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Math Help - ODE finding constants...

  1. #1
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    Thumbs down ODE finding constants...

    another problem i worked half way on and stuck on

    ODE is dx/dt = x^(3/4)

    i need to find values for constans a > 0 and b or which x(t) = (at)^b is a solution of the ODE or t>0. Is this function also a solution for t<0?

    ok, so here goes my crazy method in doing this..



    x'(t) = ba(at)^(b-1)

    put this into the ODE:

    ba(at)^(b-1) = ((at)^b)^3/4

    now what?.. no initial conditions are given. how do they expect us to find a and b =S and..... determine t? =| that looks crazy.. how you obtain a,b,t from that equation


    thank you
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by matlabnoob View Post
    another problem i worked half way on and stuck on

    ODE is dx/dt = x^(3/4)

    i need to find values for constans a > 0 and b or which x(t) = (at)^b is a solution of the ODE or t>0. Is this function also a solution for t<0?

    ok, so here goes my crazy method in doing this..



    x'(t) = ba(at)^(b-1)

    put this into the ODE:

    ba(at)^(b-1) = ((at)^b)^3/4

    now what?.. no initial conditions are given. how do they expect us to find a and b =S and..... determine t? =| that looks crazy.. how you obtain a,b,t from that equation


    thank you
    \frac{dx}{dt}=x^{3/4}\implies x^{-3/4}dx=dt\implies \int x^{-3/4}\,dx=\int\,dt \implies 4x^{1/4}=t+c\implies x=\left(\frac{1}{4}(t+c)\right)^4

    (with c being an arbitrary constant). So to model your equation to match the one given, just let c=0, and then it follows that a=\frac{1}{4} and b=4.
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  3. #3
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    Quote Originally Posted by matlabnoob View Post
    another problem i worked half way on and stuck on

    ODE is dx/dt = x^(3/4)

    i need to find values for constans a > 0 and b or which x(t) = (at)^b is a solution of the ODE or t>0. Is this function also a solution for t<0?

    ok, so here goes my crazy method in doing this..



    x'(t) = ba(at)^(b-1)

    put this into the ODE:

    ba(at)^(b-1) = ((at)^b)^3/4

    now what?.. no initial conditions are given. how do they expect us to find a and b =S and..... determine t? =| that looks crazy.. how you obtain a,b,t from that equation


    thank you
    ba(at)^{b-1}= ba(a^{b-1})t^{b-1}= ba^b t^{b-1} and ((at)^b)^{3/4}= (at)^{3b/4}= a^{3b/4}t^{3b/4}.

    In order that those be equal for all t, we must have ba^b= a^{3b/4} and b- 1= 3b/4.

    That last equation gives 4b- 4= 3b so b= 4. With that, the first equation becomes 4a^4= a^3 so either a= 0 or a= 1/4.
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  4. #4
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    Quote Originally Posted by redsoxfan325 View Post
    \frac{dx}{dt}=x^{3/4}\implies x^{-3/4}dx=dt\implies \int x^{-3/4}\,dx=\int\,dt \implies 4x^{1/4}=t+c\implies x=\left(\frac{1}{4}(t+c)\right)^4

    (with c being an arbitrary constant). So to model your equation to match the one given, just let c=0, and then it follows that a=\frac{1}{4} and b=4.

    thankks! ok..so what happens once i find a solution to this ODE... and then im told to...

    'find another solution of the ODE,different from the one i just found, which exists for all t in R (real numbers)'

    any hints, steps...anything! would be soo appreciated
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