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  1. #1
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    Help =]

    dx/dt + (x)/(20+t) = 3

    just want to make sure im doing the right thing

    ty
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  2. #2
    Junior Member Infophile's Avatar
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    Hello,

    x'(t)+\frac{x(t)}{20+t}=0\Rightarrow \frac{x'(t)}{x(t)}=-\frac{1}{20+t}\Rightarrow \ln\left(\frac{x(t)}{C}\right)=-\ln(20+t)\Rightarrow x(t)=\frac{C}{20+t}

    And notice that t\mapsto \frac{3}{2}(20+t) is solution.

    Therefore solutions of this differential equation are :

    \color{blue}\boxed{x(t)=\frac{C}{20+t}+\frac{3}{2}  (20+t)}

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  3. #3
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    Hello, Charbel!

    \frac{dx}{dt} + \frac{1}{t+20}\,x \:=\: 3

    Integrating factor: . I \;=\;e^{\int\frac{dt}{t+20}} \;=\;e^{\ln(t+20)} \;=\;t+20


    We have: . (t+20)\frac{dx}{dt} + x \:=\:3(t+20) \quad\Rightarrow\quad \frac{d}{dt}\bigg[(t+20)x\bigg] \:=\:3(t+20)


    Integrate: . (t+20)x \;=\;3\left(\frac{t^2}{2} + 20t\right) + C \;=\;\frac{3}{2}(t^2 + 40t + C)


    Therefore: . x \;=\;\frac{3(t^2 + 40t + C)}{2(t+20)}

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  4. #4
    Junior Member Infophile's Avatar
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    Our results are the same
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  5. #5
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    Quote Originally Posted by Infophile View Post
    Our results are the same
    Yes but for linear first order ODE's, one typical goes the route of finding an integrating factor as Soroban did.
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  6. #6
    Junior Member Infophile's Avatar
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    I've just learn it, in France we don't use this method, that's funny
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  7. #7
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    Quote Originally Posted by danny arrigo View Post
    Yes but for linear first order ODE's, one typical goes the route of finding an integrating factor as Soroban did.
    One may, but one doesn't necessarily. Most people learn to integrate "separable equations" before learning about "integrating factors" for linear equations.
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  8. #8
    Grand Panjandrum
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    Quote Originally Posted by Charbel View Post
    dx/dt + (x)/(20+t) = 3

    just want to make sure im doing the right thing

    ty
    The best way to do what you say you want is for you to post what you have done for us to comment on.

    CB
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