# 2y'' + y' - y = x + 1; y(0) = 1, y'(0) = 0, y_p = ?

• s3a
In summary: I'll note that I ran across this and your other thread earlier, and I skipped over them because I didn't want to bother with opening another file just to look at your work.I know some of the other HHs do the same thing.
s3a

## Homework Statement

Solve the following initial-value problem, using variation of parameters.:
2y'' + y' - y = x + 1; y(0) = 1, y'(0) = 0

## Homework Equations

Variation of parameters: u_n' = W_n/W

## The Attempt at a Solution

I'm posting this thread, because I'm getting a wrong particular solution (according to Wolfram Alpha), and I can't find the mistake after reviewing my work numerous times. It doesn't matter which column, I put the fundamental set of solutions (from the complementary solution), right?

My (typed) work is attached as MyWork.pdf.

Any help in finding my mistake(s) would be greatly appreciated!

#### Attachments

• MyWork.pdf
44.8 KB · Views: 1,466
Why are you using variation of parameters? Once you have the complementary solution, it looks like an easy undertermined coefficients.

Anyway, when you do the integral, you also need to include (x+1) in the integrand. You've missed this out and just integrated Wi/W.

I'm using variation of parameters (instead of the method of undetermined coefficients, for example), because that's what the problem wants.

As for not including (x + 1) in each integrand, why must I do this? (I still can't see why it is not the case that I had to just integrate each W_n/W, alone.) Or, is it just a mistake in one of my two integrals?

You need to include (x+1) because that's part of the technique:

$$v_1 = \int \frac{g(x)W_1(x)}{W(x)}dx$$

But, with g(x) = x+1, you only have to look for Yp = Ax + B, and save yourself a lot of calculation.

Wait, now I'm really confused. My book only says v_n' = W_n/W, and I've done previous problems by only integrating W_n/W!

I'm attaching an example of a problem where I did just that (successfully)!

Is there something I'm missing? Did I somehow do what you're saying without realizing it, in that other problem (that I'm attaching with this post)?

#### Attachments

• WorkForAnotherProblem.pdf
59.9 KB · Views: 627
Yes, I see you kept g(x) in W1/2. My mistake. I usually take it out and put (0, 1) in W1/2. To use this technique you need the coefficient of y'' to be 1. You've ended up with 2 x the correct solution.

But, the coefficients method is much easier.

Last edited:
@s3a: Since you are apparently typing your work anyway, why not just type it here? You would get more replies and it makes it easier for us to comment on various steps.

@LCKurtz:
Well, technically, I'm working on a desk where it's uncomfortable to handwrite my work, so I had them already typed up.

@PeroK, Which technique needs y'' to have a coefficient of 1? Mine? If so, could I just keep doing what I'm doing and divide my particular solution by the coefficient of the highest-order derivative of y, as a general procedure for all cases?

Well, the forum rules actually require that you type up the problem statement and work, not simply post images of them. I'll note that I ran across this and your other thread earlier, and I skipped over them because I didn't want to bother with opening another file just to look at your work. I know some of the other HHs do the same thing.

@Vela:
I remember reading that it's only bad if the image is not clear.

Having said that, I think I've found some extension to the software I use which allows me to type for myself just once, and copy over LaTex syntax into this forum, such that everybody wins.

I'll look into later for future postings.

@PeroK:
Thanks to you, I now get that all I have to do is exactly what I was doing except make sure that the coefficient of the highest-order derivative is 1, so thanks! :)

s3a said:
@LCKurtz:
Well, technically, I'm working on a desk where it's uncomfortable to handwrite my work, so I had them already typed up.

@PeroK, Which technique needs y'' to have a coefficient of 1? Mine? If so, could I just keep doing what I'm doing and divide my particular solution by the coefficient of the highest-order derivative of y, as a general procedure for all cases?

You seem to by typing you work in WORD. For Math that is a painful waste of time. It would be better to use LaTeX, which you could then cut and paste into a Forum input panel.

Nowadays there are user-friendly versions available, such as LyX; see, eg., http://www.lyx.org/ . This is used much like WORD, with easy input, etc., but with math writing orders of magnitude easier than in WORD. It produces LaTeX-consistent output files. The best part: it is free.

## 1. What is the general solution to the given differential equation?

The general solution to the differential equation 2y'' + y' - y = x + 1 is:

y(x) = C1e-x + C2ex + x + 2

where C1 and C2 are arbitrary constants.

## 2. How do you solve for the particular solution (yp) for the given initial conditions?

To solve for the particular solution (yp), we first need to find the complementary solution (yc) by setting the right side of the equation to 0 and solving for y. Then, we can use the initial conditions to find the values of the arbitrary constants C1 and C2. Finally, we can combine the complementary and particular solutions to get the general solution as well as the particular solution with the given initial conditions.

## 3. What is the purpose of the initial conditions in this differential equation?

The initial conditions, y(0) = 1 and y'(0) = 0, help us to find the particular solution (yp) for the given differential equation. They provide specific values for y and y' at the initial point (x = 0) which we can use to solve for the arbitrary constants and get the particular solution.

## 4. How can you verify that the particular solution satisfies the given differential equation?

To verify that the particular solution (yp) satisfies the given differential equation, we simply substitute the solution into the equation and check if both sides are equal. In this case, we would substitute yp = x + 2 into the equation 2y'' + y' - y = x + 1 and see if both sides are equal.

## 5. Can you graph the general solution and the particular solution for this differential equation?

Yes, we can graph both the general solution and the particular solution for this differential equation. The general solution will be a family of curves with different values for the arbitrary constants, while the particular solution will be a specific curve that satisfies the given initial conditions. By plotting these curves, we can visualize the behavior of the solutions and see how they differ from each other.

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