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Second Order Homogeneous Differential Equation – Forms and Examples
The second order homogenous differential equation is one of the first second order differential equations that you’ll learn in higher calculus. In the past, we’ve learned how to model word problems involving the first derivative of a function. To expand our ability in solving complex mathematical models, it is essential that we learn how to work with second order differential equations.
A second order homogenous differential equation is a major type of second order differential equation. These types of equations will have the highest degree of two and when all the terms are isolated on the left-hand side of the equation, the right-hand side is equal to zero.
In this article, we’ll establish the definition of second order homogenous differential equations and know the conditions we need to check before solving the equation. When working with second order homogenous linear differential equations, it is important that you know how to solve quadratic equations. Head over to our section for Algebra in case you need a refresher.
When you’re ready, let’s go ahead and dive right into the components of second order homogenous differential equations. By the end of the discussion, we hope that you’re more confident when working with these types of equations!
What Is a Second Order Homogenous Differential Equation?
The second order homogenous differential equation is one of the major types of second order differential equations that we’ll encounter and learn how to solve. Let’s explore the fundamental factors defining the second order homogenous differential equation.
- A differential equation in second order will have a differential term of at most second power.
- A second order differential equation is said to be homogenous when the terms are isolated on one side of the equation and the other side is equal to zero.
Combine this definition of second order homogenous differential equation, so it has a differential equation with a general form shown below.
\begin{aligned}y^{\prime \prime} + P(x)y^{\prime} + Q(x)y &= 0\\\dfrac{d^2y}{dx^2}+ P(x)\dfrac{dy}{dx} + Q(x)y &= 0 \end{aligned}
SECOND ORDER HOMOGENOUS DIFFERENTIAL EQUATION
Suppose that we have second order differential equation shown below. \begin{aligned}y^{\prime \prime} + P(x)y^{\prime} + Q(x)y &= f(x)\\\dfrac{d^2y}{dx^2}+ P(x)\dfrac{dy}{dx} + Q(x)y &= f(x) \end{aligned}
This second order equation is said to be homogenous when $f(x) = 0$. Consequently, when $f(x) \neq 0$, the second order differential equation is not a second order homogenous differential equation. |
One of the most common second order homogenous equation is the linear differential equation with a general form shown below.
\begin{aligned}ay^{\prime \prime} + by^{\prime}+ cy &= 0 \end{aligned}
For the homogenous linear differential equation, $a$, $b$, and $c$ must be constants, and $a$ must not be equal to zero. It’s clear to see that the latter form is simpler, so we’ll first work on second order homogenous linear differential equations and know how to find the solutions to these types of equations.
How To Solve Second Order Homogenous Linear Differential Equations?
We use an auxiliary equation when solving a second order homogenous linear differential equation. When a second order homogenous differential equation is linear, the highest exponent within the equation is the first power.
Since we’re working with second order homogenous differential equation, we expect its general solution to contain two arbitrary constants (for our discussion, we’ll label them as $C_1$ and $C_2$). Now, let’s first establish these two rules when working with second order homogenous linear differential equations:
- There exist two solutions for the differential equation. We can label them as $y_1$ and $y_2$ – we’ll use this notation throughout or discussion.
- The linear combination of these two solutions will also be a solution of the second order differential equation.
\begin{aligned}y(x) &= C_1 y_1 + C_2 y_2\end{aligned}
We’ll leave the proof for this in a later section to give you a chance to figure it out first on your own. The general solution, $y(x) = C_1 y_1 + C_2 y_2$, shows us that for $y_1$ and $y_2$ to be unique solutions, the two solutions must be linearly independent of each other.
USING AUXILIARY EQUATION TO SOLVE SECOND ORDER HOMOGENOUS LINEAR DIFFERENTIAL EQUATION
We can use the auxiliary equation to determine the general solution of the second order differential equation. We can think of $y^{\prime \prime}$, $y^{\prime}$, and $y$ as $r^2$, $r$, and the constant ($c$), respectively. \begin{aligned}ay^{\prime \prime} + &by^{\prime} + c = 0 \\&\downarrow\\ar^2 + &br + c = 0\end{aligned} The resulting quadratic equation will have two roots: $r_1$ and $r_2$. These roots will determine the general form of the differential equation’s general solution. |
As we have mentioned, the roots’ nature (or the discriminant’s sign, for that matter) will determine the form of the general solution that we’re looking for. We’ve summarized the conditions for you and use this table as a guide when working on our sample problems in the later section.
Roots’ Nature | Discriminant | Solution’s General Form |
When the roots are real and distinct. | \begin{aligned}b^2 -4ac > 0 \end{aligned} | \begin{aligned}y(x) &= C_1e^{r_1 x} + C_2e^{r_2 x} \end{aligned} |
When the two real roots are equal. \begin{aligned}r_1 = r_2 = r \end{aligned} | \begin{aligned}b^2 -4ac = 0 \end{aligned} | \begin{aligned}y(x) &= e^{rx} (C_1 + C_2 x) \end{aligned} |
When the resulting roots are complex. \begin{aligned}r_1 &= \alpha + \beta i\\ r_2 &= \alpha – \beta i\end{aligned} | \begin{aligned}b^2 -4ac < 0 \end{aligned} | \begin{aligned}y(x) &= e^{\alpha x} [C_1 \cos (\beta x) + C_2 \sin (\beta x)]\end{aligned} |
We now know the important components and factors when determining the general solution of second order homogenous linear differential equation. Before showing you an example, let’s break down the steps of finding the differential equation’s general solution:
- Write down the quadratic equation representing the second order linear differential equation’s auxiliary equation.
- Use algebraic techniques to know the nature and solve the roots of the differential equation.
- Based on the roots of the auxiliary equation, use the appropriate general form of the equation’s solution.
Let’s use these steps to solve the differential equation, $4y^{\prime \prime} + 6y^{\prime} – 4y = 0$, by first writing the auxiliary equation for the second order differential equation.
\begin{aligned}4y^{\prime \prime} + 6y^{\prime} – 4y &= 0 \rightarrow 4r^2 + 6r – 4 &= 0\end{aligned}
Solve the resulting quadratic equation to know the general form of our solution.
\begin{aligned} 4r^2 + 6r – 4 &= 0\\2r^2 + 3r – 2 &= 0\\ (2r -1)(r + 2) &= 0\\r_1 &= \dfrac{1}{2}\\r_1 &= -2\end{aligned}
These two roots are real and unique, so the general form of the solution is represented by the equation, $ y(x) = C_1e^{r_1 x} + C_2e^{r_2 x}$, where $C_1$ and $C_2$ are arbitrary constants. For our differential equation, $r_1 = \dfrac{1}{2}$ and $r_2 =- 2$.
\begin{aligned} y(x) &= C_1e^{1/2 \cdot x} + C_2e^{-2x}\\&= C_1e^{x/2} + C_2e^{-2x}\end{aligned}
This means that the second order differential equation has a general solution equal to $ y(x) = C_1e^{x/2} + C_2e^{-2x}$. Apply a similar process when working on the same types of equations. We’ve made sure that you try out more examples to master this topic, so head over to the section below when you’re ready!
Example 1
Determine whether the equations shown below are linear or nonlinear. When the equation is linear, determine whether it is homogenous or nonhomogeneous
a. $y^{\prime \prime} – 6x^3y^{\prime} + 4x^2y^2 = x^5$
b. $6y^{\prime \prime} + 2y = 4x^6$
c. $(\cos x) y^{\prime \prime} – (\sin x) y^{\prime} + 2y = 0$
Solution
Recall that for a second order differential equation to be linear, the highest exponent of the equation must be the first degree. Since the first equation, $y^{\prime \prime} – 6x^3y^{\prime} + 4x^2y^2 = x^5$, contains $y^2$ in its left-hand side, the differential equation is not linear.
a. $y^{\prime \prime} – 6x^3y^{\prime} + 4x^2y^2 = x^5$ is not linear.
Inspecting the second equation, we can see that the highest degree of $y$ is the first power, so it is indeed a linear differential equation. Now, looking at the right-hand side of the equation, $4x^6$, is a constant and not equal to zero, so it is nonhomogeneous.
b. $6y^{\prime \prime} + 2y = 4x^6$ is linear and nonhomogeneous.
Now, the third equation’s highest power (with respect to $y$) is also the first degree. This means that the differential equation is also linear. Looking at the right-hand side, we can see that it is equal to zero – satisfying the conditions for homogenous equations.
c. $(\cos x) y^{\prime \prime} – (\sin x) y^{\prime} + 2y = 0$ is linear and homogenous.
Example 2
Solve the second order differential equation, $\dfrac{d^2y}{dx^2} = 9y$.
Solution
Let’s first rewrite the equation so that it satisfies the definition of second order homogenous differential equation.
\begin{aligned}\dfrac{d^2y}{dx^2} &= 9y\\\dfrac{d^2y}{dx^2} -9y &= 0\\ y^{\prime \prime} – 9y &= 0\end{aligned}
Now that it is in the general form that we’ve established in our discussion earlier, let’s now find the auxiliary equation for the second order differential equation.
\begin{aligned} y^{\prime \prime} + 0y^{\prime} – 9y &= 0 \rightarrow r^2 – 9 &= 0\end{aligned}
Use the difference of two squares property to find the roots of the resulting quadratic equation.
\begin{aligned} r^2 – 9 &= 0\\(r – 3)(r + 3) &= 0\\r_1 &= 3\\r_2 &= -3\end{aligned}
Since the resulting roots are real and unique, the general solution will have the form, $ y(x) = C_1e^{r_1 x} + C_2e^{r_2 x}$, where $r_1 = 3$ and $r_2 = -3$. Hence, we have the general solution of the differential equation shown below.
\begin{aligned} y(x) &= C_1e^{3x} + C_2e^{-3x}\end{aligned}
Example 3
Solve the second order differential equation, $y^{\prime \prime} -4y^{\prime} +14y = 0$.
Solution
By inspection, we can see that the given equation is a second order homogenous linear differential equation. Let’s write the auxiliary equation associated with our equation by replacing $ y^{\prime \prime}$, $ y^{\prime}$, and $14y$ with $r^2$, $r$, and $14$, respectively.
\begin{aligned} y^{\prime \prime} -4y^{\prime} +14y &= 0\rightarrow r^2 – 4r+ 14 &= 0\end{aligned}
Using the coefficients of the quadratic equation, we can see that the discriminant is equal to $-40$. This means that the roots are complex and it’ll be best that we use the quadratic formula to solve for the equation’s roots.
\begin{aligned} r &= \dfrac{-(-4) \pm \sqrt{(-4)^2 – 4(1)(14)}}{2(1)}\\&= \dfrac{4 \pm \sqrt{16 – 56}}{2}\\&= \dfrac{4 \pm 2\sqrt{-10}}{2}\\\\r_1 &=2 – \sqrt{10}i\\r_2 &=2 + \sqrt{10}i\end{aligned}
Since we’re working with complex roots, we’ll use the general form, $y(x)= e^{\alpha x} [C_1 \cos (\beta x) + C_2 \sin (\beta x)]$, where $\alpha = 2$ and $\beta = \sqrt{10}$.
\begin{aligned} y(x) &= e^{\alpha x} [C_1 \cos (\beta x) + C_2 \sin (\beta x)]\\&= e^{2 x} [C_1 \cos (\sqrt{10} x) + C_2 \sin (\sqrt{10} x)]\end{aligned}
This means that the general solution to our equation is equal to $y(x) = e^{2 x} [C_1 \cos (\sqrt{10} x) + C_2 \sin (\sqrt{10} x)]$ or $y(x) = C_1 e^{2 x} \cos (\sqrt{10} x) + C_2 e^{2 x} \sin (\sqrt{10} x)$.
Example 4
Solve the initial value problem, $y^{\prime \prime} + 6y^{\prime} + 9y = 0$ with the following conditions:
\begin{aligned}y(0) &= 1\\ y^{\prime}(0) &= 2\end{aligned}
Solution
Our equation is already in the standard form for second order homogenous linear differential equations. We can proceed with writing the auxiliary equation using the coefficients of the differential equation.
\begin{aligned} y^{\prime \prime} + 6y^{\prime} + 9y &= 0 \rightarrow r^2 +6r +9&= 0\end{aligned}
The quadratic expression is a perfect square and we can rewrite it as $(r + 3)^2 =0$. This means that the first and second roots are the same and equal to $-3$. For these roots, the general solution will be equal to $y(x) = e^{rx} (C_1 + C_2 x)$, where $r =-3$.
\begin{aligned} y(x) &= e^{-3x} (C_1 + C_2 x)\end{aligned}
Now that we have the general solution, it’s time for us to use the initial conditions to find the particular solution. As we have learned in the past, we simply substitute the initial conditions into the equation to solve for the arbitrary constants’ values. We begin by using $y(0) = 1$ and solving for $C_1$.
\begin{aligned} y(0) &= e^{-3(0)} (C_1 + C_2 (0x)\\ y(0) &= C_1\\C_1 &= 1\\\\y(x) &= e^{-3x} (1 + C_2 x)\end{aligned}
We still have one more constant to work with and we find its value by finding the derivative of $y = e^{-3x} (1 + C_2 x)$ and use $y^{\prime}(0) = 2$.
\begin{aligned} y(x) &= e^{-3x} (1 + C_2 x)\\y^{\prime}(x) &= e^{-3x} [C_2(1- 3x) – 3]\\\\ y^{\prime}(0) &= e^{-3(0)}[C_2(1- 0) – 3]\\2 &= C_2 – 3\\C_2 &= 5 \end{aligned}
This means that our initial-value problem has a particular solution of $y(x) = e^{-3x} (1 + 5x)$.
Practice Questions
1. Determine whether the equations shown below are linear or nonlinear. When the equation is linear, determine whether it is homogenous or nonhomogeneous.
a. $y^{\prime \prime} + 12x^3y^{\prime} – 2x^2y^2 = x^4$
b. $2t^2x^{\prime \prime} + 6txx^{\prime} – 12x = 0$
c. $(\sin x) y^{\prime \prime} + 2 (\cos x) y^{\prime} – 6y = 0$
2. Solve the second order differential equation, $6y^{\prime \prime} + 11y^{\prime} – 35y = 0$.
3. Solve the second order differential equation, $\dfrac{d^2y}{dx^2} = 16y$.
4. Solve the second order differential equation, $y^{\prime \prime} – 5y^{\prime} + 25y = 0$.
5. Solve the initial value problem, $2y^{\prime \prime} + 8y^{\prime} + 10y = 0$ with the following conditions:
\begin{aligned}y(0) &= 0\\ y^{\prime}(0) &= 2\end{aligned}
Answer Key
1.
a. Equation is nonlinear.
b. Equation is nonlinear.
c. Equation is linear and homogenous.
2. $y(x) = C_1e^{5x/3} + C_2e^{-7x/2}$
3. $y(x) = C_1e^{4x} + C_2e^{-4x}$
4. $y(x) = e^{5x/2} \left[\sin \left(\dfrac{5\sqrt{3}x}{2}\right) + \cos\left(\dfrac{5\sqrt{3}x}{2}\right)\right]$
5. $y(x) = 2e^{-2x}\sin x$