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By the [[chain rule]], we can rewrite <math>-x^2+(1-y^2)*\frac{dy}{dx}</math> to get the equation | By the [[chain rule]], we can rewrite <math>-x^2+(1-y^2)*\frac{dy}{dx}</math> to get the equation | ||
<math>\frac{d}{dx}(-\frac{x^3}{3})+\frac{d}{dx}(y-\frac{y^3}{3})=0</math> | <math>\frac{d}{dx}(-\frac{x^3}{3})+\frac{d}{dx}(y-\frac{y^3}{3})=0</math> | ||
| − | + | which results in <math>\frac{d}{dx}(-\frac{x^3}{3}+y-\frac{y^3}{3})=0.</math> | |
| − | <math>\frac{d}{dx}(-\frac{x^3}{3}+y-\frac{y^3}{3})=0.</math> | + | |
By integrating this (and adding an arbitrary constant) the result <math>-x^3+3y-y^3=c</math> results. | By integrating this (and adding an arbitrary constant) the result <math>-x^3+3y-y^3=c</math> results. | ||
Latest revision as of 06:15, 26 January 2009
If an equation $ M(x,y)dx+N(x,y)\frac{dy}{dx}=0 $ can be written in the form $ M(x)dx+N(y)\frac{dy}{dx}=0 $ (in other words, M depends on only x, and N depends on only y) then the equation is called separable. This is because the variable can be separated.
Example (textbook example 1)
The equation $ \frac{dy}{dx}=\frac{x^2}{1-y^2} $ is separable. To see this, multiply by $ 1-y^2 $ and subtract $ x^2 $ from both sides. The result is $ -x^2+(1-y^2)*\frac{dy}{dx} $. M(x)=$ -x^2 $ and N(x)=$ 1-y^2 $.
By the chain rule, we can rewrite $ -x^2+(1-y^2)*\frac{dy}{dx} $ to get the equation $ \frac{d}{dx}(-\frac{x^3}{3})+\frac{d}{dx}(y-\frac{y^3}{3})=0 $ which results in $ \frac{d}{dx}(-\frac{x^3}{3}+y-\frac{y^3}{3})=0. $
By integrating this (and adding an arbitrary constant) the result $ -x^3+3y-y^3=c $ results.
