Trouble with particular solution of differential equation - rewrite my answer?

In summary, the particular solution of the differential equation is y = (1/2)ln(x2 - 10x + 50) satisfying the initial condition and the differential equation.
  • #1
cowmoo32
122
0
Trouble with particular solution of differential equation - [SOLVED]

Homework Statement


Find the particular solution of the differential equation
c995a54ca48d21f6dc81446bdd54881.png

satisfying the initial condition
907c69d9098d2e376491d1dfd7f0771.png



The Attempt at a Solution


[STRIKE]I end up with (1/2)ln(-x^2+10x) which does satisfy the initial conditions, yet it's coming back as incorrect. Is there another way I can write the formula?[/STRIKE]

I figured it out, check post #3
 
Last edited:
Physics news on Phys.org
  • #2
cowmoo32 said:

Homework Statement


Find the particular solution of the differential equation
c995a54ca48d21f6dc81446bdd54881.png

satisfying the initial condition
907c69d9098d2e376491d1dfd7f0771.png



The Attempt at a Solution


I end up with (1/2)ln(-x^2+10x) which does satisfy the initial conditions, yet it's coming back as incorrect. Is there another way I can write the formula?
Show how you got the result you show. I assume you mean y = (1/2)ln(-x2 + 10x).

Satisfying the initial conditions is not enough: the solution has to satisfy the differential equation, too.
 
  • #3
Scratch that above, I had a negative out of place. See below. I corrected it and it's coming back as correct.

[itex]\frac{dy}{dx}[/itex]=e-2x(x-5)

[itex]\int[/itex] (e2y)dy = [itex]\int (x-5)dx[/itex]

e2y = x2 - 10x + C

2y = ln(x2 - 10x + C)

C = 50
 
Last edited:

What is a particular solution of a differential equation?

A particular solution of a differential equation is a specific function that satisfies the given differential equation. It is obtained by solving the differential equation using initial conditions or boundary conditions.

How do I rewrite a particular solution of a differential equation?

To rewrite a particular solution of a differential equation, you can use the initial conditions or boundary conditions to solve for any arbitrary constants that may be present in the solution. This will give you an explicit expression for the particular solution.

Why is there sometimes trouble with finding a particular solution of a differential equation?

There can be trouble with finding a particular solution of a differential equation if the equation is non-linear or if the initial or boundary conditions are not well-defined. In such cases, it may be necessary to use numerical methods or approximation techniques to find a solution.

What are some techniques for finding a particular solution of a differential equation?

Some techniques for finding a particular solution of a differential equation include separation of variables, variation of parameters, and using integrating factors. These techniques can be applied to different types of differential equations, depending on their form and characteristics.

Can a particular solution of a differential equation be unique?

No, a particular solution of a differential equation is not always unique. There can be multiple solutions that satisfy the given equation and initial or boundary conditions. In some cases, it may be necessary to impose additional constraints to determine a unique particular solution.

Similar threads

  • Calculus and Beyond Homework Help
Replies
2
Views
178
  • Calculus and Beyond Homework Help
Replies
7
Views
284
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
324
Replies
12
Views
382
  • Calculus and Beyond Homework Help
Replies
0
Views
166
  • Calculus and Beyond Homework Help
Replies
2
Views
667
  • Calculus and Beyond Homework Help
Replies
7
Views
555
  • Calculus and Beyond Homework Help
Replies
5
Views
914
  • Differential Equations
2
Replies
52
Views
816
Back
Top