Lesson 9 - Homogeneous Second Order Linear Equations

The standard form for a second order linear differential equation: $a{y}^{″}+b{y}^{\prime }+cy=f(t)$. Note that this is different from the homogeneous method of substitution from Lesson 8.

The DE is homogeneous if $f(t)=0$, and $a{y}^{″}+b{y}^{\prime }+cy=0$.

We can solve this equation by using what is called a characteristic equation. The characteristic equation resembles simple algebra factoring. We start by replacing our functions $y(x)$ with $m$. Each derivative becomes raised to the power of that level derivative.

Since $y$ is not a derivative, this is replaced with $m^0$, which is why we are left with only a constant $c$.

Once we have our characteristic equation set up, we can solve for $m$, either by factoring or by using the quadratic formula.

Recall the quadratic formula:

Once we solved for $m$, the general equation of a homogeneous equation is:

The superposition principle states that if $y_1(x)$ and $y_y(x)$ are solutions to a homogeneous equation, then $C_1{y}_1(x)+C_2{y}_2(x)$ is also a solution.

This is important to note. Due to the nature of solving for the roots of an equation, we oftentimes have more than one root. These can show in the form of real and distinct roots, repeated roots, or complex solutions. We will go over each of these scenarios next.

Consider the equation from before:

$y(x)=e^{mt}$ is a solution that leads us to the characteristic equation: $a{m}^2+bm+c=0$.

For real and distinct roots $m_1,m_2$, the general solution is:

For repeated roots, we must have solutions that are linearly independent, so we increase solutions by factors of the dependent variable.

Consider the roots $m,m$. The dependent variable in $y(x)$ is $x$, so the general solution becomes:

Multiplying one of our solutions guarantees that our solutions remain linearly independent of one another.

Occasionally, we end up with a complex root situation of the form $\alpha \pm i\beta$. In this scenario, we utilize Euler's Formula for $i\theta$.

We can derive a general formula from this using our knowledge of trig identities.

Thus, our general solution becomes:

For context, this equation often represents simple harmonic motion.

Examples

1. Find the general solution of: $2y^{″}-5y^{\prime }+3y=0$

We first convert to characteristic equation:

Now we can either use the quadratic formula or factor the equation. I will factor:

The roots for this equation can be solved by setting each binomial equal to $0$. We end up with $m=\frac{3}{2},1$. Thus, our general solution is as follows:

2. Find the general solution of: $y^{″}+4y^{\prime }+4y=0$.

Convert to characteristic equation and factor:

Our roots are $m=-2,-2$. The general solution is:

1. Find the general solution of: y''+y'+y=0

Using the general solution derived from Euler's formula, we get:

Let's talk more about linear independence, since we must have linearly independence for repeated root solutions.

Two functions $y_1$ and $y_2$ are said to be linearly dependent on an interval if and only if neither is a constant multiple of the other. If one can be expressed as a multiple of the other then the two functions are said to be linearly dependent on that interval.

If $y_1(x)$ and $y_2(x)$ are two linearly independent solutions of the homogeneous differential equation: $a{y}^{″}+b{y}^{\prime }+cy=0$, then every solution of the differential equation can be written as a linear combination $C_{1}y(x)+C_2(x)$.

We can check linear independence of solutions using the Wronskian. The Wronskian is the determinant of the two linearly independent solutions.

If $W(x)\ne 0$, then the solutions are linearly independent.

Try for yourself:

1. $4w^{″}+20w^{\prime }+25w-0$
2. $y^{″}+5y^{\prime }+6y=0$, $y(0)=1$, $y^{\prime }(0)=2$
3. $3y^{″}+11y^{\prime }-7y=0$

Next Lesson: Lesson 10 - Method of Undetermined Coefficients

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