integrate dy/dx

[intenzxboi]

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

[statdad]

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.

[rock.freak667]

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'

[intenzxboi]

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

how does that become y= Ce^3x

[Mark44]

Instead of e3x + eC, you should have gotten e3x + C = e3xeC = C' e3x

(Here, C' = eC. After all, eC is just a constant.)

[intenzxboi]

o ok thanks got it.

[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.

[Mark44]

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.

[JG89]

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."

[darkmagic]

yes its ln but I type In. Sorry.