Mathematics II (BT-202) - Unit 2 Short Notes

How unit 2 is examined

This unit covers second order linear equations with variable coefficients, and the marks sit in variation of parameters (14 marks, asked almost every session) and power series (Frobenius and Legendre), with Bessel proofs at 7 marks.

Second order linear differential equations with variable coefficients

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Definition. A second order linear equation with variable coefficients has the form $y'' + P(x)y' + Q(x)y = R(x)$, where $P, Q$ are functions of $x$ and not constants.

Key points.

1. If $R = 0$ the equation is homogeneous, and its general solution is $y = c_1 y_1 + c_2 y_2$ with $y_1, y_2$ linearly independent.
2. If one solution $y_1$ is known, put $y = y_1 v$ (reduction of order) to get the second solution $y_2 = y_1\int \frac{1}{y_1^2}e^{-\int P\,dx}\,dx$.
3. With $R \ne 0$ the general solution is $y = y_c + y_p$, and $y_p$ is found by variation of parameters when $y_c$ is known.
4. If no closed form exists, a power series about an ordinary point or a Frobenius series about a regular singular point is used.

Method of variation of parameters

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Definition. Variation of parameters finds a particular integral of $y'' + Py' + Qy = X$ by replacing the constants $c_1, c_2$ of the complementary function by functions $u(x), v(x)$.

Formula. With $y_c = c_1y_1 + c_2y_2$ and Wronskian $W = y_1y_2' - y_1'y_2$:

$$y_p = u y_1 + v y_2,\quad u = -\int \frac{y_2 X}{W}\,dx,\quad v = \int \frac{y_1 X}{W}\,dx$$

Key points.

1. The equation must first be written with the coefficient of $y''$ equal to 1, so $X$ is the right-hand side after that division.
2. The complementary function comes from the auxiliary equation; for $m = \pm ai$ it is $c_1\cos ax + c_2\sin ax$.
3. The Wronskian is never zero for independent $y_1, y_2$, and for $\cos ax, \sin ax$ it equals $a$.
4. The method works for any $X$, including $\tan ax$, $\sec ax$, $\cot ax$, $e^x/x$, where undetermined coefficients fail.
5. The conditions $u'y_1 + v'y_2 = 0$ and $u'y_1' + v'y_2' = X$ give the two formulas by Cramer's rule.
6. The complete solution is $y = y_c + y_p$, and it must contain two arbitrary constants.

Example. Solve $(D^2+a^2)y = \tan ax$.

Solution: $y = c_1\cos ax + c_2\sin ax - \frac{1}{a^2}\cos ax\,\ln(\sec ax + \tan ax)$. For $(D^2+4)y=\tan 2x$ put $a=2$; for $(D^2+9)y=\tan 3x$ put $a=3$.

Example. Solve $(D^2+1)y = x$: $y_c = c_1\cos x + c_2\sin x$, $W = 1$, $u = -\int x\sin x\,dx = x\cos x - \sin x$, $v = \int x\cos x\,dx = x\sin x + \cos x$, so $y_p = x\cos^2x - \sin x\cos x + x\sin^2x + \sin x\cos x = x$.

Solution: $y = c_1\cos x + c_2\sin x + x$.

==Variation of parameters gives $y_p = uy_1 + vy_2$ with $u = -\int \frac{y_2X}{W}dx$ and $v = \int \frac{y_1X}{W}dx$, where $W = y_1y_2' - y_1'y_2$.==

Answer frame. Open with "Given equation is $(D^2+a^2)y = X$; C.F. is found from $m^2+a^2=0$"; write $y_c$, then $y_1, y_2$, then $W$; state the formulas for $u, v$; integrate each on its own line; assemble $y_p$ and close with "complete solution $y = y_c + y_p$".

Pitfall: Forgetting the minus sign in $u$, or leaving the answer as $y_p$ without adding $y_c$.

Asked: [14 marks] (Jun 2022, Nov 2022, Dec 2023, Jun 2025) Solve $(D^2+a^2)y = \tan ax$ by the method of variation of parameters (variants with $a = 2$ and $a = 3$). Asked: [7 marks] (Dec 2024) Solve by the method of variation of parameter $(D^2+1)y = x$.

Power series solutions

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Definition. A power series solution writes $y = \sum a_m x^{m}$ about an ordinary point, or the Frobenius series $y = \sum a_m x^{m+r}$ about a regular singular point $x_0$, and finds the $a_m$ by substitution.

Key points.

1. For $y'' + Py' + Qy = 0$, the point $x_0$ is ordinary if $P, Q$ are analytic there, regular singular if $(x-x_0)P$ and $(x-x_0)^2Q$ are analytic, and irregular otherwise.
2. At an ordinary point the series $\sum a_m x^m$ is used, with $a_0, a_1$ arbitrary.
3. At a regular singular point the Frobenius series $\sum a_m x^{m+r}$ is used, with $a_0 \ne 0$.
4. Equating the lowest power of $x$ gives the indicial equation, whose roots give $r$.
5. Equating the coefficient of a general power gives the recurrence relation for $a_m$.
6. Roots not differing by an integer give two Frobenius series; equal roots or roots differing by an integer may bring a $\log x$ term.

Example 1 (Frobenius). Solve $x(1-x)y'' + 2(1-2x)y' - 2y = 0$. Here $x=0$ is a regular singular point. Put $y = \sum a_m x^{m+r}$:

$$\sum a_m\left[(m+r)(m+r-1)x^{m+r-1} + 2(m+r)x^{m+r-1} - \left\{(m+r)(m+r-1) + 4(m+r) + 2\right\}x^{m+r}\right] = 0$$

Lowest power $x^{r-1}$: $a_0\,r(r+1) = 0$, so indicial roots $r = 0, -1$. Coefficient of $x^{m+r}$ gives $(m+r+1)(m+r+2)\,a_{m+1} = (m+r+1)(m+r+2)\,a_m$, so $a_{m+1} = a_m$.

• $r = 0$: all $a_m = a_0$, so $y_1 = a_0(1 + x + x^2 + \dots) = \frac{a_0}{1-x}$.
• $r = -1$: $a_1$ stays arbitrary (the factor $m+r+1$ vanishes at $m=0$), so $y = \frac{a_0}{x} + a_1(1 + x + \dots)$ with no log term.

Solution: $y = \frac{A}{x} + \frac{B}{1-x}$.

Example 2 (Legendre, $n=1$). Solve $(1-x^2)y'' - 2xy' + 2y = 0$ with $y = \sum a_m x^m$. The coefficient of $x^m$ gives $(m+2)(m+1)a_{m+2} = [m(m+1) - 2]a_m$, so

$$a_{m+2} = \frac{m-1}{m+1}\,a_m$$

Even: $a_2 = -a_0$, $a_4 = -\frac{a_0}{3}$, $a_6 = -\frac{a_0}{5}$. Odd: $a_3 = 0$, so all higher odd terms vanish.

Solution: $y = a_1 x + a_0\left[1 - x^2 - \frac{x^4}{3} - \frac{x^6}{5} - \dots\right]$.

==At a regular singular point put $y = \sum a_m x^{m+r}$, equate the lowest power of $x$ to zero for the indicial equation, and use the coefficient of the general power for the recurrence relation.==

Answer frame. Open with "$x=0$ is an ordinary (or regular singular) point, so assume $y = \sum a_m x^{m}$ (or $x^{m+r}$)"; write $y', y''$ and substitute; give the indicial equation or recurrence; tabulate the first coefficients; close with the general solution containing two constants.

Pitfall: Mixing up the index shift, so that the recurrence links $a_m$ to the wrong $a_{m+k}$.

Asked: [14 marks] (Jun 2023) Solve $x(1-x)y'' + 2(1-2x)y' - 2y = 0$ using Frobenius method. Asked: [7 marks] (Jun 2025) Solve in series Legendre's differential equation $(1-x^2)y'' - 2xy' + 2y = 0$.

Legendre polynomials

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Definition. Legendre's equation is $(1-x^2)y'' - 2xy' + n(n+1)y = 0$, and its polynomial solution for integer $n$ is $P_n(x)$, given by Rodrigues' formula $P_n(x) = \frac{1}{2^n n!}\frac{d^n}{dx^n}(x^2-1)^n$.

Key points.

1. The first polynomials are $P_0 = 1$, $P_1 = x$, $P_2 = \frac{3x^2-1}{2}$, $P_3 = \frac{5x^3-3x}{2}$.
2. $P_n(1) = 1$, and $P_n(-x) = (-1)^nP_n(x)$, so $P_n$ is even for even $n$ and odd for odd $n$.
3. Orthogonality: $\int_{-1}^{1}P_mP_n\,dx = 0$ for $m \ne n$, and $\frac{2}{2n+1}$ for $m = n$.
4. Recurrence: $(n+1)P_{n+1} = (2n+1)xP_n - nP_{n-1}$.

Bessel functions of the first kind and their properties

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Definition. Bessel's equation is $x^2y'' + xy' + (x^2-n^2)y = 0$, and its solution is the Bessel function of the first kind

$$J_n(x) = \sum_{r=0}^{\infty}\frac{(-1)^r}{r!\,\Gamma(n+r+1)}\left(\frac{x}{2}\right)^{n+2r}$$

Key points.

1. The series converges for all $x$, and $J_n(x)$ is the Frobenius solution for the indicial root $r = n$.
2. For integer $n$, $J_{-n}(x) = (-1)^nJ_n(x)$, so $J_n$ and $J_{-n}$ are dependent and the second solution is $Y_n$; for non-integer $n$ they are independent.
3. $J_0(0) = 1$ and $J_n(0) = 0$ for $n > 0$.
4. Recurrence relations: $\frac{d}{dx}[x^nJ_n] = x^nJ_{n-1}$ and $\frac{d}{dx}[x^{-n}J_n] = -x^{-n}J_{n+1}$.
5. $J_{-1/2}(x) = \sqrt{\frac{2}{\pi x}}\cos x$, and $J_{1/2}$ and $J_{-1/2}$ are elementary.

Proof of $J_{1/2}$. Put $n = \frac12$ and use $r!\,\Gamma(r+\frac32) = \frac{(2r+1)!}{2^{2r+1}}\sqrt\pi$ (check: $r=0$ gives $\Gamma(\frac32) = \frac{\sqrt\pi}{2}$, from $\Gamma(\frac12) = \sqrt\pi$):

$$J_{1/2}(x) = \sqrt{\frac x2}\sum_{r=0}^{\infty}\frac{(-1)^r\,2^{2r+1}}{(2r+1)!\sqrt\pi}\cdot\frac{x^{2r}}{2^{2r}} = \sqrt{\frac{2x}{\pi}}\cdot\frac1x\sum_{r=0}^{\infty}\frac{(-1)^rx^{2r+1}}{(2r+1)!} = \sqrt{\frac{2}{\pi x}}\sin x$$

Proof of $J_n(-x)$. In the series, $\left(\frac{-x}{2}\right)^{n+2r} = (-1)^{n+2r}\left(\frac x2\right)^{n+2r} = (-1)^n\left(\frac x2\right)^{n+2r}$ because $(-1)^{2r} = 1$ and $n$ is an integer. Every term carries the same factor $(-1)^n$, so $J_n(-x) = (-1)^nJ_n(x)$. For negative integer $n$ the terms with $r + n + 1 \le 0$ vanish since $1/\Gamma = 0$, and the rest are treated identically.

==$J_n(x) = \sum_{r\ge0}\frac{(-1)^r}{r!\,\Gamma(n+r+1)}\left(\frac x2\right)^{n+2r}$, and for integer $n$, $J_n(-x) = (-1)^nJ_n(x)$.==

Answer frame. Open with the series definition of $J_n(x)$; for $J_{1/2}$ substitute $n=\frac12$, rewrite $\Gamma(r+\frac32)$, and factor out $\sqrt{2/\pi x}$ to reach the $\sin x$ series; for $J_n(-x)$ replace $x$ by $-x$, pull out $(-1)^{n+2r}$, and close with $(-1)^n$.

Asked: [7 marks] (Jun 2023, Dec 2023) Show that $J_{1/2}(x) = \sqrt{\frac{2}{\pi x}}\sin x$. Asked: [7 marks] (Dec 2024) Show that $J_n(-x) = (-1)^nJ_n(x)$ when $n$ is positive or negative integer.

Last-minute revision

• Variation of parameters: $y_p = uy_1 + vy_2$, $u = -\int\frac{y_2X}{W}dx$, $v = \int\frac{y_1X}{W}dx$.
• $W = y_1y_2' - y_1'y_2$; for $\cos ax, \sin ax$, $W = a$.
• $(D^2+a^2)y = \tan ax$ gives $y_p = -\frac{1}{a^2}\cos ax\ln(\sec ax+\tan ax)$.
• $(D^2+1)y = x$ gives $y = c_1\cos x + c_2\sin x + x$.
• Frobenius: $y = \sum a_mx^{m+r}$, indicial equation from the lowest power of $x$.
• $x(1-x)y'' + 2(1-2x)y' - 2y = 0$: roots $0, -1$, $a_{m+1} = a_m$, $y = A/x + B/(1-x)$.
• Legendre $n=1$: $a_{m+2} = \frac{m-1}{m+1}a_m$, $y = a_1x + a_0[1 - x^2 - \frac{x^4}{3} - \dots]$.
• $P_n(x) = \frac{1}{2^nn!}\frac{d^n}{dx^n}(x^2-1)^n$, $P_n(1) = 1$.
• $J_{1/2} = \sqrt{\frac{2}{\pi x}}\sin x$, $J_{-1/2} = \sqrt{\frac{2}{\pi x}}\cos x$.
• $J_{-n} = (-1)^nJ_n$ and $J_n(-x) = (-1)^nJ_n(x)$ for integer $n$.

Memory hooks

• "Minus on $u$, plus on $v$": $u$ has $-y_2X/W$, $v$ has $+y_1X/W$.
• Wronskian is the cross-product of $y$ and $y'$: $y_1y_2' - y_1'y_2$.
• Lowest power gives the indicial equation; general power gives the recurrence.
• Legendre $n=1$: numerator $m-1$ kills $a_3$, so the odd series is just $x$.
• Bessel sign trick: $(-x)^{n+2r}$, the $2r$ vanishes and only $(-1)^n$ survives.

Coverage checklist

• Second order linear differential equations with variable coefficients: definition, reduction of order, no past questions.
• Method of variation of parameters: $\tan ax$ (14 marks, Jun 2022, Nov 2022, Dec 2023, Jun 2025), $(D^2+1)y = x$ (7 marks, Dec 2024).
• Power series solutions: Frobenius (14 marks, Jun 2023), Legendre series (7 marks, Jun 2025).
• Legendre polynomials: Rodrigues formula, first polynomials, properties, no past questions.
• Bessel functions of the first kind and their properties: $J_{1/2}$ (Jun 2023, Dec 2023), $J_n(-x)$ (Dec 2024).