Trigonometric functions as particular integrals

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(Last Updated On: May 28, 2020)

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

$a\frac{d^2y}{dx^2} + b\frac{dy}{dx} + c y = f(x)~\text{or}~ m.$

Here $a, b, c$ are constants and $f(x)$ is a function of $x$ . Also $m$ 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

$a\frac{d^2y}{dx^2} + b\frac{dy}{dx} + c y = 0.$

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

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

So, I’ll start with the $f(x)$ .

Therefore if the value of $f(x)$ is

$x^n$ , then the particular integral will be $Ax^n + Bx^{n-1}+...+Dx + M$ .
$e^{nx}$ , then the particular integral will be $Ae^{nx}$ , $n = 1, 2, ....$ .
$\sin x$ , then the particular integral will be $A \sin x + B \cos x$ .
$\cos x$ , then the particular integral will be $A \sin x + B \cos x$ .

Now I’ll solve some examples on the trigonometric functions as particular integrals in second-order ODE. And in these examples, the value of $f(x)$ 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:

$\frac{d^2 y}{dx^2}+4 \frac{d y}{d x} + 4 y = 2 \cos^2 x.$

”

Solution

Here the given differential equation is

$\frac{d^2 y}{dx^2}+4 \frac{d y}{d x} + 4 y = 2 \cos^2 x.$

I can also rewrite it as

(1) $\begin{equation*} \frac{d^2 y}{dx^2}+4 \frac{d y}{d x} + 4 y = 1 + \cos 2x. \end{equation*}$

So the homogeneous part of equation (1) is

$\frac{d^2 y}{dx^2}+4 \frac{d y}{d x} + 4 y= 0.$

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

Step 1

Therefore the auxiliary equation (characteristic equation) is

$m^2 +4m + 4 = 0.$

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

So I can also rewrite this equation as

$\begin{equation*} (m+2)^2 = 0. \end{equation*}$

This means

$m = -2, -2.$

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) $\begin{equation*}y_c = (C_1+ C_2 x)e^{-2x}.\end{equation*}$

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

Step 2

Let’s give it a name, say, $y_p$ . Here the non-homogeneous part of the differential equation is $(1 + \cos 2x).$

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

(3) $\begin{equation*}y_p = A + B \cos 2x + C\sin 2x. \end{equation*}$

Now I’ll find out the values of $A, B, C$ .

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

So it will be

(4) $\begin{equation*}\frac{dy_p}{dx} = -2B \sin 2x + 2C \cos 2x.\end{equation*}$

Next, I’ll differentiate $\cfrac{dy_p}{dx}$ with respect to $x$ .

Thus it will be

(5) $\begin{equation*}\frac{d^2y_p}{dx^2} = -4B \cos 2x - 4C \sin 2x.\end{equation*}$

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

Then the differential equation will look like

$\begin{eqnarray*} -4B \cos 2x - 4C \sin 2x + 4 (-2B \sin 2x + 2C \cos 2x) \\+ 4 (A + B \cos 2x + C\sin 2x)&=& 1+ \cos 2x\\ -4B \cos 2x - 4C \sin 2x -8B \sin 2x + 8C \cos 2x \\+4A + 4B \cos 2x + 4C\sin 2x&=& 1+ \cos 2x\\ 4A-8B \sin 2x+ 8C \cos 2x&=& 1+ \cos 2x. \end{eqnarray*}$

Now I can compare the coefficients of $\cos 2x, \sin 2x$ and constants on both sides of the equation.

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

Thus it will be

$4A = 1.$

So this gives

(6) $\begin{equation*}A = \frac{1}{4}.\end{equation*}$

Next, I’ll compare the coefficient of $\cos 2x$ on both sides.

Thus it will be

$8C = 1.$

So this gives

(7) $\begin{equation*}C = \frac{1}{8}.\end{equation*}$

At the end, I’ll compare the coefficient of $\sin 2x$ on both sides.

Thus it will be

$8B = 0.$

So this gives

(8) $\begin{equation*}B = 0.\end{equation*}$

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

$y_p = \frac{1}{4} + \frac{1}{8}\sin 2x.$

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

This means $y = (C_1 + C_2 x) e^{-2x} + \cfrac{1}{4} + \cfrac{1}{8}\sin 2x.$

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

$\frac{d^2 x}{dt^2} - 3 \frac{dx}{d t} + 2x = \sin t$

given that at $t =0, x =0$ and $\frac{dx}{dt} = 0.$ ”

Solution

Here the given differential equation is

(9) $\begin{equation*}\frac{d^2 x}{dt^2} - 3 \frac{dx}{d t} + 2x = \sin t. \end{equation*}$

So the homogeneous part of equation (9) is

$\frac{d^2 x}{dt^2} - 3 \frac{dx}{d t} + 2x = 0.$

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

Step 1

Therefore the auxiliary equation (characteristic equation) is

$m^2 -3m + 2 = 0.$

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

So I can also rewrite this equation as

$\begin{eqnarray*} m^2 -m -2m +2 &=& 0\\ (m-1)(m-2) &=& 0. \end{eqnarray*}$

This means

$m = 1, 2.$

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) $\begin{equation*}x_c = (C_1 e^{t} + C_2 e^{2t}).\end{equation*}$

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

Step 2

Let’s give it a name, say, $x_p$ . Here the non-homogeneous part of the differential equation is $\sin t.$

So I’ll choose $x_p$ as

(11) $\begin{equation*}x_p = A\cos t + B \sin t.\end{equation*}$

Now I’ll find out the values of $A, B$ .

First of all, I’ll differentiate $x_p$ with respect to $t$ .

So it will be

(12) $\begin{equation*}\frac{dx_p}{dt} = -A \sin t + B \cos t.\end{equation*}$

Next, I’ll differentiate $\cfrac{dx_p}{dt}$ with respect to $t$ .

Thus it will be

(13) $\begin{equation*}\frac{d^2x_p}{dt^2} = -A \cos t - B \sin t.\end{equation*}$

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

Then the differential equation will look like

$\begin{eqnarray*} -A \cos t - B \sin t -3(-A \sin t + B \cos t)+2(A\cos t + B \sin t) &=& \sin t\\ -A \cos t - B \sin t + 3A \sin t- 3B \cos t + 2A \cos t + 2B \sin t &=& \sin t\\ (A - 3B)\cos t + (3A + B)\sin t &=& \sin t. \end{eqnarray*}$

Now I can compare the coefficients of $\cos t, \sin t$ on both sides of the equation.

At first, I’ll compare the coefficient of $\cos t$ on both sides.

So this gives

(14) $\begin{equation*}A - 3B = 0.\end{equation*}$

Then I’ll compare the coefficient of $\sin t$ on both sides.

So this gives

(15) $\begin{equation*}3A + B = 1.\end{equation*}$

From equation (14), I can say

$A = 3B.$

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

$9B + B = 1 \Rightarrow 10B = 1 \Rightarrow B = \frac{1}{10}.$

So this means

$A = \frac{3}{10}.$

Now I’ll put back the values of $A,B$ to equation (11) to get

$x_p = \frac{3}{10} \cos t + \frac{1}{10} \sin t.$

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

Therefore it will be

(16) $\begin{equation*}x = C_1 e^{t} + C_2 e^{2t} +\frac{3}{10} \cos t + \frac{1}{10} \sin t.\end{equation*}$

Now I’ll find out the particular solution of the equation for $t = 0$ .

Step 3

Thus I’ll differentiate equation (16) with respect to $t$ to get

(17) $\begin{equation*} \frac{dx}{dt} = C_1 e^{t} +2 C_2 e^{2t} -\frac{3}{10} \sin t + \frac{1}{10} \cos t.\end{equation*}$

Here I already know from the given example that at $t = 0, x = 0$ .

Now I’ll substitute $t = 0, x = 0$ in equation (16) to get

$\begin{eqnarray*}0 &=& C_1 e^{0} + C_2 e^{2.0} +\frac{3}{10} \cos 0+ \frac{1}{10} \sin 0\\&=& C_1.1+C_2.1+ \frac{3}{10}.1+0\\&=& C_1 + C_2 + \frac{3}{10}. \end{eqnarray*}$

So I can say

(18) $\begin{equation*}C_1 + C_2 = -\frac{3}{10}.\end{equation*}$

Similarly, I already know from the given example that at $t = 0, \cfrac{dx}{dt} = 0.$

Now I’ll substitute $t = 0, \cfrac{dx}{dt} = 0$ in equation (17) to get

$\begin{eqnarray*}0 &=& C_1 e^{0} +2 C_2 e^{2.0} -\frac{3}{10} \sin 0 + \frac{1}{10} \cos 0\\&=& C_1.1 +2 C_2.1 -\frac{3}{10}.0 + \frac{1}{10}.1\\&=& C_1 + 2C_2 + \frac{1}{10}. \end{eqnarray*}$

Therefore I can say

(19) $\begin{equation*} C_1 + 2C_2 =-\frac{1}{10}.\end{equation*}$

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

$C_2 = \frac{1}{5}.$

Now I’ll substitute the value of $C_2$ to equation (18) to get the value of $C_1$ as

$C_1+\frac{1}{5} = -\frac{3}{10} \Rightarrow C_1 = -\frac{3}{10} - \frac{1}{5}\Rightarrow C_1 = - \frac{1}{2}.$

At the end, now I substitute the values of $C_1$ and $C_2$ to equation (16) to get the particular solution of $x$ as

$x = - \frac{1}{2} e^{ t} - \frac{1}{10} e^{2 t} -\frac{3}{10}\sin t + \frac{1}{10}\cos t.$

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!!

*Reference: K. A. Stroud and Dexter J. Booth (2013): Engineering Mathematics, Industrial Press, Inc.; 7th edition (March 8, 2013), Chapter: Second order differential equations, Further problems 26, p. 1025, Q. No. 7 (Example 1); Q. No. 14 (Example 2).

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January 20, 2019 at 6:17 pm
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.
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Trigonometric functions as particular integrals | Engineering mathematics

Trigonometric functions as particular integrals in second-order ODE

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

Example 1

Solution

Step 1

Step 2

Example 2

Solution

Step 1

Step 2

Step 3

*Reference: K. A. Stroud and Dexter J. Booth (2013): Engineering Mathematics, Industrial Press, Inc.; 7th edition (March 8, 2013), Chapter: Second order differential equations, Further problems 26, p. 1025, Q. No. 7 (Example 1); Q. No. 14 (Example 2).

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Trigonometric functions as particular integrals in second-order ODE

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

Example 1

Solution

Step 1

Step 2

Example 2

Solution

Step 1

Step 2

Step 3

*Reference: K. A. Stroud and Dexter J. Booth (2013): Engineering Mathematics, Industrial Press, Inc.; 7th edition (March 8, 2013), Chapter: Second order differential equations, Further problems 26, p. 1025, Q. No. 7 (Example 1); Q. No. 14 (Example 2).

About Dr. Aspriha Peters

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