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Simple integral with root

  1. Mar 3, 2017 #1
    1. The problem statement, all variables and given/known data
    $$\frac{dy}{dx}=\sqrt[3]{\frac{y}{x}},~x>0$$
    Why do i need the x>0, indeed my result is good for all x since it contains x2
    2. Relevant equations
    $$\frac{dy}{dx}=f(x)~\rightarrow~dy=f(x)dx~\rightarrow~y=\int f(x)dx$$

    3. The attempt at a solution
    $$\int \frac{dy}{y^{1/3}} = \int \frac{dy}{x^{1/3}} \rightarrow y^{-1/3}=x^{-1/3}+C$$
    $$\frac {1}{\sqrt[3]{y^2}}=\frac {1}{\sqrt[3]{x^2}}+C$$
    $$\sqrt[3]{y^2}=\frac{\sqrt[3]{x^2}}{1+C\sqrt[3]{x^2}}$$
    $$y=\frac{x^2}{(1+C\sqrt[3]{x^2})^2}$$
     
  2. jcsd
  3. Mar 3, 2017 #2

    haruspex

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    Have another go at integrating ##x^{-\frac 13}##.
    (There are also several typos.)
     
  4. Mar 4, 2017 #3
    $$\int \frac{dy}{y^{1/3}} = \int \frac{dy}{x^{1/3}} \rightarrow \int y^{-1/3}=\int x^{-1/3}+C$$
    $$\int y^{-1/3}=\frac{1}{\left( -\frac{1}{3} \right)+1}y^{\left( -\frac{1}{3} \right)}=\frac{3}{2}y^{2/3}$$
     
  5. Mar 4, 2017 #4

    haruspex

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    Right, so what relationship do you get between x and y now?
     
  6. Mar 4, 2017 #5
    $$y=(\sqrt[3]{x^2}+C)^3=...=x^3+3Cx\sqrt[3]{x}+3C^2\sqrt[3]{x^2}+C^3$$
    Besides being so complicated, and i don't know if it's true, what's the connection to x>0?
     
  7. Mar 4, 2017 #6

    haruspex

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    I think you mean y2=x2 + etc., and you have two terms on the right with x2/3.
    Looking at the original problem statement, there is clearly a difficulty at x=0. This means that the solution might be invalid if the integral is across a range including 0. I think they just specified x> 0 to avoid your having to worry about such issues.
     
  8. Mar 4, 2017 #7
    Which 2 terms on the right are with x2/3?
    $$(a+b)^3=\left( \begin{array}{m} 3 \\ 0 \end{array} \right)a^3+\left( \begin{array}{m} 3 \\ 1 \end{array} \right)a^2 b+\left( \begin{array}{m} 3 \\ 2 \end{array} \right)ab^2+\left( \begin{array}{m} 3 \\ 3 \end{array} \right)b^3$$
    There is only one term with a.
    Which difficulty is at x=0? is it because of ##~\int \frac{dy}{x^{1/3}}##? do i ignore ##~\int \frac{dy}{y^{1/3}}##? how do i treat it, since y≠0 too
     
  9. Mar 4, 2017 #8

    haruspex

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    My mistake.
    In the differential equation there is y/x.
     
  10. Mar 4, 2017 #9
    Thank you very much Haruspex
     
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