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Carl Christian
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Focus less on symbols and concentrate on the properties of the objects. A change of symbols might clarify matters. Let $y : \mathbb{R} \rightarrow \mathbb{R}$ be given by $$ y(t) = \int_0^t e^{-s^2}ds.$$ Then $y(0) = 0$ and $$ y'(t) = e^{-t^2}.$$ If we define $f : \mathbb{R} \times \mathbb{R}$ as follows, $$ f(t,y) = e^{-t^2},$$ where there is no explicit dependence on the second variable, then we can write $$ y'(t) = f(t,y).$$ You are now in a position to apply, say, the classical fourth order Runge-Kutta method.

Focus less on symbols and concentrate on the properties of the objects. A change of symbols might clarify matters. Let $y : \mathbb{R} \rightarrow \mathbb{R}$ be given by $$ y(t) = \int_0^t e^{-s^2}ds.$$ Then $y(0) = 0$ and $$ y'(t) = e^{-t^2}.$$ If we define $f : \mathbb{R} \times \mathbb{R}$ as follows, $$ f(t,y) = e^{-t^2},$$ where there is explicit dependence on the second variable, then we can write $$ y'(t) = f(t,y).$$ You are now in a position to apply, say, the classical fourth order Runge-Kutta method.

Focus less on symbols and concentrate on the properties of the objects. A change of symbols might clarify matters. Let $y : \mathbb{R} \rightarrow \mathbb{R}$ be given by $$ y(t) = \int_0^t e^{-s^2}ds.$$ Then $y(0) = 0$ and $$ y'(t) = e^{-t^2}.$$ If we define $f : \mathbb{R} \times \mathbb{R}$ as follows, $$ f(t,y) = e^{-t^2},$$ where there is no explicit dependence on the second variable, then we can write $$ y'(t) = f(t,y).$$ You are now in a position to apply, say, the classical fourth order Runge-Kutta method.

Source Link
Carl Christian
  • 13.7k
  • 2
  • 15
  • 38

Focus less on symbols and concentrate on the properties of the objects. A change of symbols might clarify matters. Let $y : \mathbb{R} \rightarrow \mathbb{R}$ be given by $$ y(t) = \int_0^t e^{-s^2}ds.$$ Then $y(0) = 0$ and $$ y'(t) = e^{-t^2}.$$ If we define $f : \mathbb{R} \times \mathbb{R}$ as follows, $$ f(t,y) = e^{-t^2},$$ where there is explicit dependence on the second variable, then we can write $$ y'(t) = f(t,y).$$ You are now in a position to apply, say, the classical fourth order Runge-Kutta method.