Find the particular solution of a differential equation

In summary, the conversation discusses finding the particular solution of a differential equation and using an initial condition to solve for the constant C. The participant also mentions integrating the equation and isolating y to find the solution.
  • #1
TrueStar
95
0

Homework Statement



Find the particular solution of the differential equation dy/dx = (x-4)e^(-2y)

Satisfying initial condition y(4)=ln(4)

Homework Equations



N/A

The Attempt at a Solution



I separated this into dy/e^(-2y) = (x-4)dx

I then integrated it to get e^(2y)/2 = x^2/2 - 4x

I then tried to isolate y and got y= 1/2ln(x^2-8x+C)

Plugging in the values, I get ln(4)=1/2ln(-16+C)

What do I do with that 1/2 out in front? If it wasn't a part of the equation, C would equal 20. That is, if I'm correct so far.
 
Last edited:
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  • #2
Put the everything inside the ln - it will make you feel happier.

[tex]e^{2y} = x^2-8x+c \Rightarrow y=\ln \sqrt{x^2-8x+c}[/tex]... now the initial condition makes sense?
 
  • #3
You add the +C after you integrate. Not at the very end. So you should have gotten e^(2y)/2 = x^2/2 - 4x + C. You can actually solve for C at that step as well. Then isolate y.
 
  • #4
Yeah I realized that, but the 1/2 is still throwing me off. I'm sure it's something simple that I've forgotten.
 
  • #5
[tex]a\ln(x)=ln(x^a)[/tex]
But solve for c while y is still in the exponential like Dick says also works.
 
  • #6
TrueStar said:
Yeah I realized that, but the 1/2 is still throwing me off. I'm sure it's something simple that I've forgotten.

I'm not sure what's throwing you off. C isn't 20. It's something else. Solve for it!
 
  • #7
I forgot the rule that Simon posted. It makes complete sense now. I will look at it the other way as well. Thank you both!
 

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 different from the general solution, which includes all possible solutions to the equation.

How do you find the particular solution of a differential equation?

To find the particular solution of a differential equation, you can use various methods such as separation of variables, variation of parameters, or using an integrating factor. The method used depends on the type of differential equation and its initial conditions.

What is the role of initial conditions in finding the particular solution of a differential equation?

The initial conditions are used to find the specific values of the arbitrary constants in the general solution, thus determining the particular solution. These conditions provide the starting point for solving the differential equation and are essential in obtaining a unique solution.

Can a differential equation have more than one particular solution?

No, a differential equation can only have one particular solution that satisfies both the equation and the initial conditions. However, the general solution may have multiple arbitrary constants, resulting in an infinite number of solutions.

How is the particular solution of a differential equation different from the general solution?

The particular solution is a specific solution that satisfies the given differential equation and initial conditions, while the general solution includes all possible solutions with arbitrary constants. The particular solution is unique, while the general solution represents a family of solutions.

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