1.3: 1.3 Appendix- Reduction of Order and Complex Roots

In this section we provide some of the details leading to the general forms for the constant coefficient and Cauchy-Euler differential equations. In the first subsection we review how the Method of Reduction of Order is used to obtain the second linearly independent solutions for the case of one repeated root. In the second subsection we review how the complex solutions can be used to produce two linearly independent real solutions.

Method of Reduction of Order

First we consider constant coefficient equations. In the case when there is a repeated real root, one has only one independent solution, \(y_{1}(x)=e^{r x}\). The question is how does one obtain the second solution? Since the solutions are independent, we must have that the ratio \(y_{2}(x) / y_{1}(x)\) is not a constant. So, we guess the form \(y_{2}(x)=v(x) y_{1}(x)=v(x) e^{r x}\). For constant coefficient second order equations, we can write the equation as

\[(D-r)^{2} y=0 \nonumber \]

where \(D=\dfrac{d}{d x}\).
We now insert \(y_{2}(x)\) into this equation. First we compute

\[(D-r) v e^{r x}=v^{\prime} e^{r x} \nonumber \]

Then,

\[(D-r)^{2} v e^{r x}=(D-r) v^{\prime} e^{r x}=v^{\prime \prime} e^{r x} \nonumber \]

So, if \(y_{2}(x)\) is to be a solution to the differential equation, \((D-r)^{2} y_{2}=0\), then \(v^{\prime \prime}(x) e^{r x}=0\) for all \(x\). So, \(v^{\prime \prime}(x)=0\), which implies that

\[v(x)=a x+b \nonumber \]

So,

\[y_{2}(x)=(a x+b) e^{r x} \nonumber \]

Without loss of generality, we can take \(b=0\) and \(a=1\) to obtain the second linearly independent solution, \(y_{2}(x)=x e^{r x}\).

Deriving the solution for Case 2 for the Cauchy-Euler equations is messier, but works in the same way. First note that for the real root, \(r=r_{1}\), the characteristic equation has to factor as \(\left(r-r_{1}\right)^{2}=0\). Expanding, we have

\[r^{2}-2 r_{1} r+r_{1}^{2}=0 \nonumber \]

The general characteristic equation is

\[a r(r-1)+b r+c=0 \nonumber \]

Rewriting this, we have

\[r^{2}+\left(\dfrac{b}{a}-1\right) r+\dfrac{c}{a}=0 \nonumber \]

Comparing equations, we find

\[\dfrac{b}{a}=1-2 r_{1}, \quad \dfrac{c}{a}=r_{1}^{2} \nonumber \]

So, the general Cauchy-Euler equation in this case takes the form

\[x^{2} y^{\prime \prime}+\left(1-2 r_{1}\right) x y^{\prime}+r_{1}^{2} y=0 \nonumber \]

Now we seek the second linearly independent solution in the form \(y_{2}(x)= v(x) x^{r_{1}}\).We first list this function and its derivatives,

\[\begin{aligned}
y_{2}(x) &=v x^{r_{1}} \\
y_{2}^{\prime}(x) &=\left(x v^{\prime}+r_{1} v\right) x^{r_{1}-1} \\
y_{2}^{\prime \prime}(x) &=\left(x^{2} v^{\prime \prime}+2 r_{1} x v^{\prime}+r_{1}\left(r_{1}-1\right) v\right) x^{r_{1}-2} .
\end{aligned} \label{1.32} \]

Inserting these forms into the differential equation, we have

\[\begin{aligned}
0 &=x^{2} y^{\prime \prime}+\left(1-2 r_{1}\right) x y^{\prime}+r_{1}^{2} y \\
&=\left(x v^{\prime \prime}+v^{\prime}\right) x^{r_{1}+1}
\end{aligned} \label{1.33} \]

Thus, we need to solve the equation

\[x v^{\prime \prime}+v^{\prime}=0 \nonumber \]

or

\[\dfrac{v^{\prime \prime}}{v^{\prime}}=-\dfrac{1}{x} \nonumber \]

Integrating, we have

\[\ln \left|v^{\prime}\right|=-\ln |x|+C. \nonumber \]

Exponentiating, we have one last differential equation to solve,

\[v^{\prime}=\dfrac{A}{x} \nonumber \]

Thus,

\[v(x)=A \ln |x|+k \nonumber \]

So, we have found that the second linearly independent equation can be written as

\[y_{2}(x)=x^{r_{1}} \ln |x| \nonumber \]

Complex Roots

When one has complex roots in the solution of constant coefficient equations, one needs to look at the solutions

\[y_{1,2}(x)=e^{(\alpha \pm i \beta) x} \nonumber \]

We make use of Euler's formula

\[e^{i \beta x}=\cos \beta x+i \sin \beta x \label{1.34} \]

Then the linear combination of \(y_{1}(x)\) and \(y_{2}(x)\) becomes

\[\begin{aligned}
A e^{(\alpha+i \beta) x}+B e^{(\alpha-i \beta) x} &=e^{\alpha x}\left[A e^{i \beta x}+B e^{-i \beta x}\right] \\
&=e^{\alpha x}[(A+B) \cos \beta x+i(A-B) \sin \beta x] \\
& \equiv e^{\alpha x}\left(c_{1} \cos \beta x+c_{2} \sin \beta x\right)
\end{aligned} \label{1.35} \]

Thus, we see that we have a linear combination of two real, linearly independent solutions, \(e^{\alpha x} \cos \beta x\) and \(e^{\alpha x} \sin \beta x\).

When dealing with the Cauchy-Euler equations, we have solutions of the form \(y(x)=x^{\alpha+i \beta}\). The key to obtaining real solutions is to first recall that

\[x^{y}=e^{\ln x^{y}}=e^{y \ln x} \nonumber \]

Thus, a power can be written as an exponential and the solution can be written as

\[y(x)=x^{\alpha+i \beta}=x^{\alpha} e^{i \beta \ln x}, \quad x>0 \nonumber \]

We can now find two real, linearly independent solutions, \(x^{\alpha} \cos (\beta \ln |x|)\) and \(x^{\alpha} \sin (\beta \ln |x|)\) following the same steps as above for the constant coefficient case.