# Solve second order linear differential equation

• I
• lriuui0x0

#### lriuui0x0

Consider the second order linear ODE with parameters ##a, b##:

$$xy'' + (b-x)y' - ay = 0$$

By considering the series solution ##y=\sum c_mx^m##, I have obtained two solutions of the following form:

\begin{aligned} y_1 &= M(x, a, b) \\ y_2 &= x^{1-b}M(x, a-b+1, 2-b) \\ \end{aligned}

Where ##M## is a function after solving the recurrence coefficients, please feel free to check if my result is correct:

$$M(x, a, b) = \sum_{m=0}^\infty \frac{\prod_{i=0}^{m-1}(a+i)}{m!\prod_{i=0}^{m-1}(b+i)}x^m$$

The question is if ##b=1##, we get repeated root and ##y_1 = y_2## hence linearly dependent. How to obtain a second independent solution? The problem has a hint of considering the following limit:

$$\lim_{b \to 1} \frac{y_2 - y_1}{b - 1}$$

I observe in case ##a=b## the solution, other than trivial y=const,. is ##y=Ce^x## which meets with your result. It shows any constant A could be multiplied. For general a, b by transforming
$$y=e^{z(x)}$$ to see differential equation of z, could you check/correct
$$y=Ce^{\frac{a}{b}x}$$
is a solution ?

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I observe in case ##a=b## the solution, other than trivial y=const,. is ##y=Ce^x## which meets with your result. It shows any constant A could be multiplied. For general a, b by transforming
$$y=e^{z(x)}$$ to see differential equation of z, could you check/correct
$$y=Ce^{\frac{a}{b}x}$$
is a solution ?
It's not a solution:

let ##y=e^{\frac{a}{b}x}##, then the LHS of the differential equation is:

$$\frac{a^2}{b^2}xe^{\frac{a}{b}x} + (b-x) \frac{a}{b}e^{\frac{a}{b}x} - ae^{\frac{a}{b}x} \ne 0$$

• anuttarasammyak
Thank you so much for correction. Following your way I got
$$c_{m+1}=\frac{a+m}{(m+1)(b+m)}c_m$$
for m=0,1,2,.. with ##c_0## given which is your ##y_1##. I do not find the way to get ##y_2##.

Consider the second order linear ODE with parameters ##a, b##:

$$xy'' + (b-x)y' - ay = 0$$

By considering the series solution ##y=\sum c_mx^m##, I have obtained two solutions of the following form:

\begin{aligned} y_1 &= M(x, a, b) \\ y_2 &= x^{1-b}M(x, a-b+1, 2-b) \\ \end{aligned}

Where ##M## is a function after solving the recurrence coefficients, please feel free to check if my result is correct:

$$M(x, a, b) = \sum_{m=0}^\infty \frac{\prod_{i=0}^{m-1}(a+i)}{m!\prod_{i=0}^{m-1}(b+i)}x^m$$

Applying Frobenius' Method and setting $y = \sum_{n=0}^\infty c_nx^{n+r}$ with $c_0 \neq 0$ I obtained the indicial equation $r(r - 1 + b) = 0$ and the recurrence relation $$c_{n+1} = \frac{n + r + a}{(n + r + 1)(n + r + b)}c_n$$ which are consistent with your result, althouh I would have used the property of the Gamma function that $\Gamma(z + 1) = z \Gamma(z)$ to evaluate the products.

(@anuttarasammyak: Just using an integer power series, which to be fair is what @lriuui0x0 said they did, won't pick up the $r = 1 - b$ solution.)

The question is if ##b=1##, we get repeated root and ##y_1 = y_2## hence linearly dependent. How to obtain a second independent solution? The problem has a hint of considering the following limit:

$$\lim_{b \to 1} \frac{y_2 - y_1}{b - 1}$$

$y_1$ and $(y_2 - y_1)/(b-1)$ are linearly independent for any $b \neq 1$; for $b = 1$ you can use l'Hopital's rule to calculate $$\lim_{b \to 1} \frac{x^{1-b}M(x,a-b+1, 2-b) - M(x,a,b)}{b-1} = - M(x,a,1)\log x - \left.\frac{\partial M}{\partial a}\right|_{(x,a,1)} - 2\left.\frac{\partial M}{\partial b}\right|_{(x,a,1)}$$ and show that the result is a solution of the ODE which is linearly independent of $M(x,a,1)$.

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• anuttarasammyak
Applying Frobenius' Method and setting $y = \sum_{n=0}^\infty c_nx^{n+r}$ with $c_0 \neq 0$ I obtained the indicial equation $r(r - 1 + b) = 0$ and the recurrence relation $$c_{n+1} = \frac{n + r + a}{(n + r + 1)(n + r + b)}c_n$$ which are consistent with your result, althouh I would have used the property of the Gamma function that $\Gamma(z + 1) = z \Gamma(z)$ to evaluate the products.

(@anuttarasammyak: Just using an integer power series, which to be fair is what @lriuui0x0 said they did, won't pick up the $r = 1 - b$ solution.)

$y_1$ and $(y_2 - y_1)/(b-1)$ are linearly independent for any $b \neq 1$; for $b = 1$ you can use l'Hopital's rule to calculate $$\lim_{b \to 1} \frac{x^{1-b}M(x,a-b+1, 2-b) - M(x,a,b)}{b-1} = - M(x,a,1)\log x - \left.\frac{\partial M}{\partial a}\right|_{(x,a,1)} - 2\left.\frac{\partial M}{\partial b}\right|_{(x,a,1)}$$ and show that the result is a solution of the ODE which is linearly independent of $M(x,a,1)$.
Just to double check on these steps. I got slightly different result compared with yours that doesn't involve partial derivative of ##a##. Does it look correct?

Expand the limit with l'Hopital's rule:

\begin{aligned} &\phantom{{}={}}\lim_{b\to 1}\frac{y_2-y_1}{b-1} \\ &= \biggl[\frac{\partial}{\partial b}(x^{1-b}M(x,a-b+1,2-b) - M(x,a,b))\biggr]_{b=1} \\ &= \biggl[-x^{1-b}\ln x M(x,a-b+1,2-b) + x^{1-b}\frac{\partial M(x,a-b+1,2-b)}{\partial b} - \frac{\partial M(x,a,b)}{\partial b}\biggr]_{b=1} \\ &= -M(x,a,1)\ln x + \biggl[\frac{\partial M(x,a-b+1,2-b)}{\partial b} - \frac{\partial M(x,a,b)}{\partial b}\biggr]_{b=1} \\ &= -M(x,a,1)\ln x + \biggl[\frac{\partial M(x,a,b)}{\partial b}(-2) - \frac{\partial M(x,a,b)}{\partial b}\biggr]_{b=1} \\ &= -M(x,a,1)\ln x - 3\frac{\partial M}{\partial b}(x,a,1) \\ \end{aligned}

Check the Wronskian of ##y_1 = M(x,a,1)## and ##y_2 = -M(x,a,1)\ln x - 3\frac{\partial M}{\partial b}(x,a,1)##:

\begin{aligned} &\phantom{{}={}} y_1y_2'-y_2y_1' \\ &= M(x,a,1)(-\frac{\partial M}{\partial x}(x,a,1)\ln x - M(x,a,1)\frac{1}{x} - 3\frac{\partial^2M}{\partial x\partial b}(x,a,1)) - \frac{\partial M}{\partial x}(x,a,1)(-M(x,a,1)\ln x - 3\frac{\partial M}{\partial b}(x,a,1)) \\ &= -M(x,a,1)^2 \frac{1}{x} -3M(x,a,1)\frac{\partial^2M}{\partial x\partial b}(x,a,1) + 3\frac{\partial M}{\partial x}(x,a,1)\frac{\partial M}{\partial b}(x,a,1) \end{aligned}

When ##x \to 0##, the Wronskian diverges because of the ##\frac{1}{x}## term, so it's non-zero. Hence ##y_1, y_2## are independent.

Just to double check on these steps. I got slightly different result compared with yours that doesn't involve partial derivative of ##a##. Does it look correct?

Expand the limit with l'Hopital's rule:

\begin{aligned} &\phantom{{}={}}\lim_{b\to 1}\frac{y_2-y_1}{b-1} \\ &= \biggl[\frac{\partial}{\partial b}(x^{1-b}M(x,a-b+1,2-b) - M(x,a,b))\biggr]_{b=1} \\ &= \biggl[-x^{1-b}\ln x M(x,a-b+1,2-b) + x^{1-b}\frac{\partial M(x,a-b+1,2-b)}{\partial b} - \frac{\partial M(x,a,b)}{\partial b}\biggr]_{b=1} \\ &= -M(x,a,1)\ln x + \biggl[\frac{\partial M(x,a-b+1,2-b)}{\partial b} - \frac{\partial M(x,a,b)}{\partial b}\biggr]_{b=1} \\ &= -M(x,a,1)\ln x + \biggl[\frac{\partial M(x,a,b)}{\partial b}(-2) - \frac{\partial M(x,a,b)}{\partial b}\biggr]_{b=1} \\ &= -M(x,a,1)\ln x - 3\frac{\partial M}{\partial b}(x,a,1) \\ \end{aligned}

You haven't checked that $-M(x,a,1) \ln x - 3\frac{\partial M}{\partial b}$ actually solves the ODE; it doesn't. If $L(y) = xy'' + (b-x)y' - ay$ then differentiating partially with respect to $b$ results in $$\frac{\partial}{\partial b}L(y) = L\left(\frac{\partial y}{\partial b}\right) + y'$$ and you can also show that $$L(y \ln x) = L(y) \ln x + 2y' - y + \frac{(b-1)y}{x}.$$ Since $L(M) = 0$ for all $a$ and $b$ you can see that $$L\left(M \ln x + 3 \frac{\partial M}{\partial b}\right) = -M' - M$$ when $b = 1$.

The problem is that you haven't calculated the derivative of $M(x,a-b+1, 2-b)$ with respect to $b$ correctly. By $\frac{\partial M}{\partial b}$ we mean the derivative of $M(x,a,b)$ with respect to $b$ with $x$ and $a$ constant. But you are looking at $f(x,a,b) = M(x, g(a,b),h(a,b))$ with $g(a,b) = a-b+1$ and $h(a,b) = 2 - b$ rather than $M(x,a,b)$. The multivariate chain rule then gives $$\frac{\partial f}{\partial b} = -\frac{\partial M}{\partial a} - \frac{\partial M}{\partial b},$$ and we must evaluate the derivatives of $M$ at $(x, a-b+1, 2-b)$ rather than at $(x,a,b)$. In the limit, of course, these are both $(x,a,1)$.

You haven't checked that $-M(x,a,1) \ln x - 3\frac{\partial M}{\partial b}$ actually solves the ODE; it doesn't. If $L(y) = xy'' + (b-x)y' - ay$ then differentiating partially with respect to $b$ results in $$\frac{\partial}{\partial b}L(y) = L\left(\frac{\partial y}{\partial b}\right) + y'$$ and you can also show that $$L(y \ln x) = L(y) \ln x + 2y' - y + \frac{(b-1)y}{x}.$$ Since $L(M) = 0$ for all $a$ and $b$ you can see that $$L\left(M \ln x + 3 \frac{\partial M}{\partial b}\right) = -M' - M$$ when $b = 1$.

The problem is that you haven't calculated the derivative of $M(x,a-b+1, 2-b)$ with respect to $b$ correctly. By $\frac{\partial M}{\partial b}$ we mean the derivative of $M(x,a,b)$ with respect to $b$ with $x$ and $a$ constant. But you are looking at $f(x,a,b) = M(x, g(a,b),h(a,b))$ with $g(a,b) = a-b+1$ and $h(a,b) = 2 - b$ rather than $M(x,a,b)$. The multivariate chain rule then gives $$\frac{\partial f}{\partial b} = -\frac{\partial M}{\partial a} - \frac{\partial M}{\partial b},$$ and we must evaluate the derivatives of $M$ at $(x, a-b+1, 2-b)$ rather than at $(x,a,b)$. In the limit, of course, these are both $(x,a,1)$.
Thanks @pasmith ! I found the error according to your correction, could you please check now? Especially if the argument on linear independence of the two solutions is valid?

Expand the limit with l'Hopital's rule:
\begin{aligned} &\phantom{{}={}}\lim_{b\to 1}\frac{y_2-y_1}{b-1} \\ &= \biggl[\frac{\partial}{\partial b}(x^{1-b}M(x,a-b+1,2-b) - M(x,a,b))\biggr]_{b=1} \\ &= \biggl[-x^{1-b}\ln x M(x,a-b+1,2-b) + x^{1-b}\frac{\partial M(x,a-b+1,2-b)}{\partial b} - \frac{\partial M(x,a,b)}{\partial b}\biggr]_{b=1} \\ &= -M(x,a,1)\ln x + \biggl[\frac{\partial M(x,a-b+1,2-b)}{\partial b} - \frac{\partial M(x,a,b)}{\partial b}\biggr]_{b=1} \\ &= -M(x,a,1)\ln x + \biggl[-\frac{\partial M(x,a,b)}{\partial a} -\frac{\partial M(x,a,b)}{\partial b} - \frac{\partial M(x,a,b)}{\partial b}\biggr]_{b=1} \\ &= -M(x,a,1)\ln x - \frac{\partial M}{\partial a}(x,a,1) - 2\frac{\partial M}{\partial b}(x,a,1) \\ \end{aligned}
Consider the differential operator ##L(y) = xy'' + (b-x)y' -ay##, then consider ##L(y\ln x)##, ##\frac{\partial}{\partial b}L(y)## and ##\frac{\partial}{\partial a}L(y)##:
\begin{aligned} &\phantom{{}={}} L(y\ln x) \\ &= x(y\ln x)'' + (b-x)(y\ln x)' - a(y\ln x) \\ &= xy''\ln x +2y' - y\frac{1}{x} + (b-x)y' \ln x + (b-x)y\frac{1}{x} - ay\ln x \\ &= \ln x(xy'' + (b-x)y' -ay) + 2y' -y + (b-1)y\frac{1}{x} \\ &= \ln x L(y) + 2y' -y + (b-1)y\frac{1}{x} \\ \end{aligned}

\begin{aligned} &\phantom{{}={}} \frac{\partial}{\partial b}L(y) \\ &= \frac{\partial}{\partial b}(xy'' + (b-x)y' - ay) \\ &= x\frac{\partial }{\partial b}y'' + y' +(b-x)\frac{\partial}{\partial b}y' - a\frac{\partial}{\partial b}y \\ &= x(\frac{\partial y}{\partial b})'' + (b-x) (\frac{\partial y}{\partial b})' - a\frac{\partial y}{\partial b} + y' \\ &= L(\frac{\partial y}{\partial b}) + y' \end{aligned}

\begin{aligned} &\phantom{{}={}} \frac{\partial}{\partial a}L(y) \\ &= \frac{\partial}{\partial a}(xy'' + (b-x)y' - ay) \\ &= x\frac{\partial }{\partial a}y'' +(b-x)\frac{\partial}{\partial b}y' - y -a\frac{\partial}{\partial a}y \\ &= x(\frac{\partial y}{\partial a})'' + (b-x) (\frac{\partial y}{\partial a})' - a\frac{\partial y}{\partial a} -y \\ &= L(\frac{\partial y}{\partial a}) - y \end{aligned}

Check ##y_2 = -M(x,a,1)\ln x - \frac{\partial M}{\partial a}(x,a,1) - 2\frac{\partial M}{\partial b}(x,a,1)## is indeed a solution to the differential equation:
\begin{aligned} &\phantom{{}={}} L(y_2) \\ &= -L(M \ln x) - L(\frac{\partial M}{\partial a}) - 2L(\frac{\partial M}{\partial b}) \\ &= -\ln xL(M) - 2M' +M -\frac{\partial}{\partial b}L(M) - M -2\frac{\partial}{\partial a}L(M) + 2M' \\ &= 0 \end{aligned}
Check the Wronskian of ##y_1 = M(x,a,1)## and ##y_2 = -M(x,a,1)\ln x - \frac{\partial M}{\partial a}(x,a,1) - 2\frac{\partial M}{\partial b}(x,a,1)##:
\begin{aligned} &\phantom{{}={}} y_1y_2'-y_2y_1' \\ &= M(-M'\ln x - M\frac{1}{x} - (\frac{\partial M}{\partial a})' -2(\frac{\partial M}{\partial b})') - M'(-M\ln x -\frac{\partial M}{\partial a} -2\frac{\partial M}{\partial b}) \\ &= -M^2 \frac{1}{x} -M(\frac{\partial M}{\partial a})' + M'\frac{\partial M}{\partial a} -2M(\frac{\partial M}{\partial a})' + 2M'\frac{\partial M}{\partial b} \end{aligned}
When ##x \to 0##, the Wronskian diverges because of the ##\frac{1}{x}## term, so it's non-zero. Hence ##y_1, y_2## are independent.

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