What is a particular integral in second order ODE? Hello friends, today I’ll talk about the particular integral in any second-order ordinary differential equation. Have a look!!

### What is a particular integral in second order ODE?

Now 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

In my earlier posts, I have already discussed about the three possible solutions.

So it can be a second-order homogeneous ODE with real and different roots.

Or, it can also be a second-order homogeneous ODE with real and equal roots.

And the last possibility is that it will be a second-order homogeneous ODE with complex conjugate roots.

Now the solution of the homogeneous part is the complementary solution of the equation. So it is .

Next, I’ll do the next part. And that 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. And in these examples, the value of will be the first one from the above-mentioned list.

So here it goes.

#### Examples of the particular integral 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:

”

##### Solution

Now 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

So I can say or .

That shows the given differential equation has complex conjugate 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

So the particular integral will be

(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

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

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

So 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 this example.

Now comes my second example.

##### Example 2

According to Stroud and Booth (2013) “Obtain the general solution of the equation:

”

##### Solution

Now 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 .

As I can see, it is not possible to factorise the characteristic equation.

So I’ll use the standard formula for the solution of a quadratic equation.

Thus the value of will be

So I can say or .

That shows the given differential equation has complex conjugate 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

So the particular integral will be

(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

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

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

Thus it will be

(14)

Next, I’ll compare the constants on both sides.

So it will be

(15)

Now I’ll put back the values from equations (14) and (15) to equation (11) to get

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

This means

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

Dear friends, this is the end of my today’s post. 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!!

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