📂Odinary Differential Equations

Solutions to Euler's Differential Equations

Definition

The differential equation of the following form is called the Euler differential equation or Euler-Cauchy equation.

$$\begin{equation} a_{2}x^{2}\frac{ d ^{2 }y}{ dx^{2} }+a_{1}x\frac{ d y}{ d x }+a_{0}y=0 \end{equation}$$

Explanation

For a non-homogeneous equation where the right side is not $0$, it can be solved by substituting it with $x=e^{z}$.

Solution

For convenience of calculation, both sides of $(1)$ are divided by $a_{2}$, and let’s call the coefficients of the other two terms $a_{1}$ and $a_{0}$, respectively. Then

$$x^{2}\frac{ d ^{2 }y}{ dx^{2} } + a_{1}x\frac{ d y}{ d x } + a_{0}y = 0$$

By observing the differential equation, it becomes $0$ by differentiating twice and multiplying by the second term, differentiating once and multiplying by the first term, and adding the original function. Thus, the solution can be stated as follows.

$$y=x^{r}$$

Substituting into the differential equation results in

$$\begin{align*} r(r-1)x^{r}+a_{1}rx^{r}+a_{0}x^{r}=0 \\ [r(r-1)+a_{1}r+a_{0}]x^{r}=0 \\ [r^{2}-(a_{1}-1)r+a_{0}]x^{r}=0 \end{align*}$$

Since $x^{r}\ne0$, it is $r^{2}-(a_{1}-1)r+a_{0}=0$. This is a simple quadratic equation whose solutions are

$$r=\frac{-(a_{1}-1)\pm \sqrt{(a_{1}-1)^{2}-4a_{0}}}{2}$$

Let’s call the two solutions $r_{1}$ and $r_{2}$, respectively. Depending on the condition of the two roots, the solution to the differential equation changes.

• Case 1. $r_{1}$ and $r_{2}$ are distinct real numbers

  The two solutions are $y_{1}=x^{r_{1}}$ and $y_{2}=x^{r_{2}}$. By checking the Wronskian,

  $$W[y_{1},y_{2}]=(r_{2}-r_{1})x^{r_{1}+r_{2}-1}$$

  Since $r_{1}\ne r_{2}$, it is proven that when $x>0$, it must be $W[x^{1},r^{2}]\ne 0$. Thus, the two solutions form a fundamental set of solutions, so the general solution is

  $$y=c_{1}x^{r_{1}}+c_{2}x^{r_{2}},\quad x>0$$

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• Case 2. $r_{1}$ and $r_{2}$ are the same real numbers

• In this case, since $y_{1}$ and $y_{2}$, a second solution must be found. The first solution is described as $y_{1}=x^{r_{1}}$, and let’s say the differential operator $L$ is as follows.

  $$L[y]=x^{2}y^{\prime \prime}+xy^{\prime}+y$$

  Then

  $$L[x^{r}]=[r^{2}-(a_{1}-1)r+a_{0}]x^{r}=0$$

  In this case, since the quadratic equation for $r$ has a repeated root, it can be expressed in a perfect square form as follows.

  $$L[x^{r}]=(r-r_{1})^{2}x^{r}=0$$

  Differentiating $0$ gives $0$, so differentiating the left side by $r$ gives

  $$\frac{ \partial }{ \partial r}L[x^{r}]=0$$

  And since the order of differentiation for $x$ and $r$ can be swapped without issue,

  $$\frac{ \partial }{ \partial r }L[x^{r}]=L\left[ \frac{ \partial x^{r}}{ \partial r }\right]=L[x^{r}\ln x]=0$$

  Thus, $y_{2}=x^{r_{1}}\ln x$ is the second solution. Calculating the Wronskian gives $W[x^{r_{1}},x^{1}\ln x]=x^{2r_{1}-1}\ne 0$, so the two solutions form a fundamental set. Therefore, the general solution is

  $$y=c_{1}x^{r_{1}}+c_{2}x^{r_{1}}\ln x,\quad x>0$$

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• Case 3. $r_{1}$ and $r_{2}$ are distinct complex numbers

  Let’s call them $r_{1}=\lambda+i\mu$ and $r_{2}=\lambda -i\mu$. Then the two solutions are

  $$y_{1}=x^{\lambda+i\mu},\quad y_{2}=x^{\lambda-i\mu}$$

  Therefore, the fundamental solution is

  $$y=c_{1}x^{\lambda+i\mu}+c_{2}x^{\lambda-i\mu}$$

  However, in the case of complex functions, it’s usual to express them in trigonometric functions. By Euler’s formula, the following equation holds:

  $$\begin{align*} x^{\lambda +i \mu}=x^{\lambda}x^{i\mu}=x^{\lambda} e^{\ln x^{i\mu}} =&\ x^{\lambda}e^{i\mu \ln x} \\ =&\ x^{\lambda}[\cos(\mu \ln x)+i\sin (\mu \ln x) ] ,\quad x>0 \end{align*}$$

  Therefore, the general solution for complex constants $c_{1}$ and $c_{2}$ is expressed as

  $$y=c_{1}x^{\lambda}\cos (\mu \ln x)+c_{2}x^{\lambda}\sin(\mu \ln x),\quad x>0$$

  Since $\cos$ and $\sin$ are independent, it is understood that the Wronskian will definitely not be $0$ without calculating it. However, if one calculates,

  $$ W[x^{\lambda}\cos (\mu \ln x),x^{\lambda}\sin(\mu \ln x)]=\mu x^{2\lambda-1}\ne 0
  $$

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