# Stuck on Variation of Parameters: Help with a Calculus Problem

• karlmartin
In summary: In other words, you are evaluating the antiderivative at two different values and subtracting (the other way to think about integration by parts is that you are using the product rule for derivatives in reverse).In summary, the conversation is about solving a differential equation using the method of variation of parameters. The person asking for help reaches a point where the equations become too difficult to integrate and is looking for help or confirmation that the equations are correct. Others suggest alternative methods, but the person is determined to master the variation of parameters technique. Eventually, a solution is found using integration by parts multiple times.
karlmartin
Hey ya'll!

This is the equation under discussion:

y'' - 2y' - 3y = x + 2

I'm asked to use the method of variation of parameters to determine a solution for this differential equation, but I reach a point where my the equations just look too ridiculous to continue.

The point I have in mind is where I finally get the second equation for the system of equations that is supposed to give me meaningful functions for the derivatives of the two variable functions.

These are my equations(the variable functions are u1 and u2):

u1'e3x + u2'e-x = 0
3u1'e3x - u2'e-x = x + 2

From these I am unable to extract equations for the variables that I would be able to integrate without resorting to some sort of external aid. Also, it seems unlikely that these overly difficult equations would really have be the result of the necessary work, considering this is from a beginner calculus course(Early Transcendentals, 6th international edition, chapter 17.3, problem 20).

I need to master this technique, but every problem I try ends up with me looking at an insane integral, puzzled beyond the twilight zone. Could someone please point out my mistake, or at least confirm that the equations are indeed correct and perhaps point out a method of simplification that doesn't result in an integral too hard to process?

Thank you very much for your time.

I have to apologize in advance in case I violated some rule, since this is a first post. Also, it is 5 AM here and it is quite possible that I'm dreaming of posting this instead of actually doing it. Good night.

karlmartin said:
I have to apologize in advance in case I violated some rule, since this is a first post. Also, it is 5 AM here and it is quite possible that I'm dreaming of posting this instead of actually doing it. Good night.

Very poetic. But you don't need to do variation of parameters. You just need to find a particular solution for your ODE. Try substituting y=ax+b and try to find a and b,

Thanks for the alternative, Dick, but I know that I could use the method of undetermined coefficients, I'm just asked not to. My goal here is to correctly use the variation of parameters and to know that technique too.

Dreaming of posting instead of actually posting, that made me fall off my chair LOL.

y'' - 2y' - 3y = x + 2

Like Dick said, you need to find a particular solution to it.

y$_{p}$ = ax + b

However, you should first find y$_{h}$; the homogeneous solution:y'' - 2y' - 3y = 0

Where you should get, y$_{h}$ = A*e^ etc. etc. you know what I mean?

I know what you mean and I've already done that. As I said, I know how to find the particular solution using the method of undetermined coefficients, as you suggested. However, my quarrel is with the method of variation of parameters. I don't care for the solution as much as I want to use the method of variation of parameters properly.

Solve the linear system of equations w.r.t. u'1, and u'2. Integrate. Substitute in the general form of the solution.

But I can't integrate the solutions. The integrals are too hard and I don't see how they would become any simpler. The solution itself according to the undetermined coefficients method is pretty simple. I feel like flipping the table, opening the windows and yelling prophecies in tongues to the people on the streets.

karlmartin said:
But I can't integrate the solutions. The integrals are too hard and I don't see how they would become any simpler. The solution itself according to the undetermined coefficients method is pretty simple. I feel like flipping the table, opening the windows and yelling prophecies in tongues to the people on the streets.

You are one funny character, lol.

Variation of parameters is definitely the hard way to do this. But it can be done.

Thank you all for the help, I think I have the technique on my toolbelt now and it's properly calibrated. It just seems the necessary integration is mostly very difficult.

NewtonianAlch, jokes are all I can offer in return. Also, manbreasts. But those aren't very noteworthy so I'll stick to the jokes.

$$\int{x \, e^{a \, x} \, dx} = \frac{d}{d a} \int{e^{a \, x} \, dx} = \frac{d}{d a} \frac{e^{a \, x}}{a} = \frac{e^{a \, x}}{a^2} (a \, x - 1)$$

In general, you can always integrate something like $\int e^{ax}x^n dx$ (even more generally, $\int f(x)x^ndx$, for f any repeatedly integrable function) by using integration by parts n times, each time taking $dv= e^{ax}dx$, $u= x^j$ so that you are repeatedly reducing the power of x until you just have $\int e^{ax}dx$.

## 1. What is the variation of parameters method in calculus?

The variation of parameters method is a technique used in differential equations to find the particular solution of a non-homogeneous equation. It involves finding a set of functions that satisfy the homogeneous equation and using them to construct a particular solution.

## 2. How do you apply the variation of parameters method?

To apply the variation of parameters method, you first need to find the general solution of the homogeneous equation. Then, you use this solution to find the particular solution by substituting it into the original non-homogeneous equation and solving for the coefficients.

## 3. What is the difference between variation of parameters and the method of undetermined coefficients?

The variation of parameters method is used for non-homogeneous equations, while the method of undetermined coefficients is used for homogeneous equations with constant coefficients. Additionally, the variation of parameters method allows for a wider range of functions to be used in the particular solution, while the method of undetermined coefficients only uses specific types of functions.

## 4. Can the variation of parameters method be used for higher-order differential equations?

Yes, the variation of parameters method can be used for higher-order differential equations. However, the process can become more complex as the order of the equation increases.

## 5. Are there any limitations to the variation of parameters method?

One limitation of the variation of parameters method is that it can only be used for linear differential equations. It also requires the homogeneous equation to have a known general solution, which may not always be the case.

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