# Integrate dy/dx

by intenzxboi
Tags: dy or dx, integrate
 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
 HW Helper P: 1,371 Simply separate the variables. \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*} You should be able to finish from here.
 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'
 P: 98 Integrate dy/dx i got y= e^3x + e^c how does that become y= Ce^3x
 Mentor 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.)
 P: 98 o ok thanks got it.
 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.
Mentor
P: 21,253
 Quote by darkmagic 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.
 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 $$y' = \alpha y$$ where $$\alpha$$ is a constant, then y has the form $$y = f(x) = ce^{\alpha x}$$ where c is also a consant; conversely, every function of the form $$ce^{\alpha x}$$ satisfies the equation $$y' = \alpha y$$. It is clear that $$y = ce^{\alpha x}$$ satisfies this equation for any arbitrary constant c. Conversely, no other function satisfies the differential equation $$y' - \alpha y = =$$. For if y is such a function, we consider the function $$u = ye^{-\alpha x}$$. We then have $$u' = y'e^{-\alpha x} - \alpha y e^{-\alpha x} = e^{-\alpha x}(y' - \alpha y}) .$$ However, the right-hand side vanishes, since we have assumed that $$y' = \alpha y$$; hence $$u' = 0$$ so that u is a constant c and $$y = ce^{\alpha x}$$ as we wished to prove."
 P: 164 yes its ln but I type In. Sorry.

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