# Find a Particular Solution of 9y''+5y'+2y=sin^2(x)

• Tom McCurdy
In summary, the method of undetermined coefficients was used to find a particular solution yp of the given differential equation, 9y''+5y'+2y=sin^2(x). The solution was found to be y_p = (1/4)+(17/1256)*cos(2*x)+(-5/1256)*sin(2*x), using the derivatives A, B, and C and solving for their corresponding values. After checking the answer by plugging it into the differential equation, it was confirmed to be correct.

#### Tom McCurdy

What did I do Wrong??

## Homework Statement

Find a particular solution yp of the differential equation
$$9y''+5y'+2y=sin^2(x)$$

using the Method of Undetermined Coefficients. Primes denote derivatives with respect to x.

## Homework Equations

$$sin^2(x) = \frac{1-cos(2x)}{2}$$

## The Attempt at a Solution

Possible derivatives
• A
• B cos(2x)
• C sin(2x)

$$y_p = A + Bcos(2x) + Csin(2x)$$

$$y_p' = -2Bsin(2x) + 2Ccos(2x)$$

$$y_p'' = -4Bcos(2x) + -4Csin(2x)$$

Sub back into original modified with cos substitution on the right

$$9y''+5y'+2y = \frac{1-cos(2x)}{2}$$

Becomes

$$9[-4Bcos(2x) + -4Csin(2x)] + 5[-2Bsin(2x) + 2Ccos(2x)] + 2[A + Bcos(2x) + Csin(2x)] = \frac{1-cos(2x)}{2}$$

From that I get
2A = 1/2 => A=1/4

-36B + 10C + 2B = -1/2
-36C + -10B + 2C = 0

Solving for B and C

$$y_p = (1/4)+(17/1056)*cos(2*x)+(5/1056)*sin(2*x)$$

Where
B = 17/1056
and
C = 5/1056

Last edited:

Did you check your answer by plugging it into the diff. eq.? (I'm sure you will see that A is wrong at least.)

I realized I type it wrong when i had A defined, but i had it correct or so I thought in the equation A=1/4

-36B + 10C + 2B = -1/2
-36C + -10B + 2C = 0

Where
B = 17/1056
and
C = 5/1056
You solved for A correctly, but B and C are wrong.

-36B + 10C + 2B = -1/2
-36C + -10B + 2C = 0

-34B + 10 C = -1/2
-10B + -34C = 0

34B + 115.6C = 0
125.6C=-1/2

C = -5/1256

Therefore B equal

-10B + -34C = 0

B = 17/1256

I am down to one final attempt at this problem can someone confirm this correct?

$$y_p = (1/4)+(17/1256)*cos(2*x)+(-5/1256)*sin(2*x)$$

That's what I get.

Algerbra FTW

... eh wow I feel stupid

it worked!

## 1. What is the general form of the particular solution for a second order linear differential equation?

The general form of the particular solution for a second order linear differential equation is y_p = A*sin(x) + B*cos(x), where A and B are constants determined by the coefficients of the equation.

## 2. How do you determine the particular solution for a specific second order linear differential equation?

To determine the particular solution for a specific second order linear differential equation, you can use the method of undetermined coefficients. This involves finding a function that satisfies the equation and substituting it into the original equation to solve for the coefficients.

## 3. What is the role of the nonhomogeneous term in finding the particular solution?

The nonhomogeneous term, in this case sin^2(x), is the forcing term that drives the solution of the differential equation. It provides the specific conditions that the particular solution must satisfy.

## 4. Can there be more than one particular solution for a given second order linear differential equation?

No, there can only be one particular solution for a given second order linear differential equation. However, there can be a general solution that includes a combination of the particular solution and the complementary solution.

## 5. How can I check if my particular solution is correct?

To check if your particular solution is correct, you can substitute it into the original differential equation and see if it satisfies the equation. You can also take the second derivative of the particular solution and see if it matches the given differential equation.