , need help again , stucked..

    2(3-x) dy/dx + y = -1 , y(2)=4


    Treat this D.E as 1st order linear, the intergrating factor is = 1/ square root of (3-x)

    I can't seem to get this answer.. the ans. I got was e ^ 1/2 * ln (3-x)
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  12. #12
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    Quote Originally Posted by aircraft View Post
    Hi Guys , need help again , stucked..

    2(3-x) dy/dx + y = -1 , y(2)=4


    Treat this D.E as 1st order linear, the intergrating factor is = 1/ square root of (3-x)

    I can't seem to get this answer.. the ans. I got was e ^ 1/2 * ln (3-x)
    It would be useful to see your working, that is, to see how you got your answer.
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  13. #13
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    trying to arrange the equation into this format- dy/dx + P * y = Q

    Integrating factor = e ^ int P dx

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

    so P = 1/ 2(3-x)

    1/ 2(3-x) = 2(3-x)^ -1 so cannot use the normal mathod to integrate, so I use ln instead.

    so 1/2 * 1/ (3-x) = 1/2 ln (3-x)

    integrating factor = e^ 1/2 in (3-x) and stucked , dunno wat to do next..
    correct answer is 1/ square root of (3-x)

    Thanks for your help..
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  14. #14
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    Quote Originally Posted by aircraft View Post
    trying to arrange the equation into this format- dy/dx + P * y = Q

    Integrating factor = e ^ int P dx

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

    so P = 1/ 2(3-x)

    1/ 2(3-x) = 2(3-x)^ -1 so cannot use the normal mathod to integrate, so I use ln instead. Mr F asks: Why are you doing this??

    so 1/2 * 1/ (3-x) = 1/2 ln (3-x)

    integrating factor = e^ 1/2 in (3-x) and stucked , dunno wat to do next..
    correct answer is 1/ square root of (3-x)

    Thanks for your help..
    Do you know how to use the given integrating factor?

    The technique boils down to y = \frac{1}{\mu} \left( \int \mu Q \, dx + C \right), where \mu = \mu(x) is the integrating factor and Q is as you've defined above.

    So in your case the solution boils down to


    y = \sqrt{3 - x} \left( -\frac{1}{2}\int \frac{1}{3 - x} \frac{1}{\sqrt{3 - x}} \, dx + C \right) = \sqrt{3 - x} \left( -\frac{1}{2}\int \frac{1}{(3 - x)^{3/2}} \, dx + C \right)


    = \sqrt{3 - x} \left( -\frac{1}{2}\int (3 - x)^{-3/2} \, dx + C \right).


    So again, getting the answer boils down to being able to integrate ......
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