Method of Undetermined Coefficients (2024)

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

d²ydx² + P(x)dydx + Q(x)y = f(x)

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

Please read Introduction to Second Order Differential Equations first, it shows how to solve the simpler "hom*ogeneous" case where f(x)=0

Example 1: d²ydx² − y = 2x² − x − 3

(For the moment trust me regarding these solutions)

The hom*ogeneous equation d²ydx² − y = 0 has a general solution

y = Ae^x + Be^-x

The non-hom*ogeneous equation d²ydx² − y = 2x² − x − 3 has a particular solution

y = −2x² + x − 1

So the complete solution of the differential equation is

y = Ae^x + Be^-x − 2x² + x − 1

Let’s check if the answer is correct:

y = Ae^x + Be^-x − 2x² + x − 1

dydx = Ae^x − Be^-x − 4x + 1

d²ydx² = Ae^x + Be^-x − 4

Putting it together:

d²ydx² − y = Ae^x + Be^-x − 4 − (Ae^x + Be^-x − 2x² + x − 1)

= Ae^x + Be^-x − 4 − Ae^x − Be^-x + 2x² − x + 1

= 2x² − x − 3

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.

Example 1 (again): Solve d²ydx² − y = 2x² − x − 3

1. Find the general solution of

d²ydx² − 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 = Ae^x + Be^-x

2. Find the particular solution of

d²ydx² − y = 2x² − x − 3

We make a guess:

Let y = ax² + bx + c

dydx = 2ax + b

d²ydx² = 2a

Substitute these values into d²ydx² − 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:

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 = Ae^x + 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 d²ydx² + pdydx + 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

6d²ydx² − 13dydx − 5y = 5x³ + 39x² − 36x − 10

1. Find the general solution of 6d²ydx² − 13dydx − 5y = 0

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

Factor: (2r − 5)(3r + 1) = 0

r = 52 or −13

So the general solution of the differential equation is

y = Ae^(5/2)x + Be^(−1/3)x

2. Find the particular solution of 6d²ydx² − 13dydx − 5y = 5x³ + 39x² − 36x − 10

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

Let y = ax³ + bx² + cx + d

dydx = 3ax² + 2bx + c

d²ydx² = 6ax + 2b

Substitute these values into 6d²ydx² − 13dydx −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:

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 d²ydx² + 3dydx − 10y = −130cos(x) + 16e^3x

In this case we need to solve three differential equations:

1. Find the general solution to d²ydx² + 3dydx − 10y = 0

2. Find the particular solution to d²ydx² + 3dydx − 10y = −130cos(x)

3. Find the particular solution to d²ydx² + 3dydx − 10y = 16e^3x

So, here’s how we do it:

1. Find the general solution to d²ydx² + 3dydx − 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 = Ae^2x+Be^-5x

2. Find the particular solution to d²ydx² + 3dydx − 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)

dydx = − asin(x) + bcos(x)

d²ydx² = − acos(x) − bsin(x)

Substitute these values into d²ydx² + 3dydx − 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:

From equation (2), a = −11b3

Substitute into equation (1)

121b3 + 3b = −130

130b3 = −130

b = −3

a = −11(−3)3 = 11

So the particular solution is:

y = 11cos⁡(x) − 3sin(x)

3. Find the particular solution to d²ydx² + 3dydx − 10y = 16e^3x

Guess.

Try y = ce^3x

dydx = 3ce^3x

d²ydx² = 9ce^3x

Substitute these values into d²ydx² + 3dydx − 10y = 16e^3x

9ce^3x + 9ce^3x − 10ce^3x = 16e^3x

8ce^3x = 16e^3x

c = 2

So the particular solution is:

y = 2e^3x

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

y = Ae^2x + Be^-5x + 11cos⁡(x) − 3sin(x) + 2e^3x

Example 4: Solve d²ydx² + 3dydx − 10y = −130cos(x) + 16e^2x

This is exactly the same as Example 3 except for the final term, which has been replaced by 16e^2x.

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

3. Find the particular solution to d²ydx² + 3dydx − 10y = 16e^2x

Guess.

Try y = ce^2x

dydx = 2ce^2x

d²ydx² = 4ce^2x

Substitute these values into d²ydx² + 3dydx − 10y = 16e^2x

4ce^2x + 6ce^2x − 10ce^2x = 16e^2x

0 = 16e^2x

Oh dear! Something seems to have gone wrong. How can 16e^2x = 0?

Well, it can’t, and there is nothing wrong here except that there is no particular solution to the differential equation d²ydx² + 3dydx − 10y = 16e^2x

...Wait a minute!
The general solution to the hom*ogeneous equation d²ydx² + 3dydx − 10y = 0, which is y = Ae^2x + Be^-5x, already has a term Ae^2x, so our guess y = ce^2x already satisfies the differential equation d²ydx² + 3dydx − 10y = 0 (it was just a different constant.)

dydx = ce^2x + 2cxe^2x

d²ydx² = 2ce^2x + 4cxe^2x + 2ce^2x = 4ce^2x + 4cxe^2x

Substitute these values into d²ydx² + 3dydx − 10y = 16e^2x

4ce^2x + 4cxe^2x + 3ce^2x + 6cxe^2x − 10cxe^2x = 16e^2x

7ce^2x = 16e^2x

c = 167

So in the present case our particular solution is

y = 167xe^2x

Thus, our final complete solution in this case is:

y = Ae^2x + Be^-5x + 11cos⁡(x) − 3sin(x) + 167xe^2x

Example 5: Solve d²ydx² − 6dydx + 9y = 5e^-2x

1. Find the general solution to d²ydx² − 6dydx + 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 = Ae^3x + Bxe^3x

2. Find the particular solution to d²ydx² − 6dydx + 9y = 5e^-2x

Guess.

Try y = ce^-2x

dydx = −2ce^-2x

d²ydx² = 4ce^-2x

Substitute these values into d²ydx² − 6dydx + 9y = 5e^-2x

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

25e^-2x = 5e^-2x

c = 15

So the particular solution is:

y= 15e^-2x

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

y= Ae^3x + Bxe^3x + 15e^-2x

Example 6: Solve d²ydx² + 6dydx + 34y = 109cos(5x)

1. Find the general solution to d²ydx² + 6dydx + 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 d²ydx² + 6dydx + 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 hom*ogeneous 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:

dydx = −5asin⁡(5x) + 5bcos(5x)

d²ydx² = −25acos⁡(5x) − 25bsin(5x)

Substitute these values into d²ydx² + 6dydx + 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):

From equation (1), b = −3a10

Substitute into equation (2)

9(−3a10) − 30a = 109

−27a − 300a = 1090

−327a = 1090

a = −103

b = 1

So the particular solution is:

y = −103cos⁡(5x) + sin(5x)

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

y = e^-3x(Acos⁡(5x) + iBsin(5x)) − 103cos⁡(5x) + sin(5x)

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