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Thread: Differential Equations General Solution

  1. #1
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    Differential Equations General Solution

    Hello!

    My differential equation is given as following:

    $\displaystyle \frac{dy}{dt}=\lambda y$

    I am then to find the general solution.

    Which is:

    $\displaystyle \int\frac{dy}{dt}\,dt=\int \lambda y\,dt \implies y+C_1 = \lambda^2+C_2 \implies \color{red}\boxed{y=\lambda^2+C}$

    Is this correct?

    Addition, if the$\displaystyle \lambda$ is positive, negative or 0 what would happen in these three cases if you let $\displaystyle t \rightarrow \infty$

    Any help would be greatly appreciated!
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  2. #2
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    Quote Originally Posted by jokke22 View Post
    Hello!

    My differential equation is given as following:

    $\displaystyle \frac{dy}{dt}=\lambda y$

    I am then to find the general solution.

    Which is:

    $\displaystyle \int\frac{dy}{dt}\,dt=\int \lambda y\,dt \implies y+C_1 = \lambda^2+C_2 \implies \color{red}\boxed{y=\lambda^2+C}$

    Is this correct?

    Addition, if the$\displaystyle \lambda$ is positive, negative or 0 what would happen in these three cases if you let $\displaystyle t \rightarrow \infty$

    Any help would be greatly appreciated!
    Sorry. $\displaystyle \lambda $ is just a parameter. Your equation is separable

    $\displaystyle \frac{dy}{y} = \lambda \,dx $ so $\displaystyle \ln|y| = \lambda \,x + \ln c$ or $\displaystyle y = ce^{\lambda x}$.
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  3. #3
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    I just plain suck at these equation.

    Thanks yet again Danny!

    If I could trouble you yet again..

    If you were to solve this one then:
    $\displaystyle \frac{dy}{dt}= y^2 cos t$

    I may be able to find a solution, but how do you show that its periodic with $\displaystyle 2\pi$

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    Quote Originally Posted by jokke22 View Post
    I just plain suck at these equation.

    Thanks yet again Danny!

    If I could trouble you yet again..

    If you were to solve this one then:
    $\displaystyle \frac{dy}{dt}= y^2 cos t$

    I may be able to find a solution, but how do you show that its periodic with $\displaystyle 2\pi$

    $\displaystyle \frac{dy}{y^2} = \cos t\,dt $ so $\displaystyle - \frac{1}{y} = \sin t - c$

    thus $\displaystyle y = \frac{1}{c - \sin t}$. Since $\displaystyle \sin t $ is $\displaystyle 2 \pi $ - periodic then so is $\displaystyle y$.
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  5. #5
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    Quote Originally Posted by jokke22 View Post
    Hello!

    My differential equation is given as following:

    $\displaystyle \frac{dy}{dt}=\lambda y$

    I am then to find the general solution.

    Which is:

    $\displaystyle \int\frac{dy}{dt}\,dt=\int \lambda y\,dt \implies y+C_1 = \lambda^2+C_2 \implies \color{red}\boxed{y=\lambda^2+C}$

    Is this correct?
    You already know this is not correct. The reason you can't do it is because on the right you have $\displaystyle \int\lambda y dt$ and that has to be integrated with respect to t. But y is an unknown function of t. Since you don't know what y is in terms of t, you can't integrate with respect to t.

    Notice that danny arigo's method gives a function of y time "dy" equal to a function of t times "dt". That means you integrate the function of y with respect to y and the function of t with respect to t.

    Addition, if the$\displaystyle \lambda$ is positive, negative or 0 what would happen in these three cases if you let $\displaystyle t \rightarrow \infty$

    Any help would be greatly appreciated![/QUOTE]
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    Grateful for all the help so far!

    I do have one more though I can't solve... Appreciate all the help I can get!

    $\displaystyle y^{/}+\frac{1}{x}y=e^{x^2}$
    where
    $\displaystyle x>0$

    which satisfy $\displaystyle y(1) = e$

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  7. #7
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    Quote Originally Posted by jokke22 View Post
    Grateful for all the help so far!

    I do have one more though I can't solve... Appreciate all the help I can get!

    $\displaystyle y^{/}+\frac{1}{x}y=e^{x^2}$
    where
    $\displaystyle x>0$

    which satisfy $\displaystyle y(1) = e$

    This is a linear ODE in standard form. The integrating factor is $\displaystyle \mu = exp \int \frac{dx}{x} = x$ so our equation comes

    $\displaystyle x y^{/} + y = xe^{x^2}$ or $\displaystyle \frac{d}{dx} (x y) = xe^{x^2}$. If you let u = xy then $\displaystyle \frac{d}{dx} u = xe^{x^2}$ which is separable. I think you can take it from here.
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  8. #8
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    I can't seem to get it right...

    $\displaystyle \int\frac{d}{dx} u = \int xe^{x^2}$

    which
    $\displaystyle \int xe^{x^2} dx$


    Sorry mate, just getting confused... Apply conditions for $\displaystyle x$ , $\displaystyle xe^{x^2}$ and $\displaystyle dx$ ?


    Go detailed since i just plain suck at this, if I could bother you yet again!
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    Okey, in stead of replacing... Could I go this way:

    $\displaystyle \frac{d}{dx}(xy)=xe^{x^2} \rightarrow xy = \int xe^{x^2}dx+C$

    How would this end up?

    Help is greatly appreciated!

    Maybe:

    $\displaystyle 2xe^{x}-e^x+C$
    Last edited by jokke22; May 22nd 2009 at 02:27 AM.
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  10. #10
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    Quote Originally Posted by jokke22 View Post
    Okey, in stead of replacing... Could I go this way:

    $\displaystyle \frac{d}{dx}(xy)=xe^{x^2} \rightarrow xy = \int xe^{x^2}dx+C$

    How would this end up?

    Help is greatly appreciated!

    Maybe:

    $\displaystyle 2xe^{x}-e^x+C$
    The left hand step is good but there's a mistake on the right.

    $\displaystyle
    \int x e^{x^2}dx = \frac{1}{2} e^{x^2} + c
    $
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