# Variation of Parameters

1. Mar 1, 2014

### vanceEE

1. The problem statement, all variables and given/known data
$$y'' - 2y = x + 1$$

2. Relevant equations
$$y_{o} = Ae^{√(2)x} + Be^{-√(2)x}$$
$$v_{1}'e^{√(2)x} + v_{2}'e^{-√(2)x}\equiv 0$$
$$√(2)v_{1}'e^{√(2)x}-√(2) - v_{2}'e^{-√(2)x} = x + 1$$

3. The attempt at a solution

$$v_{2}' = \frac{x+1}{-2√(2)e^{-√(2)x}}$$
$$v_{2} = (\frac{-x}{4}+\frac{√2}{8}-\frac{1}{4})*e^{√(2)x}$$

$$v_{1}' = (\frac{x+1}{-2√(2)e^{-√(2)x}})(e^{-2√(2)x})$$
$$v_{1} = (\frac{-x}{4}-\frac{√2 + 2}{8})*e^{-√(2)x}$$

$$y = Ae^{√(2)x} + Be^{-√(2)x} + (\frac{-x}{4}-\frac{√2 + 2}{8})*e^{-√(2)x} + (\frac{-x}{4}+\frac{√2}{8}-\frac{1}{4})*e^{√(2)x}$$

Where's my error? The correct solution is $$y = Ae^{√(2)x} + Be^{-√(2)x} - \frac{x}{2} - \frac{1}{2}$$

2. Mar 1, 2014

### vanceEE

1. The problem statement, all variables and given/known data
$$y'' - 2y = x + 1$$

2. Relevant equations
$$y_{h} = Ae^{√(2)x} + Be^{-√(2)x}$$
$$v_{1}'e^{√(2)x} + v_{2}'e^{-√(2)x}\equiv 0$$
$$√(2)v_{1}'e^{√(2)x}-√(2) - v_{2}'e^{-√(2)x} = x + 1$$

3. The attempt at a solution

$$v_{2}' = \frac{x+1}{-2√(2)e^{-√(2)x}}$$
$$v_{2} = (\frac{-x}{4}+\frac{√2}{8}-\frac{1}{4})*e^{√(2)x}$$

$$v_{1}' = (\frac{x+1}{-2√(2)e^{-√(2)x}})(e^{-2√(2)x})$$
$$v_{1} = (\frac{-x}{4}-\frac{√2 + 2}{8})*e^{-√(2)x}$$

$$y = Ae^{√(2)x} + Be^{-√(2)x} + (\frac{-x}{4}-\frac{√2 + 2}{8})*e^{-√(2)x} + (\frac{-x}{4}+\frac{√2}{8}-\frac{1}{4})*e^{√(2)x}$$

Where's my error? The correct solution is $$y = Ae^{√(2)x} + Be^{-√(2)x} - \frac{x}{2} - \frac{1}{2}$$

Last edited: Mar 1, 2014
3. Mar 1, 2014

### vanceEE

Since I am letting $$y(x) = v_{1}y_{1} + v_{2}y_{2}$$ my error is that I did not multiply v_{1} and v_{2} by the associated homogeneous solutions, correct?
$$y_{o}(x) =(\frac{-xe^{√(2)x}}{4}+\frac{√2e^{√(2)x}}{8}-\frac{e^{√(2)x}}{4})*e^{-√(2)x} + (\frac{-xe^{-√(2)x}}{4}-\frac{(√2 + 2)e^{-√(2)x}}{8})*e^{√(2)x}$$
$$= -(\frac{x}{2}+\frac{1}{2})$$

$$y(x) = Ae^{√(2)x} + Be^{-√(2)x} -(\frac{x}{2}+\frac{1}{2})$$

Last edited: Mar 1, 2014
4. Mar 1, 2014

### vela

Staff Emeritus
The particular solution is $y_p = u_1 y_1 + u_2 y_2$. You forgot the y's.

5. Mar 1, 2014

### Dick

Variation of parameters is overkill for a problem like this. You should easily be able to find a particular solution to your original ODE by inspection. But your problem is that $v_1+v_2$ isn't part of your solution. You have to multiply them by the homogeneous parts. You want $v_1 e^{\sqrt{2} x}+v_2 e^{-\sqrt{2} x}$. The exponential parts will cancel in the particular solution.

6. Mar 1, 2014

### Dick

Exactly, as I said in your other thread. Please don't double post, I assume this was an accident.

7. Mar 1, 2014

### vanceEE

What do you mean by double posting? Did I post two forums or are you referring to me editing what I've posted on my forum?

8. Mar 1, 2014

### Dick

I'm referring to the fact there are two threads called "Variation of Parameters" by vanceEE in the "Calculus and Beyond" forum. You are working on one, I was working on the other.

9. Mar 1, 2014

### vanceEE

Please explain what you mean by solving by inspection.

10. Mar 1, 2014

### vanceEE

Sorry, just noticed that. BTW, what did you mean by solving by inspection?

11. Mar 1, 2014

### Dick

Guess a form for the particular solution, say y=ax+b, then y''=0. So y''-2y=x+1 gives -2(ax+b)=x+1, so a=b=(-1/2) and y=-x/2-1/2 works. If the inhomogeneous part is a polynomial, guess a polynomal form and try to solve for it. If it works, it's usually a lot easier than the variation of parameters track.

12. Mar 1, 2014

### Dick

13. Mar 1, 2014

### vanceEE

Oh, and yes you're right; I'm aware that I could have used an easier method like undetermined constants to find this solution but the question asked me to use variation of parameters to find a solution for the equation.

Last edited: Mar 1, 2014
14. Mar 1, 2014

### Dick

I thought so, just checking that you also saw how easy it really was if you don't have to use variation of parameters.