Special Functions Questions and Answers

Q: Consider the ordinary differential equation\[ y'' - 2xy' + 4y = 0 \]and its solution \(y(x) = a + bx + cx^{2}\). Then

The coefficients satisfy \(c = -2a \ne 0\) and \(b = 0\).Substitute \(y = a + bx + cx^{2}\), \(y' = b + 2cx\), \(y'' = 2c\):\[ \begin{aligned} 2c - 2x(b + 2cx) + 4(a + bx + cx^{2}) \\ = (2c + 4a) + 2bx = 0 \end{aligned} \]The \(x^{2}\) terms cancel. For this to vanish for all \(x\), \(b = 0\) and \(c = -2a\), so the solution is \(a(1 - 2x^{2})\), proportional to the Hermite polynomial \(H_{2}\). The option with \(a = 0\) and \(c = -2b\) fails the constant term, \(b = -2a\) with \(c = 0\) fails because \(b\) must vanish, and \(c = 2a\) has the wrong sign.

Q: \(\frac{d^{2} f}{dx^{2}} + 2 \operatorname{sech}^{2} x\, f - f = 0.\)One of the solution is :

One solution is sech x.Try \(f = \operatorname{sech} x\). Using \((\operatorname{sech} x)' = -\operatorname{sech} x \tanh x\) and \((\tanh x)' = \operatorname{sech}^{2} x\):\[ \begin{aligned} f' &= -\operatorname{sech} x \tanh x \\ f'' &= \operatorname{sech} x \tanh^{2} x - \operatorname{sech}^{3} x \end{aligned} \]Substituting:\[ \begin{aligned} &f'' + 2\operatorname{sech}^{2}x\, f - f \\ &= \operatorname{sech} x\,(\tanh^{2} x + \operatorname{sech}^{2} x - 1) = 0 \end{aligned} \]since \(\tanh^{2} x + \operatorname{sech}^{2} x = 1\). The same substitution with \(\operatorname{sech}^{2} x\), \(\tanh x\) or \(\operatorname{sech}^{3} x\) leaves terms that do not cancel; for example \(f = \tanh x\) has \(f'' = -2\operatorname{sech}^{2}x\tanh x\), which cancels the middle term and leaves \(-\tanh x \neq 0\).

Consider the ordinary differential equation

\[ y'' - 2xy' + 4y = 0 \]

and its solution \(y(x) = a + bx + cx^{2}\). Then

A.\(a = 0,\ c = -2b \ne 0\)
B.\(c = -2a \ne 0,\ b = 0\)
C.\(b = -2a \ne 0,\ c = 0\)
D.\(c = 2a \ne 0,\ b = 0\)

Exam Year(s)

GATE Physics - 2022,

Answer: B

The coefficients satisfy \(c = -2a \ne 0\) and \(b = 0\).

Substitute \(y = a + bx + cx^{2}\), \(y' = b + 2cx\), \(y'' = 2c\):

\[ \begin{aligned} 2c - 2x(b + 2cx) + 4(a + bx + cx^{2}) \\ = (2c + 4a) + 2bx = 0 \end{aligned} \]

The \(x^{2}\) terms cancel. For this to vanish for all \(x\), \(b = 0\) and \(c = -2a\), so the solution is \(a(1 - 2x^{2})\), proportional to the Hermite polynomial \(H_{2}\). The option with \(a = 0\) and \(c = -2b\) fails the constant term, \(b = -2a\) with \(c = 0\) fails because \(b\) must vanish, and \(c = 2a\) has the wrong sign.

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\(\frac{d^{2} f}{dx^{2}} + 2 \operatorname{sech}^{2} x\, f - f = 0.\)

One of the solution is :

A.sech x
B.\(\operatorname{sech}^{2} x\)
C.tanh x
D.\(\operatorname{sech}^{3} x\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: A

One solution is sech x.

Try \(f = \operatorname{sech} x\). Using \((\operatorname{sech} x)' = -\operatorname{sech} x \tanh x\) and \((\tanh x)' = \operatorname{sech}^{2} x\):

\[ \begin{aligned} f' &= -\operatorname{sech} x \tanh x \\ f'' &= \operatorname{sech} x \tanh^{2} x - \operatorname{sech}^{3} x \end{aligned} \]

Substituting:

\[ \begin{aligned} &f'' + 2\operatorname{sech}^{2}x\, f - f \\ &= \operatorname{sech} x\,(\tanh^{2} x + \operatorname{sech}^{2} x - 1) = 0 \end{aligned} \]

since \(\tanh^{2} x + \operatorname{sech}^{2} x = 1\). The same substitution with \(\operatorname{sech}^{2} x\), \(\tanh x\) or \(\operatorname{sech}^{3} x\) leaves terms that do not cancel; for example \(f = \tanh x\) has \(f'' = -2\operatorname{sech}^{2}x\tanh x\), which cancels the middle term and leaves \(-\tanh x \neq 0\).

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