What happens with trigonometric functions as particular integrals? Dear friends, today I’ll talk about the trigonometric functions as particular integrals in any second-order ordinary differential equation. Have a look!!

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**Trigonometric functions as particular integrals in second-order ODE**

The standard form of any second-order ordinary differential equation is

Here are constants and is a function of . Also is a constant.

Next, to solve this equation, I’ll solve the homogeneous part first.

So this means I have to solve the second order homogeneous ODE as

Now the solution of the homogeneous part is the complementary solution of the equation. Let’s give it a name, say, .

Next job is to get the particular integral of , that is, .

So, I’ll start with the .

Therefore if the value of is

- , then the particular integral will be .
- , then the particular integral will be , .
- , then the particular integral will be .
- , then the particular integral will be .

Now I’ll solve some examples on the trigonometric functions as particular integrals in second-order ODE. And in these examples, the value of will be the last two from the above-mentioned list.

So here it goes.

**Examples of trigonometric functions as particular integrals in second-order ODE**

Disclaimer: These are not my own examples. I have chosen these from some book. I have also given the due reference at the end of the post.

Here are the examples.

**Example 1**

According to Stroud and Booth (2013) “Solve the following equation:

”

**Solution**

Here the given differential equation is

(1)

So the homogeneous part of equation (1) is

Next I’ll solve this homogeneous part to get the complementary solution of this equation.

**Step 1**

Therefore the auxiliary equation (characteristic equation) is

Now I’ll solve this equation to get the values of .

So I can also rewrite this equation as

This means

That shows the given second-order ordinary differential equation has real and equal roots.

Therefore the complementary solution of the given differential equation is

(2)

Next, I’ll find out the particular integral of .

**Step 2**

Let’s give it a name, say, . Here the non-homogeneous part of the differential equation is

Now I can see from the non-homogeneous part of the equation, there is a constant term 1. So I’ll choose as

(3)

Now I’ll find out the values of .

First of all, I’ll differentiate with respect to .

(4)

Next, I’ll differentiate with respect to .

(5)

Now I’ll put back the values from equations (3)-(5) in equation (1).

Then the differential equation will look like

Now I can compare the coefficients of and constants on both sides of the equation.

At first, I’ll compare the constants on both sides.

Thus it will be

(6)

Next, I’ll compare the coefficient of on both sides.

Thus it will be

(7)

At the end, I’ll compare the coefficient of on both sides.

Thus it will be

(8)

Now I’ll put back the values from equations (6)-(8) to equation (3) to get

So the general solution of equation (1) is a sum of equations (2) and (3).

This means

Hence I can conclude that this is the answer to the given example.

Now I’ll give you another example of the trigonometric functions as particular integrals.

**Example 2**

According to Stroud and Booth (2013) “Solve the equation

given that at and ”

**Solution**

Here the given differential equation is

(9)

So the homogeneous part of equation (9) is

Next I’ll solve this homogeneous part to get the complementary solution of this equation.

**Step 1**

Therefore the auxiliary equation (characteristic equation) is

Now I’ll solve this equation to get the values of .

So I can also rewrite this equation as

This means

That shows the given second-order ordinary differential equation has real and different roots.

Therefore the complementary solution of the given differential equation is

(10)

Next, I’ll find out the particular integral of .

**Step 2**

Let’s give it a name, say, . Here the non-homogeneous part of the differential equation is

(11)

Now I’ll find out the values of .

First of all, I’ll differentiate with respect to .

(12)

Next, I’ll differentiate with respect to .

(13)

Now I’ll put back the values from equations (11)-(13) in equation (9).

Then the differential equation will look like

Now I can compare the coefficients of on both sides of the equation.

At first, I’ll compare the coefficient of on both sides.

(14)

Then I’ll compare the coefficient of on both sides.

(15)

From equation (14), I can say

Now I’ll put back this value of in (15) to get

So this means

Now I’ll put back the values of to equation (11) to get

So the general solution of equation (9) is a sum of equations (10) and (11).

(16)

Now I’ll find out the particular solution of the equation for .

**Step 3**

Thus I’ll differentiate equation (16) with respect to to get

(17)

Here I already know from the given example that at .

Now I’ll substitute in equation (16) to get

(18)

Similarly, I already know from the given example that at

Now I’ll substitute in equation (17) to get

(19)

Next, I’ll subtract equation (18) from equation (19) to get

Now I’ll substitute the value of to equation (18) to get the value of as

At the end, now I substitute the values of and to equation (16) to get the particular solution of as

Hence I can conclude that this is the answer to the given example.

Dear friends, this is the end of today’s post on the trigonometric functions as particular integrals. Thank you very much for reading this. Please let me know how you feel about it. Soon I will be back again with a new post. Till then, bye, bye!!

Hickoryfoodfactory.com says

Howdy just wanted to give you a brief heads up and let you know a few of the pictures aren’t loading properly.

I’m not sure why but I think its a linking issue.

I’ve tried it in two different internet browsers and

both show the same outcome.

Dr. Aspriha Peters says

Hi,

The problem is sorted out now. Thanks a lot for letting me know.