Simple Integral: Solving $\frac{dy}{dx}=\sqrt[3]{\frac{y}{x}}$ for All x

[Karol]

1. Homework Statement
$$\frac{dy}{dx}=\sqrt[3]{\frac{y}{x}},~x>0$$
Why do i need the x>0, indeed my result is good for all x since it contains x²
2. Homework Equations
$$\frac{dy}{dx}=f(x)~\rightarrow~dy=f(x)dx~\rightarrow~y=\int f(x)dx$$

3. The Attempt at a Solution
$$\int \frac{dy}{y^{1/3}} = \int \frac{dy}{x^{1/3}} \rightarrow y^{-1/3}=x^{-1/3}+C$$
$$\frac {1}{\sqrt[3]{y^2}}=\frac {1}{\sqrt[3]{x^2}}+C$$
$$\sqrt[3]{y^2}=\frac{\sqrt[3]{x^2}}{1+C\sqrt[3]{x^2}}$$
$$y=\frac{x^2}{(1+C\sqrt[3]{x^2})^2}$$

 

[haruspex]

Have another go at integrating $x^{-\frac 13}$.
(There are also several typos.)

 

[Karol]

$$\int \frac{dy}{y^{1/3}} = \int \frac{dy}{x^{1/3}} \rightarrow \int y^{-1/3}=\int x^{-1/3}+C$$
$$\int y^{-1/3}=\frac{1}{\left( -\frac{1}{3} \right)+1}y^{\left( -\frac{1}{3} \right)}=\frac{3}{2}y^{2/3}$$

 

[haruspex]

$$\int \frac{dy}{y^{1/3}} = \int \frac{dy}{x^{1/3}} \rightarrow \int y^{-1/3}=\int x^{-1/3}+C$$
$$\int y^{-1/3}=\frac{1}{\left( -\frac{1}{3} \right)+1}y^{\left( -\frac{1}{3} \right)}=\frac{3}{2}y^{2/3}$$

Right, so what relationship do you get between x and y now?

 

[Karol]

$$y=(\sqrt[3]{x^2}+C)^3=...=x^3+3Cx\sqrt[3]{x}+3C^2\sqrt[3]{x^2}+C^3$$
Besides being so complicated, and i don't know if it's true, what's the connection to x>0?

 

[haruspex]

$$y=(\sqrt[3]{x^2}+C)^3=...=x^3+3Cx\sqrt[3]{x}+3C^2\sqrt[3]{x^2}+C^3$$
Besides being so complicated, and i don't know if it's true, what's the connection to x>0?

I think you mean y²=x² + etc., and you have two terms on the right with x^2/3.
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.

 

[Karol]

Which 2 terms on the right are with x^2/3?
$$(a+b)^3=\left( \begin{array}{m} 3 \\ 0 \end{array} \right)a^3+\left( \begin{array}{m} 3 \\ 1 \end{array} \right)a^2 b+\left( \begin{array}{m} 3 \\ 2 \end{array} \right)ab^2+\left( \begin{array}{m} 3 \\ 3 \end{array} \right)b^3$$
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

 

[haruspex]

Which 2 terms on the right are with x2/3?

My mistake.

Which difficulty is at x=0?

In the differential equation there is y/x.

 

[Karol]

Thank you very much Haruspex