# The Method of Undetermined Coefficients

The general solution to the linear differential equation L(y)=phi(x) is given as y=y_h+y_p where y_p denotes one solution to the differential equation and y_h is the general solution to the associated homogeneous equation L(y) = 0. For method for obtaining y_h when the differential equation has constant coefficients see method of solutions note.
Here one method will be given for obtaining a particular solution y_p once y_h is known.

The method of undetermined coefficients is applicable onh if phi(x) and all of its derivatives can be written in terms of the same finite set of linearly independent functions. which we denote by {y_1(x),\ y_2(x),\ ...,\ y_n(x)}.
The method is initiated by assuming a particular solution of the form y_p=A_1 y_1(x)+A_2 y_2(x)+...+A_n y_n(x) where A_1,\ A_2, ...,A_n denote arbitrary multiplicative constants. These arbitrary constants are then evaluated by substituting the proposed solution into the given differential equation and equating the coefficients of like terms.

Case 1. phi(x)=p_n(x) , polynomial of n-th degree. Assume solution of the form y_p=A_n x^n+A_(n-1) x^(n-1)+...+Ax+A_0 where A_j (j=0, 1, ..., n) are constants to be determined.

Example 1. Find particular solution of y''+3y'+2y=x^2+2 .

Assume that particular solution is of the form y_p=Ax^2+Bx+C .

Then y_p'=2Ax+B and y_p''=2A .

Now plug these values into differential equation:

(2A)+3(2Ax+B)+2(Ax^2+Bx+C)=2Ax^2+(6A+2B)x+(2A+3B+C) .

Equating like terms with x^2+2 gives:

{(2A=1),(6A+2B=0),(2A+3B+C=2):}

Which has solution A=1/2,B=-3/2,C=11/4 .

So, particular solution is y_p=1/2 x^2-3/2 x+11/4 .

Case 2. phi(x)=ke^(alpha x) . Assume solution of the form y_p=Ae^(alpha x) where A is a constant to be determined.

Example 2. Find particular solution of y'''-4y'+2y=3e^(7x) .

Assume that particular solution is of the form y_p=Ae^(7x) .

Then y_p'=7Ae^(7x) , y_p''=49Ae^(7x) and y_p'''=343Ae^(7x) .

Plugging this values into equation gives 343Ae^(7x)-4*7Ae^(7x)+2Ae^(7x)=317Ae^(7x) .

Equating coefficients with 3e^(7x) gives 317A=3 or A=3/317 .

So, particular solution is y_p=3/317 e^(7x) .

Case 3. phi(x)=k_1cos(beta x)+k_2 sin(beta x) . Assume solution in the form of y_p=Acos(beta x)+Bcos(beta x) (even when either k_1 or k_2 equals zero).

Example 3. Solve y''+y=cos(2x) .

We are asked to solve equation, so we first detemine solution of homogeneous equation y''+y=0 .

Characteristic equation is r^2+1=0 that has roots r_1=i,\ r_2=-i so solution of homogeneous equation (see how to solve homogeneous equation) is y_h=c_1 cos(x)+c_2 sin(x) .

Now, assume that particular solution is of the form y_p=Acos(2x)+Bsin(2x) then y_p'=-2Asin(2x)+2Bcos(2x) and y_p''=-4Acos(2x)-4Bsin(2x) .

Plugging these values into equation gives: -4Acos(2x)-4Bsin(2x)+Acos(2x)+Bsin(2x)=-3Acos(2x)-3Bsin(2x) .

Equating like terms with cos(2x) we obtain that

{(-3A=1),(-3B=0):}

So, A=-1/3, B=0 .

So, particular solution is y_p=-1/3cos(2x) .

Finally, general solution is y=y_h+y_p=c_1 cos(x)+c_2 sin(2x)-1/3 cos(2x) .

Generalizations. If phi(x) is the product of functions from above cases then assume particular solution to be product of assumed solutions of corresponding cases. For example, if phi(x)=e^(alpha x)p_n(x)sin(beta x) or phi(x)=e^(alpha x)p_n(x)cos(beta x) then assume particular solution in the form

y_p=e^(alpha x)(A_n x^n+A_(n-1)x^(n-1)+...+A_1x+A_0)cos(beta x)+

+e^(alpha x)(B_n x^n+B_(n-1)x^(n-1)+...+B_1x+B_0)cos(beta x) where A_j and B_j (j=0, 1, ..., n) are constants that still need to be determined.

If phi(x) is the sum (or difference) of terms already considered, then we take y_p to be the sum (or difference) of the corresponding assumed solutions and algebraically combine arbitrary constants where possible.

Example 4. Find particular solution of y''+y=e^(-x)cos(2x)+x^2 .

Here phi(x) is sum of two manageable functions: phi(x)=(e^(-x)cos(2x))+(x^2)

So, assume particular solution in the form of y_p=(e^(-x)(Acos(2x)+Bsin(2x)))+(Cx^2+Dx+E) .

Then y_p'=-Ae^(-x)cos(2x)-2Ae^(-x)sin(2x)-Be^(-x)sin(2x)+2Be^(-x)cos(2x)+2Cx+D and

y_p''=Ae^(-x)cos(2x)+2Ae^(-x)sin(2x)+2Ae^(-x)sin(2x)-4Ae^(-x)cos(2x)+Be^(-x)sin(2x)-2Be^(-x)cos(2x)-2Be^(-x)cos(2x)-4Be^(-x)sin(2x)+2C .

Plugging these values into equation gives

Ae^(-x)cos(2x)+2Ae^(-x)sin(2x)+2Ae^(-x)sin(2x)-4Ae^(-x)cos(2x)+Be^(-x)sin(2x)-2Be^(-x)cos(2x)-2Be^(-x)cos(2x)-4Be^(-x)sin(2x)+2C+e^(-x)(Acos(2x)+Bsin(2x))+Cx^2+Dx+E=(A-4A-2B-2B+A)e^(-x)cos(2x)+(2A+2A+B-4B+B)e^(-x)sin(2x)+Cx^2+Dx+(2C+E)=(-2A-4B)e^(-x)cos(2x)+(4A-2B)e^(-x)sin(2x)+Cx^2+Dx+(2C+E)

Equating like terms with e^(-x)cos(2x)+x^2 we obtain that

{(-2A-4B=1),(4A-2B=0),(C=1),(D=0),(2C+E=0):}

This system has following solution: A=-1/10, B=-1/5, C=1, D=0,E=-2 , so particular solution is

y_p=-1/10 e^(-x)cos(2x)-1/5 e^(-x)sin(2x)+x^2-2 .

If any term of the assumed solution, disregarding multiplicative constants, is also a term of y_h (the homogeneous solution), then the assumed solution must be modified by multiplying it by x^m , where m is the smallest positive integer such that the product of x_m with the assumed solution has no terms in common with yh.

Example 5. Find particular solution of y''+y=cos(x) .

As already known from example 3 homogeneous solution is y_h=c_1cos(x)+c_2sin(x) .

We can't take particular solution to be Acos(x)+Bsin(x) because its terms are parts of homogeneous solution so let

y_p=Axcos(x)+Bxsin(x) .

Then y_p'=Acos(x)-Axsin(x)+Bsin(x)+Bxcos(x) and y_p''=-Asin(x)-Asin(x)-Axcos(x)+Bcos(x)+Bcos(x)-Bxsin(x) .

Plugging these values into equation gives:

-Asin(x)-Asin(x)-Axcos(x)+Bcos(x)+Bcos(x)-Bxsin(x)+Axcos(x)+Bxsin(x)=-2Asin(x)+2Bcos(x) .

Equating like terms with cos(x) we obtain that

{(-2A=0),(2B=1):}

Solution is A=0, B=1/2 .

So, particular solution is y_p=1/2 xsin(x) .

Example 6. Solve y''-2y'+y=e^(x) .

First find solution of corresponding homogeneous equation y''-2y'+y=0 .

Characteristic equation is r^2-2r+1=0 or (r-1)^2=0 which has roots r_1=1, r_2=1 .

There is one root of multiplicity 2, so solution is y_h=c_1e^x+c_2 xe^x .

Now, what will be the particular solution? It isn't e^x because e^x is already in solution, it is not also xe^x . The lowest number m for which x^me^x is not in solution is m=2: x^2 e^x .

So, let y_p=Ax^2e^x.

Then y_p'=2Axe^x+Ax^2e^x=Ae^x (x^2+2x) and y_p''=2Ae^x+2Axe^x+2Axe^x+Ax^2e^x=Ae^x (x^2+4x+2)

Plugging obtained values into equation we have

Ae^x(x^2+4x+2)-2Ae^x(x^2+2x)+Ax^2 e^x=e^x.

2Ae^x=e^x.

Equating like terms with e^x gives 2A=1 or A=1/2.

So, particular soltuion is y_p=1/2 x^2e^x .

Finally, general solution is y=y_h+y_p=c_1e^x+c_2xe^x+1/2 x^2 e^x .

As can be seen method of undetermined coefficients require many calculations and solving system of linear equations. So it is time consuming.

In general, if phi(x) is not one of the types of functions considered above, or if the differential equation does not have constant coefficients then method of undetermined coefficients is unapplicable. For example, particular solution to the equation y''+y=(cos(x))/sin(x) cannot be found with method of undetermined coefficients.