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Integrate dy/dx

by intenzxboi
Tags: dy or dx, integrate
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Aug27-09, 03:57 PM
P: 98
can some one explain to me how the set of all solutions for dy/dx = 3y
y= Ce^3x
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
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Aug27-09, 04:18 PM
HW Helper
P: 1,371
Simply separate the variables.
\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

You should be able to finish from here.
Aug27-09, 04:20 PM
HW Helper
P: 6,202
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'

Aug27-09, 04:23 PM
P: 98
Integrate dy/dx

i got y= e^3x + e^c

how does that become y= Ce^3x
Aug27-09, 04:37 PM
P: 21,253
Instead of e3x + eC, you should have gotten e3x + C = e3xeC = C' e3x

(Here, C' = eC. After all, eC is just a constant.)
Aug27-09, 04:59 PM
P: 98
o ok thanks got it.
Aug27-09, 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.
Aug27-09, 06:03 PM
P: 21,253
Quote Quote by darkmagic View Post
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.
The function is LN, not IN. The letters come from Latin: logarithmus naturalis.
Aug27-09, 07:47 PM
P: 726
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."
Aug28-09, 05:31 AM
P: 164
yes its ln but I type In. Sorry.

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