How to use of the method of undetermined coefficients

[johnlemar_09]

Correct me if I'm wrong. :)

I'm starting to learn basics of the ordinary differential equations, and I have some troubles understanding the concept and method as a whole. I understand when to use the method of undetermined coefficients (MUC) as opposed from variation of parameters.

Suppose we have the equation

P = Q

Where P is all the y's with primes and derivatives, etc., while Q is all the terms with no primes, just x only.

Now, I know that when Q is composed of exponential term, the guess solution will be of the form exponential too, like f(x) = Ae^kx. If Q contains sine or cosines, then we will use f(x) = A sin kx + B cos kx, where k also corresponds the constant along the sines and cosines of Q. If we have a polynomial for Q, then we will also use a polynomial as the guess solution.

But how about for combination?
1. Exponential + Trigonometric functions
2. Polynomial + Algebraic functions.

For example, what is the guess solution if Q is, say, 5 cos 3x + 2 sin 3x + 3x - 9?

Thanks.

 

[HallsofIvy]

Because your differential equation is linear, for sums on the right side, try sums: if your "right hand side" is, as in your example, $5 cos(3x)+ 2 sin(3x)+ 3x- 9$ try $Acos(3x)+ B sin(3x)+ Cx+ D$
You can also "separate" them and do each part separately (that's the key property of linear problems). For example, looking at $5 cos(3x)+ 2 sin(3x)$ you would try $A cos(3x)+ B sin(3x)$ and for $3x- 9$ you would try $Ax+ B$ (not the same A and B, of course). Once you had determined those numbers, add the results.

More generally:

If your "right hand side" involves $e^{kx}$ then try $Ae^{kx}$.

If your "right hand side" involves either sin(kx) or cos(kx) try A cos(kx)+ B sin(kx)

If your "right hand side" involves $x^n$ try $A_nx^n+ A_{n-1}x^{n-1}+ \cdot\cdot\cdot+ A_1x+ A_0$. In other words, a polynomial of order n.

If your "right hand side" involve a product of those types, use a product of the suggested forms. For example, if the "right hand side" is $x^3e^2x$ try $(Ax^3+ Bx^2+ Cx+ D)e^{2x}$. If the "right hand side is $x^2 cos(4x)$ try $(Ax^2+ Bx+ C)(Dcos(x)+ E sin(x))$.

If your "right hand side" involves a function that is already a solution to the associated homogenous equation, multiply the proper form by x.

For example, the differential equation [itex]y''- 3y'+ 2y= 3e^{2x}[/tex] has characteristic equation $r^2- 3r+ 2= (r+ 2)(r+ 1)= 0$ and so has $e^{x}$ and $e^{2x}$ as solutions to its associated homogenous equation, y''- 3y'+ 2y= 0, so we would try a function of the form $Axe^{2x}$. The differential equation $y''- 4y'+ 4= e^{2x}$ has characteristic equation $r^2- 4r+ 4= (r- 2)^2= 0$ and so has $e^{2x}$ and $xe^{2x}$ as solutions. We would multiply $xe^{2x}$ by x and try a function of the form $Ax^2e^{2x}$ to get that right hand side.<br /> <br /> Of course, you can only do that when the right hand side involves only function that we would "expect" as solutions to a homogeneous linear d.e. with constant coefficients- exponentials, sine and cosine, polynomials, and products of them. If your right hand side was not of that form but were, say, ln(x) or tan(x), "undetermined coefficients" does not work and you would have to try "variation of parameters".[/itex]

 

[johnlemar_09]

Thanks, HallsofIvy! I finally found a way to solve this kind of problem! Thanks again!