The solutions to the SOKENDAI's general entrance exams are given below with questions. The original texts were written in Japanese, and they were translated into English for convenience. The original texts are publically available at the SOKENDAI's website.

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* Also I have pdf documents with answers commented in Japanese. If you want to look at it, don't hesitate to ask me.

2020 (R2)

[1]

Answer following questions. Note that the variable \(x\) and \(y\) are real numbers. Also, if you use an integral constant, specify it clearly.

1. Solve the following differential equation.

\[ \frac{dy}{dx} = e^{-y} (2x-4) \]

solution

\[ \frac{dy}{dx} = e^{-y} (2x-4) \] Multiply both sides by \(e^{y} dx\) \[ e^y dy = (2x-4)dx \] Solve the indefinite integrals \[ \int e^y dy = \int (2x-4) dx \] \[ e^y = x^2 - 4x + C \] Finally,

\[ y = \log{|x^2-4x+C|} \, (C \, \text{is a constant}) \]

2. Solve the following differential equations. Note that \(\omega\), \(\Omega\), \(F\) are positive real constants. Also,\(\omega \neq \Omega\)

(1) \[ \frac{d^2y}{dx^2} + \omega^2y = 0 \]

solution

Substitute \(y\) as \(e^{\lambda x}\). \[ \lambda^2e^{\lambda x} + \omega^2 e^{\lambda x} = 0 \] \[ \lambda = \pm i\omega \] Thus, \[ y = C_1 e^{i\omega x} + C_2 e^{-i\omega x} \, \, (C_1\text{and } C_2 \text{ are constants.}) \] Rearrange the equation, \[ y = C_1 (\cos{\omega x + i\sin{\omega x}}) + C_2 (\cos{\omega x} - i\sin{\omega x}) \] \[ y = (C_1 + C_2) \cos{\omega x} + (C_1-C_2)i \sin{\omega x} \] Now, substitute \(\sqrt{(C_1+C_2)^2 + (C_1-C_2)^2\cdot i^2}\) as \(A\) \[ y = A\Big(\frac{(C_1+C_2)}{A}\cos{\omega x} + \frac{(C_1-C_2)i}{A}\sin{\omega x} \Big) \]

\[ y = A \sin{(\omega x + \phi)} \]

\[ (\phi = \tan^{-1}{\frac{C_1+C_2}{(C_1- C_2)i}}) \]

(2) \[ \frac{d^2y}{dx^2} + \omega^2y = F\cos{(\Omega x)} \]

solution

The general solution is the sum of the complementary solution: (1), and particular solution. For particular solution, substitute \(y = G\cos{(\Omega x + \phi)}\) \[ -G\Omega^2\cos{(\Omega x + \phi)} + \omega^2 G \cos{(\Omega x + \phi)} = F \cos{(\Omega x)} \] \[ G(\omega^2 - \Omega^2) \cos{\Omega x + \phi} = F\cos{\Omega x} \] \[ G = \frac{F}{\omega^2 -\Omega^2}\text{, and } \phi = 0 \] Thus, \[ y = \frac{F}{\omega^2 - \Omega^2} \cos{\Omega x} \] Finally,

\[ y = A \sin{(\omega x + \phi)} + \frac{F}{\omega^2 - \Omega^2} \cos{\Omega x} \]

3. Solve the following differential equation. \[ y^{-2} \frac{dy}{dx} + 4x^{-1}y^{-1} = x^3 \]

solution

Substitute \(u = y^-1\) and \(\frac{du}{dy}=-y^{-2}\), \[ \frac{du}{dx} - 4\frac{u}{x} = -x^3 \] First, solve the homogeneous equation. \[ \frac{du}{dx} - 4\frac{u}{x} = 0 \] \[ \frac{1}{u} du = \frac{4}{x} dx \] \[ \log{|u|} = \log{x}^4 + \log{e}^{C_1} \] \[ u = C_2 x^4 (C_2 \text{is a constant}) \] Next, substitute \(u = C_2(x)x^4\) and \(\frac{du}{dx} = C_2'(x)x^4 + 4C_2'(x)x^3\). \[ C_2'(x)x^4 +4C_2(x)x^3 -4C_2(x)x^3 = -x^3 \] \[ C_2'(x) = -x^{-1} \] \[ C_2(x) = -\log{x} + C \] \[ u = x^4 (- \log{x} + C) \]

\[ y = \frac{1}{x^4(-\log{x} + C)} (C \text{is a constant} ) \]

💡Tips

For the details of calculations, use Wolfram Alpha.

[2]

1. Introduce a point particle having single degree of freedom. The system's Lagrangian is described as \[ L = L(q, \dot{q}, t). \] Assume the point particle started moving at P_1 when time is \(t=t_1\) and arrived at P_2 when time is \(t=t_2\). Let us introduce the action integral:\(S\) defined as \[ S = \int^{t_2}_{t_1} L dt. \] According to principle of least action, the motion of the point particle is constrained so that the value of S to be minimum. This implies that when the motion is to be a bit off from the proper path \((q+\delta q, \dot{q}+\delta{\dot{q}})\), the action integral does not change against any kind of \(\delta q \text{ and } \delta \dot{q} \). Thus, \[ \delta S = \delta \int^{t_2}_{t_1} L dt = 0. \] Based on the relation above, derive Euler•Lagrange equation; \[ \frac{\partial L}{\partial q} - \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}} \right). \tag{2.1} \] Note that the \(P_1\):start and \(P_2\):end are fixed and \(\delta(t_1) = \delta{t_2} = 0\). Furthermore, \(\delta q \text{ and } \delta \dot{q}\) is so small that you may ignore all terms higher than the second order.

solution

\[ \delta L = \frac{\partial{L}}{\partial{q}} \delta q + \frac{\partial L}{\partial \dot{q}} \delta \dot{q} \] Use integration by parts, \[ \int \frac{\partial L}{\partial \dot{q}} \delta \dot{q} dt = - \int \frac{d}{dt} \Big( \frac{\partial L}{\partial \dot{q}} \Big) \delta q dt + \left[ \frac{\partial L}{\partial \dot{q}} \delta q dt \right] \] Then, \[ \delta S = \int^{t_2}_{t_1} \delta L dt \] \[ = \int^{t_2}_{t_1} \frac{\partial L}{\partial q} \delta q + \frac{\partial L}{\partial \dot{q}} \delta \dot{q} dt \] \[ = \left[ \frac{\partial L}{\partial \dot{q}} \delta q \right]^{t_2}_{t_1} + \int^{t_2}_{t_1} \left\{ \frac{\partial L}{\partial q} - \frac{d}{dt}\left( \frac{\partial L}{\partial \dot{q}} \right) \right\} \delta q = 0 \] Thus, \[ \frac{\partial L}{\partial q} - \frac{d}{dt}\left( \frac{\partial L}{\partial \dot{q}} \right) = 0 \]

2. Assume a system that the Lagrangian is constant against translation of the coordinate \(q \rightarrow q + \Delta q \). In that context, \(p\) conserves. \[ p \equiv \frac{\partial L}{\partial \dot{q}} \] This can be understood from Euler•Lagrange equation; the first term of the equation is to be 0. This is also corresponds to the law of the conservation of momentum. Meanwhile, when the Lagrangian is constant against translation of time \(t \rightarrow t + \Delta t\), show the energy conservation described as following equation. Hint; solve total differential of L.

solution

The answer should satisfy \(dE = 0\). \[ dE = d (p\dot{q} - L) \] \[ = d(p\dot{q}) - dL \tag{2.2.1} \] Also, \[ dL = \frac{\partial L}{\partial q} dq + \frac{\partial{L}}{\partial \dot{q}} d\dot{q} + \frac{\partial L}{\partial t} dt \tag{2.2.2} \] Here, \[ \frac{\partial L}{\partial t} dt = 0, \, \frac{\partial L}{\partial \dot{q}} = p, \, \frac{\partial L}{\partial q} = \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}} \right) \] Thus, rearrange the equation \((2.2.2)\), \[ dL = \frac{\partial L}{\partial q} dq + \frac{\partial L}{\partial \dot{q}} d\dot{q}. \] \[ dL = \dot{p} dq + p d\dot{q} = d(p\dot{q}) \tag{2.2.3} \] Substitute \((2.2.3)\) for \((2.2.1)\), \[ dE = d(p\dot{q}) - d(p\dot{q}) = 0 \]

3. As shown in Fig. 2.1, at the cylindrical coordinate system, the point particle having mass of \(m\) moves smoothly along with the wire which is fixed to \(z = f(r)\). \(f(0) = 0\), the wire is continuous and \(f(r)\) can be differentiable.

(1) Describe this system's Lagrangian \(L\) using \(m, g, \omega, r, \dot{r}, f, f'(=\frac{df}{dr})\). Note that the Lagrangian is \(L = K - U\), where \(K\) is the kinematic energy and \(U\) is the potential energy.

solution

The location of the point particle in orthogonal coordinate system is, \[ (x, y, z) = (r\cos{\omega t}, r\sin{\omega t}, f(r)). \tag{2.3.1} \] Differentiate Eq. (2.3.1) by \(t\). \[ (\dot{x}, \dot{y}, \dot{z}) = (\dot{r}\cos{\omega t} - r\omega\sin{\omega t}, \dot{r}\sin{\omega t} + r\omega \cos{\omega t}, \dot{r}f') \] Thus, \[ v^2 = \dot{x}^2 + \dot{y}^2 + \dot{z}^2 \] \[ = \dot{r}^2\cos^2{\omega t} - 2\dot{r}r\omega\cos{\omega t}\sin{\omega t} + r^2\omega^2\sin^2{\omega t} \] \[ + \dot{r}^2\sin^2{\omega t} + 2\dot{r}r\omega\cos{\omega t}\sin{\omega t} + r^2\omega^2\cos^2{\omega t} \] \[ + \dot{r}f^{'2} \] \[ = \dot{r} + r^2\omega^2 + \dot{r}f^{'2} \]

\[ L = T - U = \frac{1}{2} m (\dot{r}^2 + r^2\omega^2 + \dot{r}f^{'2}) - mgf(r) \tag{2.3.2} \]

(2) When the point particle remain stationary (\(\dot{r} = 0\)), substitute L for Eq. \((2.1)\), then describe \(f(r)\) using \(g, \omega, r\).

solution

Substitute \(L\) of Eq.\((2.3.2)\) for \(L\) in Eq.\((2.1)\) \[ \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{r}} \right) = m\ddot{r} + m\ddot{r}f^{'2} + 2m\dot{r}^2 \frac{df(r)}{dt}\frac{d^2f(r)}{dr^2} \] When \(\dot{r} = \ddot{r} = 0\), \[ \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{r}} \right) = 0 \] \[ \frac{\partial L}{\partial r} = mr\omega^2 - mgf(r)' \] \[ 0 = r\omega^2 - g f(r)' \] \[ f(r)' = \frac{\omega^2}{g} r \] \[ \frac{df(r)}{dr} = \frac{\omega^2}{g}r \] \[ f(r) = \frac{\omega^2}{2g} r^2 + C \] From the initial condition, the constant \(C\) is \[ f(0) = 0, C = 0 \] Finally,

\[ f(r) = \frac{\omega^2}{2g} r^2 \]

[3]

Answer following questions. Note that the variable \(x\) is a real number, \(m, n\) are positive integers. \(e\) is the base of natural logarithm. The convergence argument of functional sequences and improper integrals is not necessary.

1. Following equation is a Fourier series of a real function \(f(x)\) at the interval of \(-\pi \leq x\leq \pi \). \[ f(x) = \frac{a_0}{2} + \sum^{\infty}_{n = 1} \{a_n\cos{(nx)} + b_n\sin{(nx)} \} \tag{3.1} \] Note that \(f(x)\) is continuous on the interval of \(-\pi \leq x\leq \pi \).

(1) Solve the coefficients \(a_0, a_n, b_n\) of the Fourier series. You may use the following equation. \[ \frac{1}{\pi} \int^{\pi}_{-\pi} \cos{(mx)}\cos{(nx)} dx = \delta_{mn} \tag{3.2a} \] \[ \frac{1}{\pi} \int^{\pi}_{-\pi} \sin{(mx)}\sin{(nx)} dx = \delta_{mn} \tag{3.2b} \] \[ \frac{1}{\pi} \int^{\pi}_{-\pi} \cos{(mx)}\sin{(nx)} dx = 0 \tag{3.2c} \] Here, \(\delta_{mn}\) is the Kronecker delta defined as \[ \delta_{mn} = \begin{cases} 1 (m=n)\\ 0 (m\neq n) \\ \end{cases}. \]

solution

(i) First, integrate the Eq. (3.1) with respect to \(x\) from \(-\pi\) to \(\pi\). \[ \int^\pi_\pi f(x) dx =\int^\pi_\pi \frac{a_0}{2}dx + \sum^\infty_{n=1} \int^\pi_\pi a_n \cos{(nx)} dx + \sum^\infty_{n=1} \int^\pi_\pi a_n \sin{(nx)} dx \] \[ = a_0\pi + 0 + 0 \]

\[ a_0 = \frac{1}{\pi} \int^\pi_\pi f(x) dx \tag{3.1.1a} \]

(ii) Second, multiply the both sides of Eq. (3.1) by \(\cos{(mx)}\) and integrate it with respect to \(x\) from \(-\pi\) to \(\pi\). \[ \int^\pi_\pi f(x)\cos{(mx)} dx =\int^\pi_\pi \frac{a_0}{2}\cos{(mx)}dx + \sum^\infty_{n=1} \int^\pi_\pi a_n \cos{(nx)}\cos{(mx)} dx + \sum^\infty_{n=1} \int^\pi_\pi a_n \sin{(nx)}\cos{(mx)} dx \] \[ = 0 + a_n \pi + 0 \]

\[ a_n = \frac{1}{\pi} \int^\pi_\pi f(x)\cos{(mx)} \tag{3.1.1b} \]

(iii) Finally, multiply the both sides of Eq. (3.1) by \(\sin{(mx)}\) and integrate it with respect to \(x\) from \(-\pi\) to \(\pi\). \[ \int^\pi_\pi f(x)\sin{(mx)} dx =\int^\pi_\pi \frac{a_0}{2}\sin{(mx)}dx + \sum^\infty_{n=1} \int^\pi_\pi a_n \cos{(nx)}\sin{(mx)} dx + \sum^\infty_{n=1} \int^\pi_\pi a_n \sin{(nx)}\sin{(mx)} dx \] \[ = 0 + 0 + b_n \pi \]

\[ b_n = \frac{1}{\pi} \int^\pi_\pi f(x)\sin{(mx)} \tag{3.1.1c} \]

(2) When \(f(x)\) is defined as \[ f(x) = 2x^2 - \pi^2, \tag{3.3} \] solve \(a_0, a_n, b_n\) of the Fourier series by using \(n\).

solution

Substitute \(f(x)=2x^2-\pi^2\) for \(f(x)\) of Eq. (3.1.1a). \[ a_0 = \frac{1}{\pi} \int^{\pi}_{-\pi} (2x^2-\pi^2) dx \] \[ = \frac{1}{\pi} \left[ \frac{2}{3}x^3 - \pi^2 x\right]^\pi_{-\pi} \]

\[ a_0 = -\frac{2}{3} \pi^2 \]

Next, similarly substitute\(f(x)=2x^2-\pi^2\) for \(f(x)\) of Eq. (3.1.1b). \[ a_n = \frac{1}{\pi} \int^{\pi}_{-\pi} (2x^2-\pi^2)\cos{(mx)} dx \] \[ = \frac{2}{\pi}\int^{\pi}_{-\pi} x^2 \cos{(mx)} dx - \pi \int^{\pi}_{-\pi} \cos{(mx)} dx \] \[ = - \frac{2}{m}\int^{\pi}_{-\pi} \sin{(mx)} dx \] When \(k = 1,2,3 ....\),

\[ a_n = \begin{cases} \frac{4\pi}{m^2} (m=2k)\\ - \frac{4\pi}{m^2} (m=2k-1) \\ \end{cases} \]

Finally, \(b_0 = 0\) because \(f(x)\) is an even function.

\[ b_n = 0 \]

(1) When \(x>0\), show the following equation stands. \[ \frac{1}{e^x-1} = \sum^{\infty}_{n=1} e^{-nx} \tag{3.4} \]

solution

\[ S = e^{-x} + e^{-2x} + e^{-3x} + .... \tag{3.2.2.1} \] \[ e^{^-x}S = + e^{-2x} + e^{-3x} + e^{-4x}.... \tag{3.2.2.2} \] Subtract Eq. (3.2.2.1) from Eq. (3.2.2.2). \[ S(1-e^{-x}) = e^{-x} \] \[ S = \frac{e^{-x}}{1-e^{-x}} \] \[ = \frac{1}{e^x-1} \]

(2) Use the Eq. (3.4) and show the following relation stands. \[ \int^{\infty}_{0} \frac{x}{e^x-1} dx = \sum^{\infty}_{n=1} \frac{1}{n^2} \tag{3.5} \]

solution

From Eq. (3.4), \[ \int^\infty_0 \frac{x}{e^x-1}dx = \sum^\infty_0 xe^{-nx} dx \] \[ \sum^\infty_{n=1} \left\{ -\frac{1}{n} \left[ xe^{-nx}\right]^{\infty}_0 + \frac{1}{n} \int^{\infty}_0 e^{-nx} dx \right\} \] \[ = \sum^\infty_{n=1} - \frac{1}{n^2} \left[ e^{^nx} \right]^\infty_0 \] \[ = \sum^\infty_{n=1} \frac{1}{n^2} \]

(3) Use the answer of (1) of the subsection 1, and show the integral of Eq. (3.5) without using the series.

solution

From Parseval's identity, \[ \int^{\pi}_{-\pi} \{ f(x) \}^2 dx = \frac{a_0^2}{2} + \sum^\infty_{n=1}\{a_n^2 + b_n^2\} \tag{3.2.3.1} \] When \(f(x)=x\), \[ \int^{\pi}_{-\pi} x^2 dx = \frac{2}{3}\pi^3 \tag{3.2.3.2} \] \(f(x) = x \text{is a odd function, thus} a_0, a_n = 0 \) Then, \[ b_n = \frac{1}{\pi}\int^{\pi}_{-\pi} x\sin{(nx)}dx \] \[ = \frac{1}{\pi} \left\{ -\frac{1}{n}\left[ x \cos{(nx)}\right]^{\pi}_{\pi} + \int^{\pi}_{-\pi} \right\} \] \[ b_n = \begin{cases} -\frac{2}{n} (n\text{ is an even number}) \\ \frac{2}{n} (n\text{ is an odd number}) \end{cases} \tag{3.2.3.3} \] Substitute Eq. (3.2.3.2) and Eq. (3.2.3.3) for Eq. (3.2.3.1). \[ \sum^\infty_{n=1} \frac{1}{n^2} = \frac{\pi^2}{6} \]

[4]

Consider the interaction between material and electromagnetic waves in a magnetic field that is uniformly distributed in space and independent to time. Assume a situation that electromagnetic waves having the electric field vector given as \(\boldsymbol{E}(r,t) = \boldsymbol{E_0} exp\{i(\boldsymbol{k\cdot r} -\omega t)\}\) propagate in the space having magnetic flux density \(\boldsymbol{B_0}= (B_x, B_y, B_z) = (0, 0, B_0)\) which is acting along with \(z\)-axis. Here, \(i\) is the imaginary unit, \(\boldsymbol{r}\) is the position vector, \(\boldsymbol{k}\) is the wave vector, \(\omega\) is the angular frequency, \(t\) is the time. Answer following 1 and 2 questions.

1. TBU

solution

TBU

2. TBU

solution

TBU

#	1	2	3	4	5	6
2020 (R2)	Differential Equation	Analytical Mechanics	Fourier Series	Electromagnetism	Electric Circuit	-
2019 (R1)	Fourier Transform	Probability	Analytical Mechanics	Quantum Mechanics	Electric Circuit	-

Solutions for Entrance Exams (Astronomy dept., SOKENDAI)