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 y&#039;&#039;- 3y&#039;+ 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&#039;&#039;- 3y&#039;+ 2y= 0, so we would try a function of the form Axe^{2x}. The differential equation y&amp;#039;&amp;#039;- 4y&amp;#039;+ 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 &quot;expect&quot; 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), &quot;undetermined coefficients&quot; does not work and you would have to try &quot;variation of parameters&quot;.

 

[johnlemar_09]

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