Activities for Series Solutions

Section 4.7 Activities for Series Solutions

Subsection 4.7.1

Activity 4.7.1.

Solve this second order linear equation using a Taylor series centred at zero.

\begin{equation*} 4y^{\prime \prime} + t^2y = 0 \end{equation*}

Solution.

Zero is a ordinary point of the differential equation, so I can use a Taylor series. There are no singular points at all, so I expect solutions with infinite radius of convergence. First, I need the derivatives of an unknown series.

\begin{align*} y \amp = \sum_{n=0}^\infty c_n t^n \\ y^\prime \amp = \sum_{n=1}^\infty c_n n t^{n-1}\\ y^{\prime\prime} \amp = \sum_{n=2}^\infty c_n n(n-1) t^{n-2} \end{align*}

Then I put these expressions into the differential equation.

\begin{align*} 4y^{\prime \prime} + t^2 y \amp = 0 \\ 4\sum_{n=2}^\infty c_n n(n-1) t^{n-2} + \sum_{n=0}^\infty c_n t^{n+2} \amp = 0 \end{align*}

I shift the sums so that I have the same exponent for \(t\) in each sum.

\begin{equation*} 4\sum_{n=0}^\infty c_{n+2} (n+2)(n+1) t^n + \sum_{n=2}^\infty c_{n-2} t^{n} = 0 \end{equation*}

Then I pull out terms so that the starting indices are the same for each sum. This lets me combine the sums into one sum.

\begin{align*} 8 c_2 + 24 c_3t + 4\sum_{n=2}^\infty c_{n+2} (n+2)(n+1) t^n + \sum_{n=2}^\infty c_{n-2} t^{n} \amp = 0\\ 8 c_2 + 24 c_3t + 4\sum_{n=2}^\infty \left[ c_{n+2} (n+2)(n+1) + c_{n-2} \right] t^{n} \amp = 0 \end{align*}

First I look at the isolated terms. In this case, the constant and linear terms can only be zero if \(c_2 = 0\) and \(c_3 = 0\text{.}\) Then I look at the general terms and solve for the recurrence relation.

\begin{equation*} c_{n+2} (n+2)(n+1) + c_{n-2} = 0 \implies c_{n+2} = -\frac{c_{n-2}}{(n+2)(n+1)} \end{equation*}

With the recurrence relation, I start calculating coefficients. This is a degree four recurrence relation. However, since \(c_2 = 0 \) and \(c_3 = 0\) all coefficients of the form \(n = 4k + 2\) and \(n = 4k + 3\) will be zero. For the others, I leave \(c_0\) and \(c_1\) as unknown constant (to be determined by initial conditions, if those had been provided). I’m going to work with multiples of \(4\) first.

\begin{align*} c_4 \amp = \frac{-c_0}{(4)(3)} \\ c_8 \amp = \frac{c_0}{(8)(7)(4)(3)} \\ c_{12} \amp = \frac{-c_0}{(12)(11)(8)(7)(4)(3)} \end{align*}

I can intuit a general form for this set of coefficients.

\begin{equation*} c_{4k} = \frac{(-1)^k c_0}{(4k)(4k-1)(4k-4)(4k-5)\ldots(8)(7)(4)(3)} \end{equation*}

Now I’ll repeat the process for coefficients of the form \(n = 4k+1\text{.}\)

\begin{align*} c_5 \amp = \frac{-c_1}{(5)(4)} \\ c_9 \amp = \frac{c_1}{(9)(8)(5)(4)} \\ c_{13} \amp = \frac{-c_1}{(13)(12)(9)(8)(5)(4)} \end{align*}

From these calculations, I can intuit a general pattern for this set of coefficients.

\begin{equation*} c_{4k+1} = \frac{(-1)^k c_1}{(4k+1)(4k)(4k-3)(4k-4)\ldots(9)(8)(5)(4)} \end{equation*}

I put the two general pattern into the Taylor series form, each with the matching unknown constant. With the two unknowns \(c_0\) and \(c_1\text{,}\) I get a linear combination of two linearly independent solutions.

\begin{align*} y \amp = c_0 \sum_{k=0}^{\infty} \frac{(-1)^k t^{4k}}{(4k)(4k-1)(4k-4)(4k-5)\ldots(8)(7)(4)(3)}\\ \amp + c_1 \sum_{k=0}^{\infty} \frac{(-1)^k t^{4k+1}}{(4k+1)(4k)(4k-3)(4k-4)\ldots(9)(8)(5)(4)} \end{align*}

I expected solutions with an infinite radius of convergence. I could use one of the convergence tests to verify this if I wanted.

Activity 4.7.2.

Solve this second order linear equation using a Taylor series centred at zero.

\begin{equation*} 2y^{\prime \prime} + \frac{t}{2} y^{\prime} - y = 0 \end{equation*}

Solution.

Zero is a ordinary point of the differential equation, so I can use a Taylor series. There are no singular points at all, so I expect solutions which have infinite radius of convergence. First, I need the derivatives of an unknown series.

\begin{align*} y \amp = \sum_{n=0}^\infty c_n t^n \\ y^\prime \amp = \sum_{n=1}^\infty c_n n t^{n-1}\\ y^{\prime\prime} \amp = \sum_{n=2}^\infty c_n n(n-1) t^{n-2} \end{align*}

Then I put these expressions into the differential equation.

\begin{align*} 2y^{\prime \prime} + \frac{t}{2} y^\prime - y \amp = 0 \\ 2\sum_{n=2}^\infty c_n n(n-1) t^{n-2} + \frac{1}{2} \sum_{n=1}^\infty c_n nt^{n} - \sum_{n=0}^\infty c_n t^{n} \amp = 0 \end{align*}

I shift the sums so that I have the same exponent for \(t\) in each sum.

\begin{equation*} 2\sum_{n=0}^\infty c_{n+2} (n+2)(n+1) t^n + \frac{1}{2} \sum_{n=1}^\infty c_n nt^{n} - \sum_{n=0}^\infty c_n t^{n} = 0 \end{equation*}

Then I pull out terms so that the starting indices are the same for each sum. This lets me combine the sums into one sum.

\begin{align*} 4c_2 + c_0 + 2\sum_{n=1}^\infty c_{n+2} (n+2)(n+1) t^n + \frac{1}{2} \sum_{n=1}^\infty c_n nt^{n} - \sum_{n=1}^\infty c_n t^{n} \amp = 0 \\ 4c_2 + c_0 + \sum_{n=1}^\infty \left[ c_{n+2} 2(n+2)(n+1) + \frac{1}{2} c_n n - c_n \right] t^{n} \amp = 0\\ 4c_2 + c_0 + \sum_{n=1}^\infty \left[ c_{n+2} 2(n+2)(n+1) + \left( \frac{n}{2} - 1 \right) c_n \right] t^{n} \amp = 0 \end{align*}

First I look at the isolated terms. The constant term implies that \(c_2 = \frac{-c_0}{4}\text{,}\) where \(c_0\) is unknown. Then I look at the general terms and solve for the recurrence relation.

\begin{equation*} c_{n+2} = -\frac{\left( \frac{n}{2} - 1 \right) c_n}{2(n+2)(n+1)} = -\frac{(n-2) c_n}{4(n+2)(n+1)} \end{equation*}

This is a second order recurrence relation, so \(c_0\) and \(c_1\) being unknown is sufficient. I have already calculated \(c_2\text{,}\) so the recurrence relation will start calculating at \(c_3\text{.}\) Before I start the general calculations, I can notice something interesting about the recurrence relation. the numerator will vanish when \(n=2\text{;}\) on the left, that means that \(c_4 = 0\text{.}\) After that point, all even coefficients will also be zero by the recurrence relation. Therefore, the solution starting with \(c_0\) only has a constant and linear term. I alreacy know those coefficients: \(c_0\) is arbitrary and \(c_2 = \frac{-c_0}{4}\text{.}\) Therefore, I already have one of the solutions, which is a quadratic polynomial.

\begin{equation*} y_1 = c_0 \left( 1 - \frac{t^2}{4} \right) \end{equation*}

Now I can proceed with calculations. I have already dealt with all the even coefficients, so I’ll only calculate the odd.

\begin{align*} c_3 \amp = \frac{-c_1}{4(3)(3)} \\ c_5 \amp = \frac{-c_3}{4(5)(4)} = \frac{2c_1}{4^2\cdot 5!} \\ c_7 \amp = \frac{-c_5}{4(7)(6)} = \frac{-2c_1}{4^3\cdot 7!} \end{align*}

From these coefficients, I can intuit a general pattern.

\begin{equation*} c_{2k+1} = \frac{(-1)^k 2c_1}{4^k (2k+1)!} \end{equation*}

I put this general pattern into the Taylor series form to get the second solution.

\begin{equation*} y_2 = c_1 \sum_{k=0}^{\infty} \frac{(-1)^k t^{2k+1}}{4^k (2k+1)!} \end{equation*}

The general solution is the linear combination of the two, using the two unknown coefficients from the process.

\begin{equation*} y = c_0 \left( 1 - \frac{t^2}{4} \right) + c_1 \sum_{k=0}^{\infty} \frac{(-1)^k t^{2k+1}}{4^k (2k+1)!} \end{equation*}

I expected solutions with infinite radius of convergence. Obviously, the polynomial satisfies. I could use the various tests to check the other series as well, if I wished.

Activity 4.7.3.

Solve this second order linear equation using a Taylor series centred at zero. In this quesiton, only calculate the solution up to the \(t^6\) term, since it is prohibitively difficult to establish a pattern with this recurrence relation.

\begin{equation*} (t+1)y^{\prime \prime} + 4t^2y^{\prime} + 3y = 0 \end{equation*}

Solution.

Zero is a ordinary point of the differential equation, so I can use a Taylor series. There is a singular point at \(t = -1\text{,}\) so if I found a full Taylor series solution, I would only expect convergence with radius one. First, I need the derivatives of an unknown series.

\begin{align*} y \amp = \sum_{n=0}^\infty c_n t^n \\ y^\prime \amp = \sum_{n=1}^\infty c_n n t^{n-1}\\ y^{\prime\prime} \amp = \sum_{n=2}^\infty c_n n(n-1) t^{n-2} \end{align*}

Then I put these expressions into the differential equation.

\begin{align*} (t+1)y^{\prime \prime} + 4t^2y^\prime + 3y \amp = 0 \\ \sum_{n=2}^\infty c_n n(n-1) t^{n-1} + \sum_{n=2}^\infty c_n n(n-1) t^{n-2} \\ + 4 \sum_{n=1}^\infty c_n nt^{n+1} + 3\sum_{n=0}^\infty c_n t^{n} \amp = 0 \end{align*}

I shift the sums so that I have the same exponent for \(t\) in each sum.

\begin{gather*} \sum_{n=1}^\infty c_{n+1} (n+1)(n) t^n + \sum_{n=0}^\infty c_{n+2} (n+2)(n+1) t^n \\ + 4 \sum_{n=2}^\infty c_{n-1} (n-1)t^n + 3\sum_{n=0}^\infty c_n t^{n} = 0 \end{gather*}

Then I pull out terms so that the starting indices are the same for each sum. This lets me combine the sums into one sum.

\begin{align*} \amp 2c_2 t + 2c_2 + 6c_3t + 3c_0 + 3c_1t + \sum_{n=2}^\infty c_{n+1} (n+1)(n) t^n \\ \amp + \sum_{n=2}^\infty c_{n+2} (n+2)(n+1) t^n + 4 \sum_{n=2}^\infty c_{n-1} (n-1)t^n + 3\sum_{n=2}^\infty c_n t^{n} = 0 \\ \amp (2c_2 + 3c_0) + (2c_2 + 6 c_3 + 3c_1) t \\ \amp + \sum_{n=2}^\infty \left[ c_{n+1} (n+1)(n) + c_{n+2} (n+2)(n+1) + 4 c_{n-1} (n-1) + 3c_n \right] t^{n} = 0\\ \amp = (2c_2 + 3c_0) + (2c_2 + 6 c_3 + 3c_1) t \\ \amp + \sum_{n=2}^\infty \left[ c_{n+1} (n+1)(n) + c_{n+2} (n+2)(n+1) + 4 c_{n-1} (n-1) + 3c_n \right] t^{n} = 0 \end{align*}

First I look at the isolated terms. Letting \(c_0\) and \(c_1\) be arbitrary, the constant term determined \(c_2\text{.}\)

\begin{equation*} c_2 = \frac{-3c_0}{2} \end{equation*}

Then the linear term similarly determines \(c_3\text{.}\)

\begin{equation*} c_3 = \frac{-3c_1 - 2c_2}{6} = \frac{-1}{2} c_1 - \frac{1}{3} \frac{-3c_0}{2} = \frac{-c_1}{2} + \frac{c_0}{2} \end{equation*}

Now the complexity begins to show. This involves both of the unknowns, so the series is unlikely to split nicely into two pieces, on associated with each of the original unknowns. This intermingling of the two unknowns will continue as I start calculating the coefficients. First, I need to determine the recurrence relation from the general term.

\begin{equation*} c_{n+2} = \frac{-1}{(n+2)(n+1)} \left( (n+1)(n) c_{n+1} + 3c_n + 4(n-1)c_{n-1} \right) \end{equation*}

Now I use this to calculate. As requested, I only calculate up to \(c_6\text{.}\) This is already complicated and miserable enough.

\begin{align*} c_4 \amp = \frac{-1}{12} \left( 6c_3 + 3c_2 + 4c_1 \right) \\ \amp = \frac{-1}{12} \left( -3c_1 + 3c_0 + 3\frac{-3c_0}{2} + 4c_1 \right) \\ \amp = \frac{-1}{12} \left( c_1 - \frac{3}{2} c_0 \right) = \frac{-c_1}{12} + \frac{c_0}{8} \\ c_5 \amp = \frac{-1}{20} \left( 12 c_4 + 3c_3 + 8 c_2 \right) \\ \amp = \frac{-1}{20} \left( 12 \left( \frac{-c_1}{12} + \frac{c_0}{8} \right) + 3 \left( \frac{-c_1}{2} + \frac{c_0}{2} \right) + 8 \frac{-3c_0}{2} \right) \\ \amp = \frac{-1}{20} \left( -c_1 + \frac{3c_0}{2} - \frac{3c_1}{2} + \frac{3c_0}{2} - 12 c_0 \right)\\ \amp = \frac{-1}{20} \left( \frac{-5c_1}{2} - 9 c_0 \right) = \frac{c_1}{8} + \frac{9c_0}{20} \\ c_6 \amp = \frac{-1}{30} \left( 20 c_5 + 3c_4 + 12 c_3 \right)\\ \amp = \frac{-1}{30} \left( 20 \left( \frac{c_1}{8} + \frac{9c_0}{20} \right) + 3 \left( \frac{-c_1}{12} + \frac{c_0}{8} \right) + 12 \left( \frac{-c_1}{2} + \frac{c_0}{2} \right) \right)\\ \amp = \frac{-1}{30} \left( \frac{5c_1}{2} + 9c_0 - 4c_1 + \frac{3c_0}{8} - 6c_1 + 6c_0 \right) \\ \amp = \frac{-1}{30} \left( \frac{-15c_1}{2} + \frac{123c_0}{8} \right) = \frac{c_1}{4} - \frac{41c_0}{80} \end{align*}

There is, of course, no hope of generating an expression for a pattern simply by inspection. However, I can group all the terms for each unknown together into two pieces.

\begin{align*} y_1 \amp = c_0 \left( 1 - \frac{3}{2} t^2 + \frac{1}{2}t^3 + \frac{1}{8} t^4 + \frac{9}{20} t^5 - \frac{41}{80} t^6 + \ldots \right) \\ y_2 \amp = c_1 \left( t - \frac{1}{t} t^3 - \frac{1}{12} t^4 + \frac{1}{8} t^5 + \frac{1}{5} t^6 + \ldots \right) \end{align*}

The general solution is a linear combination of these two solutions.

\begin{align*} y \amp = c_0 \left( 1 - \frac{3}{2} t^2 + \frac{1}{2}t^3 + \frac{1}{8} t^4 + \frac{9}{20} t^5 - \frac{41}{80} t^6 + \ldots \right) \\ \amp + c_1 \left( t - \frac{1}{t} t^3 - \frac{1}{12} t^4 + \frac{1}{8} t^5 + \frac{1}{5} t^6 + \ldots \right) \end{align*}

These are approximate solutions: Taylor polynomials. However, Taylor polynomials are often pretty good approximations for functions, so this may be enough for some use of these functions in application.

Subsection 4.7.2 Method of Frobenius

Activity 4.7.4.

Solve this second order linear equation using the Method of Frobenius at the regular singular point \(t = 0\text{.}\)

\begin{equation*} t^2y^{\prime \prime} + ty^\prime - y = 0 \end{equation*}

Solution.

First let me write the coefficient fuctions (which I get after clearing the coefficient of \(y^{\prime \prime}\)) and use the coefficient functions to determine \(p_{-1}\) and \(q_{-2}\text{.}\) In this case, since the functions are simply rational functions with poles, I can read the coefficients off directly.

\begin{align*} P(t) \amp = \frac{1}{t} \\ Q(t) \amp = \frac{-1}{t^2} \\ p_{-1} \amp = 1\\ q_{-2} \amp = -1 \end{align*}

Then I’ll write the indicial equation.

\begin{equation*} r(r-1) + p_{-1}r + q_{-2} = r(r-1) + r - 1 = r^2 - r + r - 1 = r^2 - 1 \end{equation*}

The roots of the inidial equation are \(r = \pm 1\) Since one root is a positive integer and one is not, I’ll work separately from the start. I’ll start with the positive root. First I need the derivatives of the series.

\begin{align*} y \amp = t \sum_{n=0}^\infty c_n t^n = \sum_{n=0}^\infty c_n t^{n+1} \\ y^\prime \amp = \sum_{n=0}^\infty c_n (n+1) t^n \\ y^{\prime\prime} \amp = \sum_{n=1}^\infty c_n (n+1)(n) t^{n-1} \end{align*}

Then I put these into the differential equation. I follow the standard steps of matching exponents by shifting, pulling out terms to match starting indices, and grouping the results.

\begin{align*} t^2y^{\prime \prime} + ty^\prime - y \amp = 0 \\ t^2 \sum_{n=1}^\infty c_n (n+1)(n) t^{n-1} + t \sum_{n=0}^\infty c_n (n+1) t^n - \sum_{n=0}^\infty c_n t^{n+1} \amp = 0 \\ \sum_{n=1}^\infty c_n (n+1)(n) t^{n+1} + \sum_{n=0}^\infty c_n (n+1) t^{n+1} - \sum_{n=0}^\infty c_n t^{n+1} \amp = 0 \\ c_0 t - c_0 t + \sum_{n=1}^\infty c_n (n+1)(n) t^{n+1} + \sum_{n=1}^\infty c_n (n+1) t^{n+1} - \sum_{n=1}^\infty c_n t^{n+1}\\ \sum_{n=1}^\infty c_n \left[ (n+1)(n) + (n+1) - 1 \right] t^{n+1} \amp = 0 \\ \sum_{n=1}^\infty c_n \left[ n^2 + 2n \right] t^{n+1} = 0 \end{align*}

Here \(c_0\) is arbitrary, since the linear term cancels and the equation puts no restriction on \(c_0\text{.}\) However, for all the other terms, the ‘recurrence relation’ simply translates into a simple equation.

\begin{equation*} c_n = 0 \end{equation*}

This is true for all \(n \geq 1\text{,}\) so only \(c_0\) remains. The first solution is the solution with that coefficient. (Since \(r = 1\text{,}\) this is no actually a constant.)

\begin{equation*} y_1 = t (c_0) = c_0 t \end{equation*}

Now I’ll move on to the negative root. Again, I need to start with the derivatives of series. I have to be quite careful with this derivative to make sure that constant terms go to zero and are removed from the sum.

\begin{align*} y \amp = \frac{1}{t} \sum_{n=0}^\infty c_n t^n = \sum_{n=0}^\infty c_n t^{n-1} = \frac{c_0}{t} + c_1 + \sum_{n=2}^\infty c_n t^{n-1} \\ y^\prime \amp = \frac{-c_0}{t^2} + \sum_{n=2}^\infty c_n (n-1) t^{n-2} \\ y^{\prime\prime} \amp = \frac{2c_0}{t^3} + \sum_{n=3}^\infty c_n (n-1)(n-2) t^{n-3} \end{align*}

Then I put these into the differential equation. After, I follow the standard steps of matching exponents, matching starting indices, and grouping.

\begin{align*} t^2y^{\prime \prime} + ty^\prime - y \amp = 0 \\ t^2 \left[ \frac{2c_0}{t^3} + \sum_{n=3}^\infty c_n (n-1)(n-2) t^{n-3} \right] + t \left[ \frac{-c_0}{t^2} + \sum_{n=2}^\infty c_n (n-1) t^{n-2} \right] \amp \\ - \left[ \frac{c_0}{t} + c_1 + \sum_{n=2}^\infty c_n t^{n-1} \right] \amp = 0 \\ \left[ \frac{2c_0}{t} + \frac{-c_0}{t} - \frac{c_0}{t} - c_ 1 \right] + \sum_{n=3}^\infty c_n (n-1)(n-2) t^{n-1}\\ + \sum_{n=2}^\infty c_n (n-1) t^{n-1} - \sum_{n=2}^\infty c_n t^{n-1} \amp = 0 \\ -c_1 + + c_2 t - c_2 t + \sum_{n=3}^\infty c_n \left[ (n-1)(n-2) + (n-1) - 1 \right] t^{n-1} \amp = 0 \\ -c_1 + \sum_{n=3}^\infty c_n \left[ n^2 -2n -1 \right] t^{n-1} \amp = 0 \end{align*}

From this, I conclude that \(c_0\) and \(c_2\) are arbitrary, \(c_1\) must be zero, and \(c_n = 0\) for all \(n \geq 3\text{.}\) Again, this is a polynomial solution, but this time, multiplied by \(\frac{1}{t}\text{.}\)

\begin{equation*} y_2 = \frac{1}{t} \left( c_0 + c_2 t^2 \right) = \frac{c_0}{t} + c_2 t \end{equation*}

This solution has two free paramaters, which is strange. However, I can observe that the second piece, \(c_2t\text{,}\) is the same as \(y_1\text{.}\) The linearly independent part of \(y_2\) consists only of the first piece.

\begin{equation*} y_2 = \frac{c_0}{t} \end{equation*}

I found two linearly independent solution. The complete family of solutions comes from putting these two together.

\begin{equation*} y = A \frac{1}{t} + B t \end{equation*}

Of interest, in this example, both solutions had series which truncated instead of continuing with infinitly many terms. This can happen. It’s also an examples where both solutions are relatively simple non-series functions; again this can happen.

Activity 4.7.5.

Solve this second order linear equation using the Method of Frobenius at the regular singular point \(t = 0\text{.}\)

\begin{equation*} t y^{\prime \prime} + 4y^\prime + ty = 0 \end{equation*}

Solution.

First let me write the coefficient fuctions (which I get after clearing the coefficient of \(y^{\prime \prime}\)) and use the coefficient functions to determine \(p_{-1}\) and \(q_{-2}\text{.}\)

\begin{align*} P(t) \amp = \frac{4}{t} \\ Q(t) \amp = 1 \\ p_{-1} \amp = 4\\ q_{-2} \amp = 0 \end{align*}

Then I’ll write the indicial equation.

\begin{equation*} r(r-1) + p_{-1}r + q_{-2} = r(r-1) + 4r = r^2 - r + 4r = r(r+3) \end{equation*}

The roots of the inidial equation are Since one root is a positive integer and one is not, I’ll work separately from the start. I’ll start with the positive root. First I need the derivatives of the series.

\begin{align*} y \amp = \sum_{n=0}^\infty c_n t^n \\ y^\prime \amp = \sum_{n=1}^\infty c_n (n) t^{n-1} \\ y^{\prime\prime} \amp = \sum_{n=2}^\infty c_n (n)(n-1) t^{n-2} \end{align*}

Then I put these into the differential equation and follow the standard procedure of matching exponents, matching starting indices and grouping terms.

\begin{align*} ty^{\prime \prime} + 4y^\prime + ty \amp = 0 \\ t \sum_{n=2}^\infty c_n (n-1)(n) t^{n-2} + 4 \sum_{n=1}^\infty c_n (n) t^{n-1} + t \sum_{n=0}^\infty c_n t^n \amp = 0 \\ \sum_{n=2}^\infty c_n (n-1)(n) t^{n-1} + \sum_{n=1}^\infty c_n (4n) t^{n-1} + \sum_{n=0}^\infty c_n t^{n+1} \amp = 0\\ \sum_{n=2}^\infty c_n (n-1)(n) t^{n-1} + \sum_{n=1}^\infty c_n (4n) t^{n-1} + \sum_{n=2}^\infty c_{n-2} t^{n-1} \amp = 0\\ 4c_1 + \sum_{n=2}^\infty \left[ c_n \left( (n-1)(n) + (4n) \right) + c_{n-2} \right] t^{n-1} \amp = 0 \end{align*}

At the start here, \(c_0\) is arbitrary, \(c_1=0\) from the constant term, and there is a second-order recurrence relation.

\begin{equation*} c_n = \frac{-c_{n-2}}{(n-1)(n) + 4n} = \frac{-c_{n-2}}{n^2 + 3n} = \frac{-c_{n-2}}{n(n+3)} \end{equation*}

Since \(c_1 = 0\text{,}\) all the odd \(c_n\) will be zero from the recurrence relation. I calculate some of the even terms to find the pattern.

\begin{align*} c_2 \amp = \frac{-c_0}{(2)(5)} \\ c_4 \amp = \frac{-c_2}{(4)(7)} = \frac{c_0}{(2)(4)(5)(7)} \\ c_6 \amp = \frac{-c_4}{(6)(9)} = \frac{-c_0}{(2)(4)(6)(5)(7)(9)} \\ c_8 \amp = \frac{-c_6}{(8)(11)} = \frac{c_0}{(2)(4)(6)(8)(5)(7)(9)(11)} \end{align*}

From this, I intuit a general form for the even coefficients.

\begin{equation*} c_{2k} = \frac{(-1)^k c_0}{(2)(4)(6) \ldots (2k)(5)(7)(9)(11) \ldots (2k+3)} \end{equation*}

I can do some factorial tricks to write this in terms of factorials. For the even numbers multiplied together, I can factor out \(2\) from each term (there are \(k\) terms). For the odd number multiplies together, I can include the missing even terms in the numerator and denominator to complete the factorial.

\begin{align*} c_{2k} \amp = (-1)^k c_0 \frac{1}{2^k(1)(2)(3)(4)\ldots(k)} \frac{(2)(3)(4)(5)(6)(8)(10)(12)(14) \ldots (2k+2)}{(2)(3)(4)(5)(6)\ldots (2k+3)} \\ \amp = (-1)^k c_0 \frac{1}{2^k k!} \frac{(3)(5) \left[ (2)(4)(6)(8) \ldots (2k+2) \right] }{(2k+3)!} \\ \amp = (-1)^k \frac{1}{2^k k!} \frac{(15)(2^{k+2})(k+1)!}{(2k+3)!} c_0 = \frac{(-1)^k (15) 2^{k+2} (k+1)!}{2^k (k!)(2k+3)!} c_0\\ \amp = \frac{(-1)^k (15) (4) (k+1)}{(2k+3)!} c_0 = \frac{(-1)^k (60)(k+1)}{(2k+3)!} c_0 \end{align*}

Now I can write this first solution for \(r=0\text{.}\)

\begin{equation*} y_1 = c_0 \left[ 1 - \frac{t^2}{10} + \sum_{k=2}^{\infty} \frac{(-1)^k (60)(k+1)t^{2k}}{(2k+3)!} \right] \end{equation*}

Now I’ll move on to the negative root. Again, I need to start with the derivatives of series.

\begin{align*} y \amp = \frac{c_0}{t^3} + \frac{c_1}{t^2} + \frac{c_2}{t} + \sum_{n=0}^\infty c_{n+3} t^n \\ y^\prime \amp = \frac{-3c_0}{t^4} + \frac{-2c_1}{t^3} + \frac{-c_2}{t^2} + \sum_{n=1}^\infty c_{n+3} (n) t^{n-1}\\ y^{\prime \prime} \amp = \frac{12c_0}{t^5} + \frac{6c_1}{t^4} + \frac{2c_2}{t^3} + \sum_{n=2}^\infty c_{n+3} (n)(n-1) t^{n-2} \end{align*}

Then I put these into the differential equation and follow the standard procedure of matching exponents, matching starting indices, and grouping the result.

\begin{align*} t \left[ \frac{12c_0}{t^5} + \frac{6c_1}{t^4} + \frac{2c_2}{t^3} + \sum_{n=2}^\infty c_{n+3} (n)(n-1) t^{n-2} \right] \amp \\ + 4 \left[ \frac{-3c_0}{t^4} + \frac{-2c_1}{t^3} + \frac{-c_2}{t^2} + \sum_{n=1}^\infty c_{n+3} (n) t^{n-1} \right] \amp\\ + t \left[ \frac{c_0}{t^3} + \frac{c_1}{t^2} + \frac{c_2}{t} + \sum_{n=0}^\infty c_{n+3} t^n \right] \amp = 0\\ \frac{12c_0}{t^4} + \frac{6c_1}{t^3} + \frac{2c_2}{t^2} + \sum_{n=2}^\infty c_{n+3} (n)(n-1) t^{n-1} \amp\\ + \frac{-12c_0}{t^4} + \frac{-8c_1}{t^3} + \frac{-4c_2}{t^2} + \sum_{n=1}^\infty 4c_{n+3} (n) t^{n-1} \amp\\ + \frac{c_0}{t^2} + \frac{c_1}{t} + c_2 + \sum_{n=0}^\infty c_{n+3} t^{n+1} \amp = 0\\ \frac{-2c_1}{t^3} + \frac{c_0-2c_2}{t^2} + \frac{c_1}{t} + c_2 + \sum_{n=1}^\infty c_{n+4} (n+1)(n) t^n \amp\\ + \sum_{n=0}^\infty 4c_{n+4} (n+1) t^n + \sum_{n=1}^\infty c_{n+2} t^n \amp = 0\\ \frac{-2c_1}{t^3} + \frac{c_0-2c_2}{t^2} + \frac{c_1}{t} + (c_2 + 4c_4) \amp\\ + \sum_{n=1}^\infty \left[ c_{n+4} (n+1)(n) + 4c_{n+4} (n+1) + c_{n+2} \right] t^n \amp = 0 \end{align*}

From comparing this to zero, I can determine the first few coefficients and the recurrence relation. \(c_0\) is arbitrary. \(c_1\) must be zero. \(c_2\) must be \(\frac{1}{2} c_0\text{.}\) \(c_4\) must be \(\frac{-c_2}{4} = \frac{-c_0}{8}\text{.}\) I get the recurrence relation from the term inside the sum. Since it is a second order relation and \(c_1\) is zero, all the odd terms will be zero. I calculate some even terms.

\begin{align*} c_{n+4} \amp = \frac{-c_{n+2}}{(n+1)(n) + 4(n+1)} = \frac{-c_{n+2}}{(n+1)(n+4)}\\ c_6 \amp = \frac{-c_4}{(3)(6)} = \frac{c_0}{(3)(2)(4)(6)} \\ c_8 \amp = \frac{-c_6}{(5)(8)} = \frac{-c_0}{(2)(3)(4)(5)(6)(8)} \\ c_{10} \amp = \frac{-c_8}{(7)(10)} = \frac{c_0}{(2)(3)(4)(5)(6)(7)(8)(10)} \end{align*}

I can use these calculations to intuit a pattern.

\begin{equation*} c_{2k} = \frac{(-1)^{k+1} c_0}{(2k-2)!(2k)} \end{equation*}

Now I can write the second solution.

\begin{align*} y_2 \amp = c_0 \left[ \frac{1}{t^3} \left[ 1 + \frac{t^2}{2} - \frac{t^4}{8} + \sum_{k=3}^\infty \frac{(-1)^{k+1} t^{2k}}{(2k-2)!(2k)} \right] \right] \\ \amp = c_0 \left[ \frac{1}{t^3} + \frac{1}{2t} - \frac{t}{8} + \sum_{k=3}^\infty \frac{(-1)^{k+1} t^{2k-3}}{(2k-2)!(2k)} \right] \end{align*}

The complete solution is a linear combination of the two solutions.

\begin{align*} y \amp = A \left[ 1 - \frac{t^2}{10} + \sum_{k=2}^{\infty} \frac{(-1)^k (60)(k+1)(k+2)t^{2k}}{(2k+5)!} \right]\\ \amp + B \left[ \frac{1}{t^3} + \frac{1}{2t} - \frac{t}{8} + \sum_{k=3}^\infty \frac{(-1)^{k+1} t^{2k-3}}{(2k-2)!(2k)} \right] \end{align*}

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