# Simple integral with root

1. Mar 3, 2017

### Karol

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. Mar 3, 2017

### haruspex

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

3. Mar 4, 2017

### Karol

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

4. Mar 4, 2017

### haruspex

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

5. Mar 4, 2017

### 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?

6. Mar 4, 2017

### haruspex

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.

7. Mar 4, 2017

### Karol

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

8. Mar 4, 2017

### haruspex

My mistake.
In the differential equation there is y/x.

9. Mar 4, 2017

### Karol

Thank you very much Haruspex