- #1
Karol
- 1,380
- 22
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
$$\int \frac{dy}{y^{1/3}} = \int \frac{dy}{x^{1/3}} \rightarrow y^{-1/3}=x^{-1/3}+C$$
$$\frac {1}{\sqrt[3]{y^2}}=\frac {1}{\sqrt[3]{x^2}}+C$$
$$\sqrt[3]{y^2}=\frac{\sqrt[3]{x^2}}{1+C\sqrt[3]{x^2}}$$
$$y=\frac{x^2}{(1+C\sqrt[3]{x^2})^2}$$
$$\frac{dy}{dx}=\sqrt[3]{\frac{y}{x}},~x>0$$
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
$$\int \frac{dy}{y^{1/3}} = \int \frac{dy}{x^{1/3}} \rightarrow y^{-1/3}=x^{-1/3}+C$$
$$\frac {1}{\sqrt[3]{y^2}}=\frac {1}{\sqrt[3]{x^2}}+C$$
$$\sqrt[3]{y^2}=\frac{\sqrt[3]{x^2}}{1+C\sqrt[3]{x^2}}$$
$$y=\frac{x^2}{(1+C\sqrt[3]{x^2})^2}$$