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

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In summary, the conversation discusses how to find the set of all solutions for a differential equation dy/dx = 3y. The technique used is separation of variables, where the equation is rearranged to the form f(y)dy = g(x)dx and then integrated on both sides. The resulting function is in the form y = Ce^3x, with C being a constant. The conversation also mentions that the natural logarithm function is used in this process. The proof of this method is also provided.
  • #1
intenzxboi
98
0
can some one explain to me how the set of all solutions for dy/dx = 3y
is.
y= Ce^3x
 
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  • #2
Simply separate the variables.
[tex]
\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*}
[/tex]

You should be able to finish from here.
 
  • #3
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
i got y= e^3x + e^c

how does that become y= Ce^3x
 
  • #5
Instead of e3x + eC, you should have gotten e3x + C = e3xeC = C' e3x

(Here, C' = eC. After all, eC is just a constant.)
 
  • #6
o ok thanks got it.
 
  • #7
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
darkmagic said:
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.
 
  • #9
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."
 
  • #10
yes its ln but I type In. Sorry.
 

What is "Integrate dy/dx"?

"Integrate dy/dx" is a mathematical process used to find the antiderivative of a function. It is also known as integration, and is the inverse operation of differentiation. It involves finding a function whose derivative is the original function.

Why is "Integrate dy/dx" important?

"Integrate dy/dx" is important because it allows us to find the area under a curve, which has many practical applications in fields such as physics, engineering, and economics. It also helps us solve differential equations, which are used to model many real-world phenomena.

How is "Integrate dy/dx" different from "Differentiate dy/dx"?

"Integrate dy/dx" and "Differentiate dy/dx" are two different mathematical operations. Integration involves finding the antiderivative of a function, while differentiation involves finding the derivative of a function. In other words, integration is the reverse of differentiation.

What are the different methods for "Integrate dy/dx"?

There are several methods for "Integrate dy/dx" including the power rule, u-substitution, integration by parts, and trigonometric substitution. The method used depends on the complexity of the function and the desired outcome.

How can I improve my skills in "Integrate dy/dx"?

To improve your skills in "Integrate dy/dx", it is important to practice regularly and familiarize yourself with different integration techniques. You can also seek help from a tutor or refer to online resources and textbooks for additional practice problems and examples.

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