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Math Help - Basic diff equations problem

  1. #1
    Member garymarkhov's Avatar
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    Basic diff equations problem

    Consider \dot x = x+1. Find its solution x(t) satisfying the initial condition x(0)=0.

    How should I attack this problem?
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  2. #2
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    Quote Originally Posted by garymarkhov View Post
    Consider \dot x = x+1. Find its solution x(t) satisfying the initial condition x(0)=0.

    How should I attack this problem?

    Remember that \dot x =\frac{dx}{dt}

    This gives a first order seperable ODE.

    \frac{dx}{dt}=x+1 \iff \frac{dx}{x+1}=dt

    Integrating both sides we get

    \ln|x+1|=t+c solving for x gives

    x+1=Ae^{t} \iff x(t)=Ae^{t}-1

    From here just plug in your intial condition to find A.

    Good luck.
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  3. #3
    Member garymarkhov's Avatar
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    Quote Originally Posted by TheEmptySet View Post
    Remember that \dot x =\frac{dx}{dt}

    This gives a first order seperable ODE.

    \frac{dx}{dt}=x+1 \iff \frac{dx}{x+1}=dt

    Integrating both sides we get

    \ln|x+1|=t+c solving for x gives

    x+1=Ae^{t} \iff x(t)=Ae^{t}-1

    From here just plug in your intial condition to find A.

    Good luck.
    Cool, so A = 1 and x(t)=e^t right?
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  4. #4
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    Quote Originally Posted by garymarkhov View Post
    Cool, so A = 1 and x(t)=e^t right?
    You are correct that A=1 but then

    x(t)=e^t-1

    If you try your solution

    x(0)=e^{0}=1 \ne 0 as is needed by you intial condition
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  5. #5
    Member garymarkhov's Avatar
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    Quote Originally Posted by TheEmptySet View Post
    You are correct that A=1 but then

    x(t)=e^t-1

    If you try your solution

    x(0)=e^{0}=1 \ne 0 as is needed by you intial condition
    Good thinking. Can you comment on how the x in x+1 turns into the function x(t)? It's making me uneasy.
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