(New page: Category:MA366 If an equation <math>M(x,y)dx+N(x,y)\frac{dy}{dx}=0</math> can be written in the form <math>M(x)dx+N(y)\frac{dy}{dx}=0</math> (in other words, M depends on only x, and ...) |
|||
| Line 7: | Line 7: | ||
The equation <math>\frac{dy}{dx}=\frac{x^2}{1-y^2}</math> is separable. To see this, multiply by <math>1-y^2</math> and subtract <math>x^2</math> from both sides. The result is <math>-x^2+(1-y^2)*\frac{dy}{dx}</math>. M(x)=<math>-x^2</math> and N(x)=<math>1-y^2</math>. | The equation <math>\frac{dy}{dx}=\frac{x^2}{1-y^2}</math> is separable. To see this, multiply by <math>1-y^2</math> and subtract <math>x^2</math> from both sides. The result is <math>-x^2+(1-y^2)*\frac{dy}{dx}</math>. M(x)=<math>-x^2</math> and N(x)=<math>1-y^2</math>. | ||
| − | By the [[chain rule]], we can rewrite <math>-x^2+(1-y^2)*\frac{dy}{dx}</math> as <math>\frac{d}{dx}(-\frac{x^3}{3})+\frac{d}{dx}(y-\frac{y^3}{3})=0=\frac{d}{dx}(-\frac{x^3}{3}+y-\frac{y^3}{3}). | + | By the [[chain rule]], we can rewrite <math>-x^2+(1-y^2)*\frac{dy}{dx}</math> as <math>\frac{d}{dx}(-\frac{x^3}{3})+\frac{d}{dx}(y-\frac{y^3}{3})=0=\frac{d}{dx}(-\frac{x^3}{3}+y-\frac{y^3}{3}).</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. | ||
Revision as of 07:12, 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} $ as $ \frac{d}{dx}(-\frac{x^3}{3})+\frac{d}{dx}(y-\frac{y^3}{3})=0=\frac{d}{dx}(-\frac{x^3}{3}+y-\frac{y^3}{3}). $
By integrating this (and adding an arbitrary constant) the result $ -x^3+3y-y^3=c $ results.
