Learning Differential equations, week 2 level material

In summary: Thank you for correcting the thread title, and for everything else, I am truly grateful.In summary, the problem is to find the general solution to the linear first order ODE (dy/dx) - y - e^3x=0. The attempt at a solution involves rewriting the equation as dy/dx - y = e^3x and identifying P(x) as 1 and f(x) as e^3x. The integrating factor is then calculated to be e^x and used to set up the equation as d/dx (ye^x) = e^x * e^3x. However, there is a mistake in the next step where the integral and derivative
  • #1
SYoungblood
64
1

Homework Statement



Find the general solution to the equation.

Homework Equations


(dy/dx) - y - e^3x=0

The Attempt at a Solution


[/B]
I rewrote this as dy/dx - y = e^3x

This is a linear first order ODE, in the form dy/dx + P(x)y = f(x)

P(x) = 1; f(x) = e^3x

The integrating factor = e^(integral)P(x) dx = e^(integral) 1* dx = e^x

Set the LHS equal to the integrating factor, e^x * y, and multiply the RHS by the integrating factor e^x)

d/dx (ye^x) = e^x e^3x

(integral) d/dx (ye^x) =(integral) e^(3x^2) dx

On the LHS, the integral and d/dx cancel out, but I am getting a ridiculously complicated integral that I just know isn't right on the RHS, and I am not sure where I went astray.

Thank you for your time,

SY
 
Physics news on Phys.org
  • #2
[itex]e^xe^{3x}= e^{4x}[/itex], NOT [itex]e^{3x^2}[/itex].
 
  • Like
Likes SYoungblood
  • #3
SYoungblood said:
(dy/dx) - y - e^3x=0
.
.
.
I rewrote this as dy/dx - y = e^3x
Please use parentheses where needed. e^3x would normally be interpreted as ##e^3 \cdot x##. In your first equation above, you have parentheses around dy/dx, where they aren't needed, but don't have them around the exponent 3x, where they are needed.

BTW, I changed the thread title to say "... week 2 level material". I think that's what you intended.
 
  • Like
Likes SYoungblood
  • #4
will do, thank you
 
  • #5
HallsofIvy said:
[itex]e^xe^{3x}= e^{4x}[/itex], NOT [itex]e^{3x^2}[/itex].

Thank you, that is not the first silly mistake I will make, and will not be the last...
 

1. What are the main topics covered in week 2 of learning differential equations?

In week 2, you will learn about second-order differential equations, homogeneous and non-homogeneous equations, and the method of undetermined coefficients. You will also cover the concept of linear independence and Wronskian determinants.

2. How can I solve second-order differential equations?

There are several methods for solving second-order differential equations, including the substitution method, the method of undetermined coefficients, and the method of variation of parameters. Each method has its own advantages and is useful in different situations. It is important to understand the concepts behind each method in order to choose the most appropriate one for a given problem.

3. What is the difference between homogeneous and non-homogeneous equations?

A homogeneous equation has all terms containing the dependent variable and its derivatives, while a non-homogeneous equation has additional terms that do not contain the dependent variable or its derivatives. Homogeneous equations are generally easier to solve as they have a constant coefficient, while non-homogeneous equations require additional methods such as the method of undetermined coefficients.

4. How do I determine if a set of functions is linearly independent?

To determine linear independence, you can use the Wronskian determinant. If the determinant is non-zero, the set of functions is linearly independent. If the determinant is zero, the set of functions is linearly dependent. Keep in mind that this method only applies to sets of continuous functions.

5. What are the applications of differential equations?

Differential equations have many real-world applications in fields such as engineering, physics, economics, and biology. They are used to model and analyze various physical systems, such as the motion of objects, heat transfer, population growth, and electrical circuits. They are also essential in understanding and predicting natural phenomena, making them an important tool for scientists and researchers.

Similar threads

  • Calculus and Beyond Homework Help
Replies
8
Views
750
  • Calculus and Beyond Homework Help
Replies
5
Views
609
  • Calculus and Beyond Homework Help
Replies
7
Views
688
  • Calculus and Beyond Homework Help
Replies
2
Views
129
  • Calculus and Beyond Homework Help
Replies
3
Views
844
  • Calculus and Beyond Homework Help
Replies
5
Views
891
  • Calculus and Beyond Homework Help
Replies
18
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
817
  • Calculus and Beyond Homework Help
Replies
4
Views
910
  • Calculus and Beyond Homework Help
Replies
7
Views
673
Back
Top