
#1
Aug2709, 03:57 PM

P: 98

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 



#2
Aug2709, 04:18 PM

HW Helper
P: 1,344

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. 



#3
Aug2709, 04:20 PM

HW Helper
P: 6,214

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' 



#4
Aug2709, 04:23 PM

P: 98

integrate dy/dx
i got y= e^3x + e^c
how does that become y= Ce^3x 



#5
Aug2709, 04:37 PM

Mentor
P: 21,069

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.) 



#6
Aug2709, 04:59 PM

P: 98

o ok thanks got it.




#7
Aug2709, 05:19 PM

P: 164

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.




#8
Aug2709, 06:03 PM

Mentor
P: 21,069





#9
Aug2709, 07:47 PM

P: 724

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 righthand 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." 



#10
Aug2809, 05:31 AM

P: 164

yes its ln but I type In. Sorry.



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