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### Homogeneous Equations

#### Discussion

As the results in the first section showed, solving the homogeneous equation plays a special role in solving the general problem. One way to solve such problems is to factor the equation. Consider the example $$ y''-y=0$$ We factor the equation to a pair of first order equations. Let $D=d/dx$. Then $D^2=DD=(d/dx)(d/dx)=d^2/dx^2$. Write the equation $y''-y=0$ as $$ \begin{align} (D^2 - 1)y &= 0 \\ (D-1)(D+1)y &= 0 \\ (D-1)y = 0\qquad &\text{OR}\qquad (D+1)y = 0 \\ y = Ce^x\qquad&\text{OR}\qquad y = Ke^{-x} \\ \text{General Solution:}\quad y &= Ce^x + Ke^{-x} \end{align} $$ where we use the fact from the previous section that the sum of solutions to a linear homogeneous equation is also a solution. We can then check this solution by computing $$ \eqalign {y&=Ce^x+Ke^{-x} \cr y'&=Ce^x-Ke^{-x} \cr y''&=Ce^x+Ke^{-x} \cr} $$ and so $y''-y=0$ as desired. So we have solved a linear constant coefficient homogeneous equation by treating it as a polynomial with a "variable" $D$. If the equation was not constant coefficient, the method would become much harder because we wouldn't be able to factor in the same fashion we are used to. This is because the algebra of differential operators is different from the usual algebra learned in high school. For example $$ \begin{align} (D+x)(D+1)y &= y'' + (x+1)y' + xy \\ (D+1)(D+x)y &= y'' + (x+1)y' + (x+1)y. \end{align} $$ So we see $(D+x)(D+1)$ is not equal to $(D+1)(D+x)$. In other words, the commutative law fails for differential operators. Differential Algebra is a very interesting subject but it quickly gets very difficult. That is why we restrict our attention here to the constant coefficient case. For the constant coefficient case everything works the same as in high school algebra and let's be thankful for it. There are three points to remember in solving these equations by factoring:

- The solution to $(D-r)y = 0$ is $y = Ce^{rx}$.
- The solution to $(D-r)^2 y = 0$ is $y = Cxe^{rx} + Ke^{rx}$.
- The solution to $(D-r)^n y = 0$ is $y=C_{n-1}x^{n-1}e^{rx}+C_{n-2}x^{n-2}e^{rx}+\cdots+C_0e^{rx}$.

#### Paradigm

Find the general solution of $y''+3y'-10y=0$
*Step 1:* Write in operator form (with $D$ for $d/dx$)
$$(D^2+3D-10)y=0$$
*Step 2:* Factor and find roots.

$(D-2)(D+5)y=0$ so the roots are $2$, $-5$.

*Step 3:* Write the general solution.
$$y(x)=Ce^{2x}+Ke^{-5x}$$

*EXAMPLE 1:*$$y'' + 4y' + 3y = 0 \qquad y(0)=1,\quad y'(0)=1$$ FIRST: Find the general solution.

Step 1: $(D^2 + 4D + 3)y = 0 $

Step 2: $ (D+3)(D+1)y = 0 $ so the roots are $-3$, $-1$

Step 3: $ y = Ce^{-3x} + Ke^{-x} $

SECOND: Plug in the initial values and solve for the constants. $$ \begin{align} y(0) &= Ce^{-3\times0} + Ke^{-1\times0} = C + K \buildrel{\text{set}}\over{=} 1 \\ y'(x) &= -3Ce^{-3x} - Ke^{-x} \\ y'(0) &= -3Ce^{-3\times0} - Ke^{-1\times0} = -3C - K \buildrel{\text{set}}\over{=} 1 \end{align} $$ Taking 3 times the first equation and adding it to the second we find $2K=4$ so $K=2$ and $C=-1$. This gives us the solution $$y=-e^{-3x}+2e^{-x}$$

*EXAMPLE 2:*$$ y'' + 10y' + 25y = 0$$ Step 1: $(D^2 + 10D + 25)y = 0$

Step 2:$ (D+5)^2 y = 0$, double root of $-5$

Step 3: $y = Cxe^{-5x} + Ke^{-5x}$

Now we consider the equation $y''+y=0$. If we try to solve this equation
using our new techniques, we obtain the operator $D^2+1$ which
has roots $\pm i$, so the general solution is $c_1e^{ix}+c_2e^{-ix}$.
This is in fact the correct general *complex* solution, provided
we allow $c_1$ and $c_2$ to be arbitrary complex constants.
But since
we started with a real problem we probably want a real answer. One
way to deal with that is to obtain the general complex solution as
above and find all the complex solutions which are actually real
solutions. The algebra involved in that is rather messy though.
Fortunately, there is a quicker way that will give us two linearly
independent real solutions from a single complex solution.

*Theorem:* If $e^{(\alpha+\beta i)x}$ is a solution of the equation
$ay''+by'+cy=0$ where $a$, $b$ and $c$ are all real, then
$e^{\alpha x}\cos(\beta x)$ and $e^{\alpha x}\sin(\beta x)$ are two
linearly independent solutions of $ay''+by'+cy=0$.

*Proof:* We are given that
$$
a(e^{(\alpha+\beta i)x})''+
b(e^{(\alpha+\beta i)x})'+c(e^{(\alpha+\beta i)x})=0
$$
Since $\Re[y'(x)]=(\Re[y(x)])'$ and $\Im[y'(x)]=(\Im[y(x)])'$,
it follows that
$$ \eqalign {
a(\Re[e^{(\alpha+\beta i)x}])''+
b(\Re[e^{\alpha+\beta i)x}])'+c(\Re[e^{(\alpha+\beta i)x}])&=0 \cr
a(\Im[e^{(\alpha+\beta i)x}])''+
b(\Im[e^{\alpha+\beta i)x}])'+c(\Im[e^{(\alpha+\beta i)x}])&=0 \cr}
$$
So $e^{\alpha x}\cos(\beta x)$ and $e^{\alpha x}\sin(\beta x)$ are
both real solutions to the given equation. Linear independence can be
quickly checked using the Wronskian.

Using this theorem we can now handle complex roots efficiently. One point that often bothers students at this stage is that we only need one complex solution to find the complete general solution. What happens to the other complex solution? It is a theorem that complex roots to a real polynomial always come in complex conjugate pairs, $\alpha\pm\beta i$. So since the operator has real coefficients, if $\alpha+\beta i$ is one root of the polynomial equation, $\alpha-\beta i$ must be another root of the equation, and $\alpha-\beta i$ will give rise to the same pair of real roots. So we really get our pair of real roots from a pair of complex roots.

Another point to be noted is that we can write the pair of real solutions we get from our complex conjugate pair of roots $\alpha\pm\beta i$ as either $C_1e^{\alpha x}\cos(\beta x) + C_2e^{\alpha x}\sin(\beta x)$ or as $Ae^{\alpha x}\cos(\beta x-\phi)$ (with arbitrary constants $A$ and $\phi$) using the techniques of section 3. The first form is usually the easiest for solving initial value problems while the second form is usually the easiest for sketching solutions. And, of course, in initial value problems where you have to sketch solutions, you may want to solve first in one form to solve the initial value problem and then switch to the second form to draw your sketch.

*EXAMPLE 3:* $$ y'' + 2y' + 2y = 0$$
Step 1: Write the equation in operator form
$$(D^2 + 2D + 2)y = 0$$
Step 2: Find the roots of the equation

roots are $-1\pm i$

Step 3: The general solution is $Ae^{-x}\cos(x-\phi)$ (or $C_1e^{-x}\cos(x)+C_2e^{-x}\sin(x)$ in the alternate form).

*EXAMPLE 4:* $$y''+y=0,\qquad y(0)=1,\quad y'(0)=1$$
FIRST: Find the general solution.

Step 1: $(D^2+1)y=0$

Step 2: roots are $\pm i$

Step 3: $y(x)=C\cos(x)+K\sin(x)$ (the easiest form for initial value problems).

SECOND: Plug in the initial values and solve for the constants $$ \begin{align} y(0)&=C \buildrel{\text{set}}\over{=} 1 \\ y'(x)&=-C\sin(x)+K\cos(x) \\ y'(0)&=K \buildrel{\text{set}}\over{=} 1 \end{align} $$ So the solution to the initial value problem is $y(x)=\cos(x)+\sin(x)$.

*EXAMPLE 5:* $$y'''+y=0$$
Step 1: $(D^3+1)y=0$

Step 2: $D^3+1=(D+1)(D^2-D+1)$ so the roots are $-1$ and $1/2\pm(\sqrt{3}/2)i$.

Step 3: The solution corresponding to the root $-1$ is $e^{-x}$. The solutions corresponding to the complex conjugate pair $1/2\pm\sqrt{3}/2$ are $Ae^{x/2}\cos(\sqrt{3}x/2-\phi)$. So the general solution is $$y(x)=C_1e^{-x}+Ae^{x/2}\cos(\sqrt{3}x/2-\phi).$$

*EXAMPLE 6:* $$y''''+8y'''+27y''+38y'+26y=0$$
Step 1: $(D^4+8D^3+27D^2+38D+26)y=0$

Step 2: $D^4+8D^3+27D^2+38D+26=(D^2+2D+2)(D^2+6D+13)=0$ so the roots are $-1\pm i$ and $-3\pm 2i$.

Step 3: $y=Ae^{-x}\cos(x-\phi)+Be^{-3x}\cos(2x-\psi)$, or written in the alternate form $y=C_1e^{-x}\cos(x)+C_2e^{-x}\sin(x)+C_3e^{-3x}\cos(2x)+ C_4e^{-3x}\sin(2x)$.

*EXAMPLE 7:* $$y''''+4y'''+24y''+40y'+100y=0$$
Step 1: $(D^4+4D^3+24D^2+40D+100)y=0$

Step 2: $(D^2+2D+10)^2=0$ so we have *double* roots of $-1\pm 3i$.

Step 3: We handled double real roots by using both the usual solution and the solution multiplied by $x$. We do exactly the same thing with double complex roots. So the solution is $y=Ae^{-x}\cos(3x-\phi)+Bxe^{-x}\cos(3x-\psi)$ or in the alternate form $y=C_1e^{-x}\cos(3x)+C_2e^{-x}\sin(3x)+C_3xe^{-x}\cos(3x)+C_4xe^{-x}\sin(3x)$.

You can generate additional examples of second-order constant-coefficient linear homogeneous initial value problems here.

If you have any problems with this page, please contact bennett@math.ksu.edu.

©2010, 2014 Andrew G. Bennett