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

**Exponential 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. And in these examples, the value of will be the second one from the above-mentioned list.

So here it goes.

**Examples of the exponential 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:

”

**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 different 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

As I can see from equation (2), the complementary solution has already a term with , I’ll choose as

(3)

Now I’ll find out the value of .

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

(4)

Next, I’ll differentiate with respect to .

Thus it will be

(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 on both sides of the equation.

So it will be

(6)

Now I’ll put back the value of from equation (6) 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 I’ll give another example.

**Example 2**

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

given that at and .”

**Solution**

Here the given differential equation is

(7)

So the homogeneous part of equation (7) 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

(8)

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

As I can see from equation (8), the complementary solution has already a term with , I’ll choose as

(9)

Now I’ll find out the value of .

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

(10)

Next, I’ll differentiate with respect to .

Thus it will be

(11)

Now I’ll put back the values from equations (9)-(11) in equation (7).

Then the differential equation will look like

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

So it will be

(12)

Now I’ll put back the value of from equation (12) to equation (9) to get

So the general solution of equation (7) is a sum of equations (8) and (9).

(13)

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

**Step 3**

Thus I’ll differentiate equation (13) with respect to to get

(14)

Here I already know from the given example that at and .

Now I’ll substitute in equation (13) to get

(15)

Similarly, I already know from the given example that at

Now I’ll substitute in equation (14) to get

(16)

Next, I’ll subtract equation (15) from equation (16) to get

(17)

Now I’ll substitute the value of from equation (17) to equation (15) to get the value of as

(18)

At the end, now I substitute the values of and from equations (18) and (17) respectively to equation (13) to get the particular solution of as

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

Dear friends, this is the end of 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!!

Best Maritime universities says

Great EXPONENTIAL FUNCTIONS AS PARTICULAR INTEGRALS