Kansas State University

Discussion

Paradigm

The roots of $r^2+3r+2=0$ are $r=-1$ and $r=-2$. So two solutions are $x^{-1}$ and $x^{-2}$.

Step 3: If you have found two distinct real roots, $r_1$ and $r_2$, then the general solution is $y(x)=c_1x^{r_1}+c_2x^{r_2}$.

The general solution is therefore $y(x)=c_1x^{-1}+c_2x^{-2}$.

EXAMPLE: Find the general solution to $$ x^2y''+xy'-n^2y=0 $$ ($n$ is an integer constant. This equation arises in solving Laplace's Equation $\Delta u=\partial^2u/\partial x^2+\partial^2u/\partial y^2=0$ on the disk. This should seem reasonable if you recall that in polar coordinates $\Delta=\partial^2/\partial r^2 + (1/r)\partial/\partial r + (1/r^2)\partial/\partial\theta^2$.)

Step 1: $$ \eqalign { x^2y''+xy'-n^2y&=x^2r(r-1)x^{r-2}+xrx^{r-1}-n^2x^r \cr &=r^2x^r-n^2x^r \cr &=(r^2-n^2)x^r {\buildrel \text{set}\over =} 0 \cr } $$ Step 2: The roots of $r^2-n^2=0$ are $r=\pm n$. So the two solutions are $x^n$ and $x^{-n}$.

Step 3: The general solution is $y(x)=c_1x^n+c_2x^{-n}$.

You can check using the Wronskian that $x^{r_1}$ and $x^{r_2}$ are linearly independent if $r_1\ne r_2$. If the equation is of higher order than second, this paradigm still works, you just need $n$ distinct roots of the equation to get $n$ linearly independent functions to build the general solution of an $n^{th}$ order equation. Now what if the roots are not real and distinct? In the case where the roots are complex conjugates we can treat them as we have always treated complex roots. We find a complex solution and take its real and imaginary parts to obtain two real solutions. Of course, then we have to work with quantities like $x^i$. But this isn't as bad as it seems, we just use $x^i=e^{i\log x}$. (Actually, taking a logarithm in the complex plane can be a dangerous thing, but we won't worry about that now. You should take the introductory complex variables class sometime though and learn all about it.) But what if the roots are repeated? In that case we have no obvious guess for how to find a second root. There are several ways to deal with this. The way we will use is to make a change of variables to reduce the original equation to a constant coefficient equation. This technique will work on any Euler equation, not just those with repeated roots. But the first paradigm is more efficient when it works.

Paradigm (Take 2)

The crucial calculations here are how the derivatives of $y$ with respect to $x$ convert to derivatives of $y$ with respect to $z$. We first compute $$ \frac{dy}{dx}=\frac{dy}{dz}\frac{dz}{dx}=\frac1x\frac{dy}{dz} $$ where we have used the chain rule. So $xdy/dx=dy/dz$. Next we tackle the second derivative. Here we must remember we want to write the derivative with respect to $x$ of the derivative with respect to $x$ in terms of the derivative with respect to $z$ of the derivative with respect to $z$. $$ \eqalign { \frac{d^2y}{dx^2}&=\frac{d}{dx}\left(\frac{dy}{dx}\right) \cr &=\frac1x\frac{d}{dz}\left(\frac1x\frac{dy}{dz}\right) \cr &=\frac1x\left(\frac{d(1/x)}{dz}\frac{dy}{dz} +\frac1x\frac{d^2y}{dz^2}\right) \cr &=\frac1x\left(-\frac1x\frac{dy}{dz} +\frac1x\frac{d^2y}{dz^2}\right) \cr &=\frac{1}{x^2}\left(\frac{d^2y}{dz^2}-\frac{dy}{dz}\right) \cr } $$ where we used the product rule in the third line. So $x^2d^2y/dx^2= d^2y/dz^2-dy/dz$. Applying this rule, along with our computation for $dy/dx$ we find the equation becomes $$ \eqalign { &\frac{d^2y}{dz^2}-\frac{dy}{dz}+7\frac{dy}{dz}+9y \cr =\quad &\frac{d^2y}{dz^2}+6\frac{dy}{dz}+9y=0. \cr } $$ Step 2: Solve the transformed equation.

Since $D^2+6D+9=0$ has a double root of $-3$, the solution is $c_1e^{-3z}+c_2ze^{-3z}$.

Step 3: Back transform to find the answer in terms of the original variable. $$ \eqalign { c_1e^{-3z}+c_2ze^{-3z}&=c_1e^{-3\log x}+c_2(\log x)e^{-3\log x} \cr &=c_1x^{-3}+c_2(\log x)x^{-3} \cr } $$ The crucial step to note here is that while $xdy/dx=dy/dz$, $x^2d^2y/dx^2=d^2y/dz^2-dy/dz$. So the leading coefficient and the constant coefficient will remain unchanged when the equation is transformed, but the coefficient of $dy/dx$ will be changed because it picks up not only the $dy/dx$ term but also part of the $d^2y/dx^2$ term. The two easiest errors to make in solving Euler equations using the second paradigm is either to forget to change the middle coefficient in transforming the equation or to forget to back substitute to get the answer in terms of the original variable.

Of course we have only dealt with the homogeneous case in these paradigms. To handle the inhomogeneous equation, we can either use the second paradigm, make the change of variables on the right hand side as well and hope we can apply undetermined coefficients (which isn't all that likely), or we can just use variation of parameters.

EXAMPLE: $x^2y''+5xy'+3y=e^x$

We will use the first paradigm and variation of parameters.

Step 1: $y(x)=x^r$, so $xy'(x)=rx^r$ and $x^2y''(x)=r(r-1)x^r$. Plugging these into the equation we obtain $(r^2+4r+3)x^r=0$

Step 2: The roots of $r^2+4r+3=0$ are $r=-3$ and $r=-1$.

Step 3: Two linearly independent solutions of the homogeneous equation are $x^{-3}$ and $x^{-1}$.

Now that we have two linearly independent homogeneous solutions, we want to use the variation of parameters formula to find the general solution. First we note that the formula in our theorem on variation of parameters only applies to equations where the coefficient of the leading term is 1. So we divide through our Euler equation by $x^2$ to put it in the correct form, $$y''+5x^{-1}y'+3x^{-2}y=x^{-2}e^x.$$ Our linearly independent solutions of the homogeneous equation, as noted in step 3, are $y_1(x)=x^{-3}$ and $y_2(x)=x^{-1}$. We compute $W(x^{-3},x^{-1})(x)=-x^{-5}+3x^{-5}=2x^{-5}$. The right hand side is $g(x)=x^{-2}e^x$. From our formula for variation of parameters we then find the general solution is $$ \eqalign { y(x)&=-x^{-3}\int_0^x \frac{s^{-1}s^{-2}e^s}{2s^{-5}}\,ds+x^{-1}\int_0^x \frac{s^{-3}s^{-2}e^{s}}{2s^{-5}}\,ds+C_1x^{-3}+C_2x^{-1} \cr &=\frac{-x^{-3}}{2}\int_0^x s^2e^s\,ds +\frac{x^{-1}}{2}\int_0^x e^s\,ds+C_1x^{-3}+C_2x^{-1} \cr &=\frac{-x^{-3}}{2}(x^2e^x-2xe^x+2e^x-2) +\frac{x^{-1}}{2}(e^x-1)+C_1x^{-3}+C_2x^{-1} \cr &=\frac{-x^{-1}}{2}e^x+x^{-2}e^x-x^{-3}e^x+\frac{x^{-1}}{2}e^x+(C_1+1)x^{-3} +(C_2-1/2)x^{-1} \cr &=x^{-2}e^x-x^{-3}e^x+c_1x^{-3}+c_2x^{-1} \cr } $$ where $c_1=C_1+1$ and $c_2=C_2-1/2$.