# Variation of parameters

1. ### flyingpig

1. The problem statement, all variables and given/known data

Solve for general solution with variation of parameter

$$y'''(x) - y'(x) = x$$

3. The attempt at a solution

I initially looked at $$y'''(x) - y'(x) = x$$ only and I foudn my answer to be

$$y(x) = C_1e^{x} + C_2e^{-x} + 1 - x$$

Now i looked through my book and it says it works for ay'' + by' + c = f(t) only (second order).

So I "integrated" $$y'''(x) - y'(x) = x$$

And I got $$y''(x) - y(x) = \frac{x^2}{2} + C$$, solving I got

$$y(x) = C_1e^{x} + C_2e^{-x} + C_3 - \frac{x^2}{2}$$

Using the computer, it gave me

http://www.wolframalpha.com/input/?i=Solve[y%27%27%27+-+y%27+%3D+x]

Why does computer have negative sign??

2. ### HallsofIvy

41,264
Staff Emeritus
I have no idea what you mean when you say you "looked at y'''- y'= x only". That is the entire problem! Do you mean the associated homogeneous equation, y'''- y'= 0? That has characteristic equation $r^3- r= r(r^2- 1)= r(r- 1)(r+ 1)= 0$ so the general solution to the associated homogeneous equation is $y_h(x)= C_1e^x+ C_2e^{-x}+ C_3$

Now, you have title this "variation of parameters". To use that method for this differential equation ("undetermined coefficients" would be simpler but I presume this is for practice of variation of parameters specifically), look for a solution of the form
$$y(x)= u(x)e^x+ v(x)e^{-x}+ w(x)$$
That is, we treat those constants (parameters) as variables- that is the reason for the name "variation of parameters". Differentiating, $y'= u'e^x+ ue^x+ v'e^{-x}- ve^{-x}+ w'$. There will be many different possible functions that will work- we limit our search by requiring that $u'e^x+ v'e^x+ w'= 0$

That means that we have $y'= ue^x- ve^{-x}$. Differentiating again, $y''= u'e^x+ ue^x- v'e^{-x}+ ve^{-x}$. Once again, we "limit our search" (and simplify the equation) by requiring that $u'e^x- v'e^{-x}= 0$.

That gives, now, $y''= ue^{x}+ ve^{-x}$ and, differentiating one more time, that $y'''= u'e^{x}+ ue^{x}+ v'e^{-x}- ve^{-x}$. Putting that and the formula for y' into the equation,
$$u'e^x+ ue^x+ v'e^{-x}- ve^{-x}- (ue^x- ve^{-x})= u'e^x+ v'e^{-x}= x$$

You see what has happened? We have no derivative of u, v, and w higher than first because of our "requirements" and we do not have u, v, and w themselves because they satify the homogenous equation and cancel. We have, instead the three equations
$$u'e^x+ v'e^x+ w'= 0$$
$$u'e^x- v'e^{-x}= 0$$
and
$$u'e^x+ v'e^{-x}= x$$
which we can treat as three linear equations in u', v', and w'. Solve for those and integrate to find u, v, and w.

3. ### vela

12,971
Staff Emeritus
If you're asking about the negative sign in front of c2e-x, it doesn't matter because the sign could be pulled into the arbitrary constant c2.

4. ### flyingpig

HallsofIvy, my book did the same thing, they imposed on the requirement that

$$u'e^x+ v'e^x+ w'= 0$$

Why? Was I right though initially when I 'integrated' $$y''' - y' = x$$

5. ### HallsofIvy

41,264
Staff Emeritus
Yes, because there is no "y" in the original equation you can integrate $y'''- y'= x$ to get $y''- y= (1/2)x^2+ C_1$. It will give you the same result.

However, you titled this "variation of parameters" which will be of about the same difficulty for either equation.

6. ### flyingpig

But why this condition

$$u'e^x+ v'e^x+ w'= 0$$

My book threw this (similar, except they used y_1 and y_2 for general derivation) out of nowhere and I am left blank minded.

How do they know it's going to work out? How did you know (don't say intuition...) the condition u'e^x+ v'e^x+ w'= 0 will enable to you solve the problem?

7. ### flyingpig

Oh okay I see, because if "y" was there, then we can't assume what y really is? Is that why?

Why does it make it equally correct that I "integrated" through $$y''' - y' = x$$ because y wasn't there? Why is it okay then that I don't need to know y', yet I would get the right answer?

By the way, why do you use [/itex]? What does the i do?