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Math Help - First-order Separable ODE

  1. #1
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    First-order Separable ODE

    Hi
    Can you tell me if i am correct in my method?
    thanks
    John

    Separate the variables and find the indefinite integrals.

     \frac{dy}{dx}=y(x^3-x^\frac{1}{2})

    re-arrange to get y on left hand side and x on right hand side

     dy=y(x^3-x^\frac{1}{2})\cdot\  dx

      \frac{1}{y} \cdot\ dy=(x^3-x^\frac{1}{2})\cdot\  dx

    Now integrate both sides:

     \int \frac{1}{y} \cdot\ dy=\int (x^3-x^\frac{1}{2})\cdot\  dx

     lny=\frac{x^4}{4}-\frac{2}{3}x^\frac{3}{2} +C

     y=e^{\frac{x^4}{4}-\frac{2}{3}x^\frac{3}{2} +C }

     y= e^{\frac{x^4}{4}}-e^{\frac{2}{3}x^\frac{3}{2}}+e^{c}

    As  e^C=C

     y= e^{\frac{x^4}{4}}-e^{\frac{2}{3}x^\frac{3}{2}}+C

     y=e^{\frac{x^4}{4}-\frac{2}{3}x^\frac{3}{2}} +C

    Can you tell me if i am on track?
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  2. #2
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    Re: Differential equation

    If you want to check whether your answer is correct, you can always substitute your solution in the original differential equation.

    I can save you some time in this instance though. Your work is correct up until the following:

    y = e^{\frac{x^4}{4}}-e^{\frac{2}{3}x^\frac{3}{2}}+e^C

    This is incorrect. Do you see why?
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  3. #3
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    Re: Differential equation

    Quote Originally Posted by harbottle View Post
    If you want to check whether your answer is correct, you can always substitute your solution in the original differential equation.

    I can save you some time in this instance though. Your work is correct up until the following:

    y = e^{\frac{x^4}{4}}-e^{\frac{2}{3}x^\frac{3}{2}}+e^C

    This is incorrect. Do you see why?
    No i don't see it?

    I have the ln function on the left and to bring it across i get the inverse Ln or e function of the right hand side?

    Can you tell me what's incorrect?
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  4. #4
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    Re: Differential equation

    You have y = e^{\frac{x^4}{4} - \frac{2}{3}x^{\frac{3}{2}}+C}, which is correct.

    It does not follow that y = e^{\frac{x^4}{4}} - e^{ \frac{2}{3}x^{\frac{3}{2}}} + e^C, because that's not how exponents work. Instead the next line is:

    y = e^C\cdot e^{\frac{x^4}{4} - \frac{2}{3}x^{\frac{3}{2}}}
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  5. #5
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    Re: Differential equation

    I need to work on my simplifying of exponents,

    An example on another site used the same steps as i used above(which is wrong in this case). In that case the expression was a polynomial-maybe thats the difference?


    thanks
    John
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  6. #6
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    Re: First-order Separable ODE

    Quote Originally Posted by celtic1234 View Post
    Hi
    Can you tell me if i am correct in my method?
    thanks
    John

    Separate the variables and find the indefinite integrals.

     \frac{dy}{dx}=y(x^3-x^\frac{1}{2})

    re-arrange to get y on left hand side and x on right hand side

     dy=y(x^3-x^\frac{1}{2})\cdot\  dx

      \frac{1}{y} \cdot\ dy=(x^3-x^\frac{1}{2})\cdot\  dx

    Now integrate both sides:

     \int \frac{1}{y} \cdot\ dy=\int (x^3-x^\frac{1}{2})\cdot\  dx

     lny=\frac{x^4}{4}-\frac{2}{3}x^\frac{3}{2} +C

     y=e^{\frac{x^4}{4}-\frac{2}{3}x^\frac{3}{2} +C }

     y= e^{\frac{x^4}{4}}-e^{\frac{2}{3}x^\frac{3}{2}}+e^{c}

    As  e^C=C

     y= e^{\frac{x^4}{4}}-e^{\frac{2}{3}x^\frac{3}{2}}+C

     y=e^{\frac{x^4}{4}-\frac{2}{3}x^\frac{3}{2}} +C

    Can you tell me if i am on track?
    First of all, \displaystyle \int{\frac{1}{y}\,dy} = \ln{|y|} + C_1, not \displaystyle \ln{y}...
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