Series Solutions and Special Functions Questions and Answers

Q: If \(y = \sum_{m=0}^{\infty} c_{m} x^{r+m}\) is assumed to be a solution of the differential equation\[ x^{2}y'' - xy' - 3(1 + x^{2})y = 0, \]then the values of \(r\) are

The exponents are -1 and 3.Substituting \(y = \sum c_{m}x^{r+m}\) and collecting the lowest power \(x^{r}\) gives the indicial equation from the terms \(x^{2}y''\), \(-xy'\) and \(-3y\) (the \(-3x^{2}y\) term raises the power by two and does not contribute):\[ \begin{aligned} r(r-1) - r - 3 &= 0 \\ r^{2} - 2r - 3 &= 0 \\ (r - 3)(r + 1) &= 0 \end{aligned} \]So \(r = 3\) or \(r = -1\). The other pairs come from sign errors: '1 and 3' and '1 and -3' change the sign of one root, and '-1 and -3' solves \(r^{2} + 4r + 3 = 0\), which is not the indicial equation here.

Q: Let the Legendre equation \((1 - x^{2})y'' - 2xy' + n(n+1)y = 0\) have \(n^{th}\) degree polynomial solution \(y_{n}(x)\) such that \(y_{n}(1) = 3\). If \(\int_{-1}^{1} \left(y_{n}^{2}(x) + y_{n-1}^{2}(x)\right) dx = \frac{144}{15}\), then \(n\) is

The value is \(n = 2\).The polynomial solutions are multiples of the Legendre polynomials, and \(P_{n}(1) = 1\), so \(y_{n} = 3P_{n}\); the paper's normalisation applies to every degree, so \(y_{n-1} = 3P_{n-1}\) as well. Using \(\int_{-1}^{1}P_{n}^{2}\,dx = \frac{2}{2n+1}\):\[ \begin{aligned} 9\left(\tfrac{2}{2n+1} + \tfrac{2}{2n-1}\right) &= \tfrac{144}{15} \\ \tfrac{72n}{4n^{2} - 1} &= \tfrac{144}{15} \end{aligned} \]For \(n = 2\): \(\frac{144}{15}\), as required. Check the others: \(n = 1\) gives \(\frac{72}{3} = 24\), \(n = 3\) gives \(\frac{216}{35}\), and \(n = 4\) gives \(\frac{288}{63}\). So only \(n = 2\) works.

Q: Let \(p_{n}(x)\) be the polynomial solution of the differential equation\[ \frac{d}{dx}[(1 - x^{2})y'] + n(n+1)y = 0 \]with \(p_{n}(1) = 1\) for \(n = 1, 2, 3, \ldots\). If\[ \frac{d}{dx}[p_{n+2}(x) - p_{n}(x)] = \alpha_{n} p_{n+1}(x), \]then \(\alpha_{n}\) is

\(\alpha_{n} = 2n + 3\).The equation is Legendre's equation, and its polynomial solution with \(p_{n}(1) = 1\) is the Legendre polynomial \(P_{n}\). The Legendre polynomials satisfy the recurrence\[ P'_{m+1}(x) - P'_{m-1}(x) = (2m + 1)P_{m}(x) \]Take \(m = n + 1\):\[ P'_{n+2}(x) - P'_{n}(x) = (2n + 3)P_{n+1}(x) \]Check with \(n = 1\), where \(P_{3} = \frac{1}{2}(5x^{3} - 3x)\):\[ \begin{aligned} P'_{3} - P'_{1} &= \frac{15}{2}x^{2} - \frac{3}{2} - 1 \\ &= \frac{5}{2}(3x^{2} - 1) = 5P_{2} \end{aligned} \]and \(2(1) + 3 = 5\). The values \(2n + 1\), \(2n\) and \(2n + 2\) give 3, 2 and 4 here, so they fail.

If \(y = \sum_{m=0}^{\infty} c_{m} x^{r+m}\) is assumed to be a solution of the differential equation

\[ x^{2}y'' - xy' - 3(1 + x^{2})y = 0, \]

then the values of \(r\) are

A.1 and 3
B.-1 and 3
C.1 and -3
D.-1 and -3

Exam Year(s)

GATE Mathematics - 2012,

Answer: B

The exponents are -1 and 3.

Substituting \(y = \sum c_{m}x^{r+m}\) and collecting the lowest power \(x^{r}\) gives the indicial equation from the terms \(x^{2}y''\), \(-xy'\) and \(-3y\) (the \(-3x^{2}y\) term raises the power by two and does not contribute):

\[ \begin{aligned} r(r-1) - r - 3 &= 0 \\ r^{2} - 2r - 3 &= 0 \\ (r - 3)(r + 1) &= 0 \end{aligned} \]

So \(r = 3\) or \(r = -1\). The other pairs come from sign errors: '1 and 3' and '1 and -3' change the sign of one root, and '-1 and -3' solves \(r^{2} + 4r + 3 = 0\), which is not the indicial equation here.

Discuss in Forum

Let the Legendre equation \((1 - x^{2})y'' - 2xy' + n(n+1)y = 0\) have \(n^{th}\) degree polynomial solution \(y_{n}(x)\) such that \(y_{n}(1) = 3\). If \(\int_{-1}^{1} \left(y_{n}^{2}(x) + y_{n-1}^{2}(x)\right) dx = \frac{144}{15}\), then \(n\) is

Exam Year(s)

GATE Mathematics - 2012,

Answer: B

The value is \(n = 2\).

The polynomial solutions are multiples of the Legendre polynomials, and \(P_{n}(1) = 1\), so \(y_{n} = 3P_{n}\); the paper's normalisation applies to every degree, so \(y_{n-1} = 3P_{n-1}\) as well. Using \(\int_{-1}^{1}P_{n}^{2}\,dx = \frac{2}{2n+1}\):

\[ \begin{aligned} 9\left(\tfrac{2}{2n+1} + \tfrac{2}{2n-1}\right) &= \tfrac{144}{15} \\ \tfrac{72n}{4n^{2} - 1} &= \tfrac{144}{15} \end{aligned} \]

For \(n = 2\): \(\frac{144}{15}\), as required. Check the others: \(n = 1\) gives \(\frac{72}{3} = 24\), \(n = 3\) gives \(\frac{216}{35}\), and \(n = 4\) gives \(\frac{288}{63}\). So only \(n = 2\) works.

Discuss in Forum

Let \(p_{n}(x)\) be the polynomial solution of the differential equation

\[ \frac{d}{dx}[(1 - x^{2})y'] + n(n+1)y = 0 \]

with \(p_{n}(1) = 1\) for \(n = 1, 2, 3, \ldots\). If

\[ \frac{d}{dx}[p_{n+2}(x) - p_{n}(x)] = \alpha_{n} p_{n+1}(x), \]

then \(\alpha_{n}\) is

A.\(2n\)
B.\(2n + 1\)
C.\(2n + 2\)
D.\(2n + 3\)

Exam Year(s)

GATE Mathematics - 2018,

Answer: D

\(\alpha_{n} = 2n + 3\).

The equation is Legendre's equation, and its polynomial solution with \(p_{n}(1) = 1\) is the Legendre polynomial \(P_{n}\). The Legendre polynomials satisfy the recurrence

\[ P'_{m+1}(x) - P'_{m-1}(x) = (2m + 1)P_{m}(x) \]

Take \(m = n + 1\):

\[ P'_{n+2}(x) - P'_{n}(x) = (2n + 3)P_{n+1}(x) \]

Check with \(n = 1\), where \(P_{3} = \frac{1}{2}(5x^{3} - 3x)\):

\[ \begin{aligned} P'_{3} - P'_{1} &= \frac{15}{2}x^{2} - \frac{3}{2} - 1 \\ &= \frac{5}{2}(3x^{2} - 1) = 5P_{2} \end{aligned} \]

and \(2(1) + 3 = 5\). The values \(2n + 1\), \(2n\) and \(2n + 2\) give 3, 2 and 4 here, so they fail.

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