What is the solution set for dy/dx = 3y?

[intenzxboi]

can some one explain to me how the set of all solutions for dy/dx = 3y
is.
y= Ce^3x

 

[statdad]

Simply separate the variables.
$$\begin{align*}<br /> \frac{dy}{dx} & = 3y \\<br /> \frac 1 y \frac dy dx & = 3 \\<br /> \int \left(\frac 1 y\right) \left(\frac{dy}{dx}\right) \, dx & =\int 3 \, dx<br /> \end{align*}$$

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 e^3x + e^C, you should have gotten e^{3x + C} = e^3xe^C = C' e^3x

(Here, C' = e^C. After all, e^C 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.