can some one explain to me how the set of all solutions for dy/dx = 3y is. y= Ce^3x 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution
Simply separate the variables. [tex] \begin{align*} \frac{dy}{dx} & = 3y \\ \frac 1 y \frac dy dx & = 3 \\ \int \left(\frac 1 y\right) \left(\frac{dy}{dx}\right) \, dx & =\int 3 \, dx \end{align*} [/tex] You should be able to finish from here.
Try rearranging your equation to get it in the form f(y)dy=g(x) dx. Then integrate both sides. This type of differential equation technique is called 'separation of variables'
Instead of e^{3x} + e^{C}, you should have gotten e^{3x + C} = e^{3x}e^{C} = C' e^{3x} (Here, C' = e^{C}. After all, e^{C} is just a constant.)
this is variable separable. c is constant, then you can make it In c so that when using the e function, In c become only c.
Here's a proof of the general case that I got from Courant's Introduction to Calculus and Analysis. This is one of my favourite proofs: "If a function y = f(x) satisfies an equation of the form [tex] y' = \alpha y [/tex] where [tex] \alpha [/tex] is a constant, then y has the form [tex] y = f(x) = ce^{\alpha x} [/tex] where c is also a consant; conversely, every function of the form [tex] ce^{\alpha x} [/tex] satisfies the equation [tex] y' = \alpha y [/tex]. It is clear that [tex] y = ce^{\alpha x} [/tex] satisfies this equation for any arbitrary constant c. Conversely, no other function satisfies the differential equation [tex] y' - \alpha y = = [/tex]. For if y is such a function, we consider the function [tex] u = ye^{-\alpha x} [/tex]. We then have [tex] u' = y'e^{-\alpha x} - \alpha y e^{-\alpha x} = e^{-\alpha x}(y' - \alpha y}) .[/tex] However, the right-hand side vanishes, since we have assumed that [tex] y' = \alpha y [/tex]; hence [tex] u' = 0 [/tex] so that u is a constant c and [tex] y = ce^{\alpha x} [/tex] as we wished to prove."