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Unformatted text preview: Seperable Equations Examples–ANSWERS 1 . dy dx = 3 x 2 e- y , y (0) = 1 . Since f ( x, y ) = 3 x 2 e- y and e- y 6 = 0 for any y , there are no equilibrium solutions. Sepa- rating variables, we have e y ( x ) dy ( x ) dx = 3 x 2 and integrating each side with respect to the independent variable, we have Z x e y ( s ) dy ( s ) ds ds = Z x 3 s 2 ds or e y ( x ) = x 3 + C, which gives the solution y = y ( x ) in implicit form. The arbitrary constant of integration can be determined at this stage by invoking the initial condition y (0) = 1. This yields e 1 = 0 + C, or C = e giving e y = e y ( x ) = x 3 + e. This last equation can be solved for y ( x ) to give an explicit form of the solution ln e y ( x ) = ln ( x 3 + e ) , or y ( x ) = ln ( x 3 + e ) . Note that there are no absolute value signs occuring in the argument of the logarithmic function since the initial condition is positive. However this function is defined only for positive arguments hence the given solution is valid only on the semi-infinite interval x >- e 1 3 . 2 . dy dx =- y 2 x , x 6 = 0 . There is a constant solution y ( x ) ≡ 0 which is defined on any interval which does not contain x = 0 since the differential equation is not defined at x = 0....
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- Spring '08
- Equations, Constant of integration, initial condition, dy