Method of Undetermined Coefficients

This page is about second order differential equations of this type:

d2y dx2 + P(x) dy dx + Q(x)y = f(x)

where P(x), Q(x) and f(x) are functions of x.

Two Methods

There are two main methods to solve these equations:

Undetermined Coefficients (that we learn here) which only works when f(x) is a polynomial, exponential, sine, cosine or a linear combination of those.

Variation of Parameters which is a little messier but works on a wider range of functions.

Undetermined Coefficients

To keep things simple, we only look at the case:

d2y dx2 + p dy dx + qy = f(x)

where p and q are constants.

The complete solution to such an equation can be found by combining two types of solution:

  1. The general solution of the homogeneous equation

    2. d2y dx2 + p dy dx + qy = 0

  2. Particular solutions of the non-homogeneous equation

    3. d2y dx2 + p dy dx + qy = f(x)

Note that f(x) could be a single function or a sum of two or more functions.

Once we have found the general solution and all the particular solutions, then the final complete solution is found by adding all the solutions together.

Example 1: d2y dx2 − y = 2x2 − x − 3

(For the moment trust me regarding these solutions)

The homogeneous equation d2y dx2 − y = 0 has a general solution

y = Aex + Be-x

The non-homogeneous equation d2y dx2 − y = 2x2 − x − 3 has a particular solution

y = −2x2 + x − 1

So the complete solution of the differential equation is

y = Aex + Be-x − 2x2 + x − 1

Let's check if the answer is correct:

y = Aex + Be-x − 2x2 + x − 1

dy dx = Aex − Be-x − 4x + 1

d2y dx2 = Aex + Be-x − 4

Putting it together:

d2y dx2 − y = Aex + Be-x − 4 − (Aex + Be-x − 2x2 + x − 1)

= Aex + Be-x − 4 − Aex − Be-x + 2x2 − x + 1

= 2x2 − x − 3

So in this case we have shown that the answer is correct, but how do we find the particular solutions?

We can try guessing ... !

This method is only easy to apply if f(x) is one of the following:

Either: f(x) is a polynomial function.

Or: f(x) is a linear combination of sine and cosine functions.

Or: f(x) is an exponential function.

And here is a guide to help us with a guess:

f(x) y(x) guess
aebx Aebx
a cos(cx) + b sin(cx) A cos(cx) + B sin(cx)
kxn (n=0, 1, 2,...) Anxn + An−1xn−1 + … + A0

But there is one important rule that must be applied:

You must first find the general solution to the homogeneous equation.

You will see why as we continue on.

Example 1 (again): Solve d2y dx2 − y = 2x2 − x − 3

1. Find the general solution of

d2y dx2 − y = 0

The characteristic equation is: r2 − 1 = 0

Factor: (r − 1)(r + 1) = 0

r = 1 or −1

So the general solution of the differential equation is

y = Aex + Be-x

2. Find the particular solution of

d2y dx2 − y = 2x2 − x − 3

We make a guess:

Let y = ax2 + bx + c

dy dx = 2ax + b

d2y dx2 = 2a

Substitute these values into d2y dx2 − y = 2x2 − x − 3

2a − (ax2 + bx + c) = 2x2 − x − 3

2a − ax2 − bx − c = 2x2 − x − 3

− ax2 − bx + (2a − c) = 2x2 − x − 3

Equate coefficients:

x2 coefficients: −a = 2 ⇒ a = −2   ...   (1)
x coefficients: −b = −1 ⇒ b = 1   ...   (2)
Constant coefficients: 2a − c = −3   ...   (3)

Substitute a = −2 from (1) into (3)

−4 − c = −3

c = −1

a = −2, b = 1 and c = −1, so the particular solution of the differential equation is

y = − 2x2 + x − 1

Finally, we combine our two answers to get the complete solution:

y = Aex + Be-x − 2x2 + x − 1

Why did we guess y = ax2 + bx + c (a quadratic function) and not include a cubic term (or higher)?

The answer is simple. The function f(x) on the right side of the differential equation has no cubic term (or higher); so, if y did have a cubic term, its coefficient would have to be zero.

Hence, for a differential equation of the type d2y dx2 + p dy dx + qy = f(x) where f(x) is a polynomial of degree n, our guess for y will also be a polynomial of degree n.


Example 2: Solve

6 d2y dx2 − 13 dy dx − 5y = 5x3 + 39x2 − 36x − 10

1. Find the general solution of 6 d2y dx2 − 13 dy dx − 5y = 0

The characteristic equation is: 6r2 − 13r − 5 = 0

Factor: (2r − 5)(3r + 1) = 0

r = 5 2 or − 1 3

So the general solution of the differential equation is

y = Ae(5/2)x + Be(−1/3)x

2. Find the particular solution of 6 d2y dx2 − 13 dy dx − 5y = 5x3 + 39x2 − 36x − 10

Guess a cubic polynomial because 5x3 + 39x2 − 36x − 10 is cubic.

Let y = ax3 + bx2 + cx + d

dy dx = 3ax2 + 2bx + c

d2y dx2 = 6ax + 2b

Substitute these values into 6 d2y dx2 − 13 dy dx −5y = 5x3 + 39x2 −36x −10

6(6ax + 2b) − 13(3ax2 + 2bx + c) − 5(ax3 + bx2 + cx + d) = 5x3 + 39x2 − 36x − 10

36ax + 12b − 39ax2 − 26bx − 13c − 5ax3 − 5bx2 − 5cx − 5d = 5x3 + 39x2 − 36x − 10

−5ax3 + (−39a − 5b)x2 + (36a − 26b − 5c)x + (12b − 13c − 5d) = 5x3 + 39x2 − 36x − 10

Equate coefficients:

x3 coefficients: −5a = 5 ⇒ a = −1
x2 coefficients: −39a −5b = 39 ⇒ b = 0
x coefficients: 36a −26b −5c = −36 ⇒ c = 0
Constant coefficients: 12b − 13c −5d = −10 ⇒ d = 2

So the particular solution is:

y = −x3 + 2

Finally, we combine our two answers to get the complete solution:

y = Ae(5/2)x + Be(−1/3)x − x3 + 2

And here are some sample curves:


Example 3: Solve d2y dx2 + 3 dy dx − 10y = −130cos(x) + 16e3x

In this case we need to solve three differential equations:

1. Find the general solution to d2y dx2 + 3 dy dx − 10y = 0

2. Find the particular solution to d2y dx2 + 3 dy dx − 10y = −130cos(x)

3. Find the particular solution to d2y dx2 + 3 dy dx − 10y = 16e3x

So, here's how we do it:

1. Find the general solution to d2y dx2 + 3 dy dx − 10y = 0

The characteristic equation is: r2 + 3r − 10 = 0

Factor: (r − 2)(r + 5) = 0

r = 2 or −5

So the general solution of the differential equation is:

y = Ae2x+Be-5x

2. Find the particular solution to d2y dx2 + 3 dy dx − 10y = −130cos(x)

Guess. Since f(x) is a cosine function, we guess that y is a linear combination of sine and cosine functions:

Try y = acos⁡(x) + bsin(x)

dy dx = − asin(x) + bcos(x)

d2y dx2 = − acos(x) − bsin(x)

Substitute these values into d2y dx2 + 3 dy dx − 10y = −130cos(x)

−acos⁡(x) − bsin(x) + 3[−asin⁡(x) + bcos(x)] − 10[acos⁡(x)+bsin(x)] = −130cos(x)

cos(x)[−a + 3b − 10a] + sin(x)[−b − 3a − 10b] = −130cos(x)

cos(x)[−11a + 3b] + sin(x)[−11b − 3a] = −130cos(x)

Equate coefficients:

Coefficients of cos(x): −11a + 3b = −130   ...   (1)
Coefficients of sin(x): −11b − 3a = 0   ...   (2)

From equation (2), a = − 11b 3

Substitute into equation (1)

121b 3 + 3b = −130

130b 3 = −130

b = −3

a = − 11(−3) 3 = 11

So the particular solution is:

y = 11cos⁡(x) − 3sin(x)

3. Find the particular solution to d2y dx2 + 3 dy dx − 10y = 16e3x

Guess.

Try y = ce3x

dy dx = 3ce3x

d2y dx2 = 9ce3x

Substitute these values into d2y dx2 + 3 dy dx − 10y = 16e3x

9ce3x + 9ce3x − 10ce3x = 16e3x

8ce3x = 16e3x

c = 2

So the particular solution is:

y = 2e3x

Finally, we combine our three answers to get the complete solution:

y = Ae2x + Be-5x + 11cos⁡(x) − 3sin(x) + 2e3x


Example 4: Solve d2y dx2 + 3 dy dx − 10y = −130cos(x) + 16e2x

This is exactly the same as Example 3 except for the final term, which has been replaced by 16e2x.

So Steps 1 and 2 are exactly the same. On to step 3:

3. Find the particular solution to d2y dx2 + 3 dy dx − 10y = 16e2x

Guess.

Try y = ce2x

dy dx = 2ce2x

d2y dx2 = 4ce2x

Substitute these values into d2y dx2 + 3 dy dx − 10y = 16e2x

4ce2x + 6ce2x − 10ce2x = 16e2x

0 = 16e2x

Oh dear! Something seems to have gone wrong. How can 16e2x = 0?

Well, it can't, and there is nothing wrong here except that there is no particular solution to the differential equation d2y dx2 + 3 dy dx − 10y = 16e2x

...Wait a minute!
The general solution to the homogeneous equation d2y dx2 + 3 dy dx − 10y = 0, which is y = Ae2x + Be-5x, already has a term Ae2x, so our guess y = ce2x already satisfies the differential equation d2y dx2 + 3 dy dx − 10y = 0 (it was just a different constant.)

So we must guess y = cxe2x

Let's see what happens:

dy dx = ce2x + 2cxe2x

d2y dx2 = 2ce2x + 4cxe2x + 2ce2x = 4ce2x + 4cxe2x

Substitute these values into d2y dx2 + 3 dy dx − 10y = 16e2x

4ce2x + 4cxe2x + 3ce2x + 6cxe2x − 10cxe2x = 16e2x

7ce2x = 16e2x

c = 167

So in the present case our particular solution is

y = 16 7 xe2x

Thus, our final complete solution in this case is:

y = Ae2x + Be-5x + 11cos⁡(x) − 3sin(x) + 16 7 xe2x


Example 5: Solve d2y dx2 − 6 dy dx + 9y = 5e-2x

1. Find the general solution to d2y dx2 − 6 dy dx + 9y = 0

The characteristic equation is: r2 − 6r + 9 = 0

(r − 3)2 = 0

r = 3, which is a repeated root.

Then the general solution of the differential equation is y = Ae3x + Bxe3x

2. Find the particular solution to d2y dx2 − 6 dy dx + 9y = 5e-2x

Guess.

Try y = ce-2x

dy dx = −2ce-2x

d2y dx2 = 4ce-2x

Substitute these values into d2y dx2 − 6 dy dx + 9y = 5e-2x

4ce-2x + 12ce-2x + 9ce-2x = 5e-2x

25e-2x = 5e-2x

c = 1 5

So the particular solution is:

y= 1 5 e-2x

Finally, we combine our two answers to get the complete solution:

y= Ae3x + Bxe3x + 1 5 e-2x


Example 6: Solve d2y dx2 + 6 dy dx + 34y = 109cos(5x)

1. Find the general solution to d2y dx2 + 6 dy dx + 34y = 0

The characteristic equation is: r2 + 6r + 34 = 0

Use the quadratic equation formula

r = −b ± √(b2 − 4ac) 2a

with a = 1, b = 6 and c = 34

So

r = −6 ± √[62 − 4(1)(34)] 2(1)

r = −6 ± √(36−136) 2

r = −6 ± √(−100) 2

r = −3 ± 5i

And we get:

y =e-3x(Acos⁡(5x) + iBsin(5x))

2. Find the particular solution to d2y dx2 + 6 dy dx + 34y = 109sin(5x)

Since f(x) is a sine function, we assume that y is a linear combination of sine and cosine functions:

Guess.

Try y = acos⁡(5x) + bsin(5x)

Note: since we do not have sin(5x) or cos(5x) in the solution to the homogeneous equation (we have e-3xcos(5x) and e-3xsin(5x), which are different functions), our guess should work.

Let's continue and see what happens:

dy dx = −5asin⁡(5x) + 5bcos(5x)

d2y dx2 = −25acos⁡(5x) − 25bsin(5x)

Substitute these values into d2y dx2 + 6 dy dx + 34y = 109sin(5x)

−25acos⁡(5x) − 25bsin(5x) + 6[−5asin⁡(5x) + 5bcos(5x)] + 34[acos⁡(5x) + bsin(5x)] = 109sin(5x)

cos(5x)[−25a + 30b + 34a] + sin(5x)[−25b − 30a + 34b] = 109sin(5x)

cos(5x)[9a + 30b] + sin(5x)[9b − 30a] = 109sin(5x)

Equate coefficients of cos(5x) and sin(5x):

Coefficients of cos(5x): 9a + 30b = 109   ...   (1)
Coefficients of sin(5x): 9b − 30a = 0   ...   (2)

From equation (2), a = 3b 10

Substitute into equation (1)

9( 3b 10 ) + 30b = 109

327b = 1090

b = 10 3

a = 1

So the particular solution is:

y = cos⁡(5x) + 10 3 sin(5x)

Finally, we combine our answers to get the complete solution:

y = e-3x(Acos⁡(5x) + iBsin(5x)) + cos⁡(5x) + 10 3 sin(5x)

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