Differential equation right hand function

In summary, the question requires the use of variation of parameters to solve for the particular solution to the right-hand function in the given auxiliary equation. The complementary function is solved as yc = c1e-x + c2e-2x, but there is a mistake in the variation of parameters formula given. The integral ∫e3x/x cannot be evaluated using elementary methods.
  • #1
Doubell
29
0

Homework Statement


The question specifies the auxiliary equation given is (D^2 + D - 2) = (e^x)/(x)
the method of variation of parameter must be used to find the particular solution to the right hand function. then finally the general soultion should be stated.

Homework Equations


variation of paremeters formula.
yp = -y1∫y2 * g(x)/ (w(y1 , y2))
+ y2∫y1 * g(x)/ (w(y1 , y2))

The Attempt at a Solution


i had solved the complementary function and had gotten yc = c1e-x + c2e-2x. then after applying the formula for variation of parameters i had gotten e-x*lnx/3 - e-2x/3 * ∫e3x / x
i cannot obtain an integral for ∫e3x / x i don't think it can be done have tried various mathods eg by parts, there are no suitable substitutions is it that there is no integral for the expression? i.e. it cannot be integrated.
 
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  • #2
Doubell said:

Homework Statement


The question specifies the auxiliary equation given is (D^2 + D - 2) = (e^x)/(x)
the method of variation of parameter must be used to find the particular solution to the right hand function. then finally the general soultion should be stated.


Homework Equations


variation of paremeters formula.
yp = -y1∫y2 * g(x)/ (w(y11 , y2))
+ y2∫y1 * g(x)/ (w(y11 , y2))

The Attempt at a Solution


i had solved the complementary function and had gotten yc = c1e-x + c2e-2x. then after applying the formula for variation of parameters i had gotten e-x*lnx/3 - e-2x/3 * ∫e3x / x
i cannot obtain an integral for ∫e3x / x i don't think it can be done have tried various mathods eg by parts, there are no suitable substitutions is it that there is no integral for the expression? i.e. it cannot be integrated.

I haven't worked through this problem, but you are correct that ##\frac {e^{3x}} x## does not have an elementary antiderivative.
 
  • #3
Doubell said:

Homework Statement


The question specifies the auxiliary equation given is (D^2 + D - 2) = (e^x)/(x)
the method of variation of parameter must be used to find the particular solution to the right hand function. then finally the general soultion should be stated.


Homework Equations


variation of paremeters formula.
yp = -y1∫y2 * g(x)/ (w(y1 , y2))
+ y2∫y1 * g(x)/ (w(y1 , y2))

The Attempt at a Solution


i had solved the complementary function and had gotten yc = c1e-x + c2e-2x.
You have a mistake. What you have as e-x should be e+x. The other one is fine.
Doubell said:
then after applying the formula for variation of parameters i had gotten e-x*lnx/3 - e-2x/3 * ∫e3x / x
i cannot obtain an integral for ∫e3x / x i don't think it can be done have tried various mathods eg by parts, there are no suitable substitutions is it that there is no integral for the expression? i.e. it cannot be integrated.
 
  • #4
Mark44 said:
You have a mistake. What you have as e-x should be e+x. The other one is fine.
yes it is as u say but the problem still remains the same the∫e3x/x would still have to be determined
 
  • #5
Doubell said:
yes it is as u say but the problem still remains the same the∫e3x/x would still have to be determined
But if you have an incorrect function in your variation of parameters formula, you'll definitely get the wrong answer.

Also, "textspeak" (such as "u" for "you") isn't permitted here at PF.

Please show your work in this formula:
Doubell said:
yp = -y1∫y2 * g(x)/ (w(y1 , y2))
+ y2∫y1 * g(x)/ (w(y1 , y2))
 

Related to Differential equation right hand function

1. What is a right-hand function in a differential equation?

A right-hand function in a differential equation represents the independent variable, typically denoted by the letter "t", and is used to describe how the dependent variable changes with respect to the independent variable. It is often referred to as the "input" of the equation.

2. How do you solve a differential equation with a right-hand function?

To solve a differential equation with a right-hand function, you typically use integration techniques to find the general solution. The right-hand function will be included in the integrated expression, and you can then solve for the constant of integration using initial conditions or boundary conditions.

3. Can a right-hand function be a constant?

Yes, a right-hand function can be a constant. In this case, the differential equation is considered an ordinary differential equation (ODE) with a constant coefficient. Solving for the constant is still necessary to find the particular solution.

4. What is the purpose of a right-hand function in a differential equation?

The right-hand function in a differential equation is essential for modeling real-world phenomena and understanding how variables change over time. It allows us to describe the behavior of a system and make predictions about its future state.

5. Are there any restrictions on the type of functions that can be used as a right-hand function?

No, there are no restrictions on the type of functions that can be used as a right-hand function. It can be any continuous function, including polynomials, trigonometric functions, exponential functions, etc. The choice of function depends on the specific problem and the variables involved.

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