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Chris L T521 · #5 May 22nd, 2008, 10:25 PM

Cauchy-Euler Equations:

A second order homogeneous Cauchy-Euler Equation takes the form:

$ax^2\frac{d^2y}{dx^2}+bx\frac{dy}{dx}+cy=0$ .

This time, we assume that a solution to the equation has the form $y=x^r$ . Substituting this in for y, we get the new DE:

$ax^2\left[r(r-1)x^{r-2}\right]+bx\left[rx^{r-1}\right]+cx^r=0$

Simplifying, we get:

$a(r^2-r)x^{r}+brx^{r}+cx^{r}=0$

Pulling out a common factor of $x^r$ , we get:

$\left(a(r^2-r)+br+c\right)x^{r}=0$

Assuming that $x^r\neq 0$ , we divide both sides by $x^r$ and get:

$ar^2-ar+br+c=0 \implies ar^2+(b-a)r+c=0$ .

This is the characteristic equation for the Cauchy-Euler Equation.

Again, just with the second order homogeneous DE with constant coefficients, there are three general cases:

1) Real distinct roots.
2) Real repeated roots.
3) Complex conjugate roots.

Case 1:

If we have two real and distinct roots $r=r_1 \ and \ r=r_2$ , then the solutions to the DE are $y_1=x^{r_1} \ and \ y_2=x^{r_2}$ . Thus the general solution would be:

$\color{red}\boxed{y=c_1x^{r_1}+c_2x^{r_2}}$

Example 13:

Solve $x^2\frac{d^2y}{dx^2}+7x\frac{dy}{dx}+5y=0$ .

Assuming a solution of $y=x^r$ , we get the characteristic equation:

$r^2+(7-1)r+5=0 \implies r^2+6r+5=0 \implies (r+1)(r+5)=0 \implies r=-1 \ or \ r=-5$ .

Therefore, $y_1=x^{-1} \ and \ y_2=x^{-5}$

$\color{red}\boxed{\therefore y=c_1x^{-1}+c_2x^{-5}}$

Case 2:

If we have real repeated roots, then $r=r_1=r_2$ . Thus, the solutions take the form $y_1=x^{r} \ and \ y_2=u(x)x^{r}$ where $u(x)=\int\frac{e^{-\int P(x)\,dx}}{y_{1}^{2}}\,dx$ . This is the case because $y_1 \ \text{and} \ y_2$ must be linearly independent solutions. To use the reduction of order formula, we need to manipulate the original DE:

$ax^2\frac{d^2y}{dx^2}+bx\frac{dy}{dx}+cy=0 \implies \frac{d^2y}{dx^2}+\frac{b}{ax}\frac{dy}{dx}+\frac{c}{ax^2}=0$ .

Now apply Reduction of order:

$u(x)=\int\frac{e^{\int \frac{b}{ax} \,dx}}{x^{2r}}\,dx$ where $r=\frac{-(b-a)}{2a}$ . Thus, we get:

$u(x)=\int\frac{e^{-\int \frac{b}{ax} \,dx}}{x^{2\frac{a-b}{2a}}}\,dx=\int\frac{x^{-\frac{b}{a}}}{x^{1}x^{-\frac{b}{a}}}\,dx=\int\frac{1}{x}\,dx=\color{red}\boxed{\ln x}$

Thus, the solutions are $y_1=x^r \ \text{and} \ y_2=x^r \ln x$ .

Therefore, the general solution is:

$\color{red}\boxed{y=c_1x^r+c_2x^r \ln x}$ .

Example 14:

Solve $x^2\frac{d^2y}{dx^2}-3x\frac{dy}{dx}+4y=0$ .

Assuming a solution of $y=x^r$ , we get:

$r^2+(-3-1)r+4=0 \implies r^2-4r+4=0 \implies (r-2)^2=0$
$\implies r=2\ w. \ \text{multiplicity 2}$

Thus, the solutions are $y_1=x^2 \ \text{and} \ y_2=x^2 \ln x$ .

Therefore, the general solution is:

$\color{red}\boxed{y=c_1x^2+c_2x^2 \ln x}$ .

Case 3:

Whenever we have complex conjugate roots $r = \alpha \pm \beta i$ , we see that our two solutions to the DE take on the form: $y_1 = x^{\alpha + \beta i} {\text{ and }}y_2 = x^{\alpha - \beta i}$ . However, it doesn't look pleasant with complex numbers in the solution. To take care of this, we will use Euler's Formula, which states that:

$e^{i\theta } = \cos \theta + i\sin \theta$

Since $x^r=e^{r\ln x}$ , we can say that

$\begin{gathered} y_1 = e^{r_1 \ln x} = e^{\left( {\alpha + \beta i} \right)\ln x} = e^{\alpha \ln x} e^{\left( {\beta \ln x} \right)i} \hfill \\ {\text{and }} \hfill \\ y_2 = e^{r_2 \ln x} = e^{\left( {\alpha - \beta i} \right)\ln x} = e^{\alpha \ln x} e^{ - \left( {\beta \ln x} \right)i}. \hfill \\ \end{gathered}$

Now applying Euler's Formula, we get:

$\begin{gathered} e^{\left( {\beta \ln x} \right)i} = \cos \left( {\beta \ln x} \right) + i\sin \left( {\beta \ln x} \right) \hfill \\ e^{ - \left( {\beta \ln x} \right)i} = \cos \left( { - \beta \ln x} \right) + i\sin \left( { - \beta \ln x} \right). \hfill \\ \end{gathered}$

Due to the even and odd properties of Cosine and Sine, we get that

$e^{ - \left( {\beta \ln x} \right)i} = \cos \left( {\beta \ln x} \right) - i\sin \left( {\beta \ln x} \right)$

Thus, the general solution in this case would be:

$y = c_1 y_1 + c_2 y_2 = x^{\alpha} \left[ {c_1 \left( {\cos \left( {\beta \ln x} \right) + i\sin \left( {\beta \ln x} \right)} \right) + c_2 \left( {\cos \left( {\beta \ln x} \right) - i\sin \left( {\beta \ln x} \right)} \right)} \right]$

$\Rightarrow y = x^{\alpha} \left[ {\left( {c_1 + c_2 } \right)\cos \left( {\beta \ln x} \right) + \left( {c_1 - c_2 } \right)i\sin \left( {\beta \ln x} \right)} \right]$

Letting $\left( {c_1 + c_2 } \right) = C_1$ and $\left( {c_1 - c_2 } \right)i = C_2$ , we have the general solution:

$\color{red}\boxed{y = x^{\alpha} \left[ {C_1 \cos \left( {\beta \ln x} \right) + C_2 \sin \left( {\beta \ln x} \right)} \right]}$

Example 15:

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

Assuming a solution of the form $y=x^r$ , we get the characteristic equation:

$r^2 + r + 1 = 0 \Rightarrow r = \frac{{ - 1 \pm \sqrt {1 - 4} }}{2} \Rightarrow r = - \frac{1}{2} \pm \frac{{\sqrt 3 }}{2}i$

$\color{red}\boxed{\therefore y = x^{ - \frac{1}{2}} \left[ {c_1 \cos \left( {\frac{{\sqrt 3 }}{2}\ln x} \right) + c_2 \sin \left( {\frac{{\sqrt 3 }}{2}\ln x} \right)} \right]}$

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