Differential Equation - Solve by substitution

In summary, the conversation involves solving a differential equation using substitution. The solution is obtained by eliminating dx and dy in the original equation and rearranging to solve by separation of variables. The final answer is (x + 2y)^2 - 3(x + 2y) = c.
  • #1
JJBladester
Gold Member
286
2

Homework Statement


Solve the following differential equation by using substitution.


Homework Equations



(x + 2y - 1)dx + (2x + 4y - 3)dy = 0

Let u = x + 2y

The Attempt at a Solution



u = x + 2y
du/dx = 1 + 2(dy/dx)

(x + 2y - 1)dx + (2x + 4y - 3)dy = 0

(u - 1)dx + (2u - 3)dy = 0

The solution the teacher gave is as follows but I didn't know how to go from my line above to the answer below. I believe it has something to do with getting the dx and dy out of the new equation involving u and then rearranging to solve by separation variables.

K = x + (x + 2y)2 - 3(x + 2y)
 
Physics news on Phys.org
  • #2
Use this...

JJBladester said:
du/dx = 1 + 2(dy/dx)

...to eliminate dy in this

(u - 1)dx + (2u - 3)dy = 0
 
  • #3
If u = x + 2y

then du/dx = 1 + 2(dy/dx)

and dy/dx = (du/dx - 1)/2

So...

dy/dx + (x + 2y -1)/(2x + 4y - 2) = 0

(du/dx - 1)/2 + (u - 1)/(2u - 3) = 0

du/dx - 1 + (2u - 2)/(2u - 3) = 0

If I divide (2u - 2) by (2u - 3), I get:

1 + 1/(2u - 3)

Then I have:

du/dx - 1 + 1 + 1/(2u - 3) = 0

du/dx + 1/(2u - 3) = 0

(2u - 3)du + dx = 0

The integration gives me:

u2 - 3u + x = c

So, my answer (after putting (x + 2y) back in for u) is:

(x + 2y)2 -3(x + 2y) = c

Got it, Tom. Thanks for the push in the right direction!
 
Last edited:

1. What is the concept of substitution in differential equations?

The concept of substitution in differential equations involves replacing a variable with a new variable or expression in order to simplify the equation and make it easier to solve. This is often done by introducing a new function or variable to represent the original variable.

2. How do you solve a differential equation by substitution?

To solve a differential equation by substitution, you first need to choose a new variable or expression to substitute for the original variable. Then, you plug in this substitution into the equation and solve for the new variable. Finally, you use the substitution to find the solution for the original variable.

3. When should substitution be used to solve a differential equation?

Substitution is typically used to solve differential equations that are difficult to solve using other methods, such as separation of variables or integration. It can also be used to solve equations with complicated expressions or multiple variables.

4. What are the benefits of solving a differential equation by substitution?

Solving a differential equation by substitution can help simplify the equation and make it easier to solve. It can also provide a more general solution compared to other methods and allow for the use of different techniques, such as integration or differentiation.

5. What are some common challenges when using substitution to solve a differential equation?

One challenge when using substitution is choosing the right substitution to make the equation more manageable. Another challenge is ensuring that the substitution does not introduce any extraneous solutions. It may also be difficult to find the inverse substitution to obtain the solution for the original variable.

Similar threads

  • Calculus and Beyond Homework Help
Replies
6
Views
851
  • Calculus and Beyond Homework Help
Replies
5
Views
729
  • Calculus and Beyond Homework Help
Replies
15
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
491
  • Calculus and Beyond Homework Help
Replies
2
Views
731
  • Calculus and Beyond Homework Help
Replies
3
Views
901
  • Calculus and Beyond Homework Help
Replies
18
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
742
  • Calculus and Beyond Homework Help
Replies
24
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
541
Back
Top