# Homework Help: Integrate dy/dx

1. Aug 27, 2009

### 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

2. Aug 27, 2009

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.

3. Aug 27, 2009

### 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'

4. Aug 27, 2009

### intenzxboi

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

how does that become y= Ce^3x

5. Aug 27, 2009

### Staff: Mentor

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

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

6. Aug 27, 2009

### intenzxboi

o ok thanks got it.

7. Aug 27, 2009

### 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.

8. Aug 27, 2009

### Staff: Mentor

The function is LN, not IN. The letters come from Latin: logarithmus naturalis.

9. Aug 27, 2009

### 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."

10. Aug 28, 2009

### darkmagic

yes its ln but I type In. Sorry.