Solution:

Relevant definitions and result

An element a\in G has order 2 if a^2=e and a\ne e. We are given that a is the only element of order 2 in G.

To prove ax=xa for every x\in G, it is enough to prove that xax^{-1}=a.

Step 1: Consider a conjugate of a

Let x\in G be arbitrary and put

b=xax^{-1}.

We calculate the square of b:

b^2=(xax^{-1})(xax^{-1}) =xa(x^{-1}x)ax^{-1} =xa^2x^{-1}.

Since a has order 2, a^2=e. Hence

b^2=xex^{-1}=xx^{-1}=e.

Step 2: Show that b\ne e

If b=e, then

xax^{-1}=e.

Multiplying on the left by x^{-1} and on the right by x,

a=x^{-1}ex=e,

which contradicts the fact that a has order 2. Therefore b\ne e.

Thus b=xax^{-1} is also an element of order 2.

Step 3: Use uniqueness of the element of order 2

By hypothesis, a is the only element of order 2 in G. Therefore

xax^{-1}=a.

Multiplying this equality on the right by x, we obtain

xa=ax.

Since x was arbitrary, a commutes with every element of G. Equivalently, a\in Z(G), the centre of the group.

Final Answer:

\boxed{ax=xa\quad\text{for every }x\in G.}

Solution:

Relevant definitions

Let f be a bounded real-valued function on [a,b]. For a partition P, the lower and upper Darboux sums are denoted by L(f,P) and U(f,P).

We use two standard facts:

1. If P' is a refinement of P, then L(f,P)\le L(f,P') and U(f,P')\le U(f,P).
2. For every partition P, L(f,P)\le U(f,P).

Step 1: Form a common refinement

Let

P=p_1\cup p_2.

After arranging the points in increasing order, P is a partition of [a,b]. It contains every point of both p_1 and p_2. Hence P is a refinement of both.

Step 2: Prove L(f,p_1)\le U(f,p_2)

Because P refines p_1,

L(f,p_1)\le L(f,P).

For the same partition P,

L(f,P)\le U(f,P).

Because P also refines p_2,

U(f,P)\le U(f,p_2).

Combining the three inequalities,

L(f,p_1)\le L(f,P)\le U(f,P)\le U(f,p_2).

Therefore,

L(f,p_1)\le U(f,p_2).

Step 3: Interchange the roles of p_1 and p_2

Similarly,

L(f,p_2)\le L(f,P)\le U(f,P)\le U(f,p_1),

and hence

L(f,p_2)\le U(f,p_1).

Final Answer:

\boxed{L(f,p_1)\le U(f,p_2)\quad\text{and}\quad L(f,p_2)\le U(f,p_1).}

Solution:

Given

f(x)=(x+2)^2e^x.

Relevant Laplace-transform results

For s>0,

L\{1\}=\frac1s,\qquad L\{x\}=\frac1{s^2},\qquad L\{x^2\}=\frac{2}{s^3}.

The first shifting theorem states that if L\{g(x)\}=G(s), then

L\{e^{ax}g(x)\}=G(s-a).

Step 1: Expand the polynomial

(x+2)^2=x^2+4x+4.

Therefore

f(x)=e^x(x^2+4x+4).

Step 2: Transform the polynomial part

Let g(x)=x^2+4x+4. By linearity,

G(s)=L\{g(x)\} =L\{x^2\}+4L\{x\}+4L\{1\}.

Hence

G(s)=\frac{2}{s^3}+\frac{4}{s^2}+\frac{4}{s}.

Step 3: Apply the first shifting theorem

Since the multiplying exponential is e^x=e^{1\cdot x}, replace s by s-1:

L\{(x+2)^2e^x\} = \frac{2}{(s-1)^3} +\frac{4}{(s-1)^2} +\frac{4}{s-1}.

The transform exists for \operatorname{Re}s>1.

Final Answer:

\boxed{ L\{(x+2)^2e^x\} = \frac{2}{(s-1)^3} +\frac{4}{(s-1)^2} +\frac{4}{s-1} },\qquad \operatorname{Re}s>1.

Solution:

Given functional

H[y]=\int_0^1\left[(y')^2+12xy\right]dx,

with boundary conditions

y(0)=0,\qquad y(1)=1.

Euler-Lagrange equation

For a functional \int F(x,y,y')\,dx, an extremal satisfies

\frac{\partial F}{\partial y} -\frac{d}{dx}\left(\frac{\partial F}{\partial y'}\right)=0.

Here

F(x,y,y')=(y')^2+12xy.

Step 1: Calculate the required partial derivatives

\frac{\partial F}{\partial y}=12x,

and

\frac{\partial F}{\partial y'}=2y'.

Therefore

\frac{d}{dx}\left(\frac{\partial F}{\partial y'}\right) =\frac{d}{dx}(2y') =2y''.

Step 2: Form the Euler-Lagrange differential equation

12x-2y''=0.

Thus

y''=6x.

Step 3: Integrate twice

Integrating once,

y'=\int 6x\,dx=3x^2+C_1.

Integrating again,

y=x^3+C_1x+C_2.

Step 4: Apply the boundary conditions

From y(0)=0,

0=0+C_1(0)+C_2,

so

C_2=0.

From y(1)=1,

1=1+C_1,

hence

C_1=0.

Therefore the extremal is

y=x^3.

Final Answer:

\boxed{y(x)=x^3}.

Solution:

Important wording point

Strictly speaking, every group has the trivial subgroup \{e\}, which is proper whenever the group has more than one element. Thus the standard theorem intended here is: a finite group of prime order has no non-trivial proper subgroup.

Lagrange’s theorem

If H is a subgroup of a finite group G, then

|H|\mid |G|.

Proof

Let |G|=p, where p is prime, and let H\le G. By Lagrange’s theorem, |H| must divide p.

The only positive divisors of a prime number p are

1\quad\text{and}\quad p.

Therefore there are only two possibilities.

Case 1: |H|=1. Then

H=\{e\},

the trivial subgroup.

Case 2: |H|=p=|G|. Since H\subseteq G and both finite sets have the same number of elements,

H=G.

There is no third possibility. Hence no subgroup can satisfy

\{e\}\subsetneq H\subsetneq G.

Final Answer:

A finite group of prime order has no non-trivial proper subgroup; its only subgroups are

\boxed{\{e\}\ \text{and}\ G}.

Solution:

Definition of kernel

Let f:G\to G' be a group homomorphism. Its kernel is

K=\ker f=\{g\in G:f(g)=e'\},

where e' is the identity element of G'.

We shall first verify that K is a subgroup and then prove that it is normal.

Step 1: K is non-empty

Since f is a homomorphism,

f(e)=e'.

Therefore e\in K, so K\ne\varnothing.

Step 2: Apply the subgroup test

Take a,b\in K. Then

f(a)=e',\qquad f(b)=e'.

Now

f(ab^{-1}) =f(a)f(b^{-1}) =f(a)f(b)^{-1} =e'(e')^{-1} =e'.

Hence ab^{-1}\in K. By the subgroup criterion,

K\le G.

Step 3: Prove normality

Let g\in G and k\in K. Since f(k)=e',

f(gkg^{-1}) =f(g)f(k)f(g^{-1}) =f(g)e'f(g)^{-1}.

Therefore

f(gkg^{-1})=f(g)f(g)^{-1}=e'.

Thus

gkg^{-1}\in K.

Since this holds for every g\in G and k\in K,

gKg^{-1}\subseteq K.

Replacing g by g^{-1} gives the reverse inclusion, so

gKg^{-1}=K.

Final Answer:

\boxed{\ker f\trianglelefteq G}.

Hence the kernel of every group homomorphism is a normal subgroup of its domain.

Solution:

Given

f(x)=x,\qquad 0\le x\le1.

We shall prove Riemann integrability directly by upper and lower Darboux sums and then calculate the integral.

Step 1: Choose a uniform partition

For a positive integer n, let

P_n=\left\{0,\frac1n,\frac2n,\ldots,\frac{n-1}{n},1\right\}.

Each subinterval has length

\Delta x=\frac1n.

Since f(x)=x is increasing, on the interval

I_i=\left[\frac{i-1}{n},\frac{i}{n}\right],

we have

m_i=\frac{i-1}{n},\qquad M_i=\frac{i}{n}.

Step 2: Calculate the lower Darboux sum

L(f,P_n) =\sum_{i=1}^{n}m_i\Delta x =\sum_{i=1}^{n}\frac{i-1}{n}\cdot\frac1n.

Therefore

L(f,P_n) =\frac1{n^2}\sum_{i=1}^{n}(i-1) =\frac1{n^2}\sum_{j=0}^{n-1}j.

Using \sum_{j=0}^{n-1}j=\frac{n(n-1)}2,

L(f,P_n) =\frac{n(n-1)}{2n^2} =\frac{n-1}{2n}.

Step 3: Calculate the upper Darboux sum

U(f,P_n) =\sum_{i=1}^{n}M_i\Delta x =\sum_{i=1}^{n}\frac{i}{n}\cdot\frac1n =\frac1{n^2}\sum_{i=1}^{n}i.

Since \sum_{i=1}^{n}i=\frac{n(n+1)}2,

U(f,P_n) =\frac{n(n+1)}{2n^2} =\frac{n+1}{2n}.

Step 4: Verify the Riemann integrability criterion

U(f,P_n)-L(f,P_n) =\frac{n+1}{2n}-\frac{n-1}{2n} =\frac1n.

Given any \varepsilon>0, choose n>1/\varepsilon. Then

U(f,P_n)-L(f,P_n)=\frac1n<\varepsilon. [/katex] <p>Hence [katex]f is Riemann integrable on [0,1].

Step 5: Evaluate the integral

As n\to\infty,

L(f,P_n)=\frac{n-1}{2n}\longrightarrow\frac12,

and

U(f,P_n)=\frac{n+1}{2n}\longrightarrow\frac12.

Therefore the common value of the lower and upper integrals is 1/2.

Final Answer:

\boxed{f(x)=x\text{ is Riemann integrable on }[0,1],\qquad \int_0^1x\,dx=\frac12.}

Solution:

Integral to be tested

I(a)=\int_0^\infty e^{-ax}\frac{\sin x}{x}\,dx,\qquad a\ge0.

There are two possible sources of impropriety: x=0 and x=\infty. We examine them separately.

1. Behaviour near x=0

Using the standard limit

\lim_{x\to0}\frac{\sin x}{x}=1

and \lim_{x\to0}e^{-ax}=1, we get

\lim_{x\to0}e^{-ax}\frac{\sin x}{x}=1.

Thus the apparent singularity at x=0 is removable, and the integral over any interval (0,A] is convergent.

2. Case a>0

For x\ge1,

\left|e^{-ax}\frac{\sin x}{x}\right| \le \frac{e^{-ax}}{x} \le e^{-ax}.

But

\int_1^\infty e^{-ax}\,dx = \left[-\frac1a e^{-ax}\right]_1^\infty =\frac{e^{-a}}{a}<\infty. [/katex] <p>Hence by comparison, the given integral is <strong>absolutely convergent</strong> when [katex]a>0.

3. Case a=0

The integral becomes

\int_0^\infty\frac{\sin x}{x}\,dx.

On [1,\infty), the primitive of \sin x is bounded:

\int_1^X\sin x\,dx=\cos1-\cos X,

and 1/x is positive, monotone decreasing, and tends to zero. Therefore Dirichlet's test gives convergence of

\int_1^\infty\frac{\sin x}{x}\,dx.

It is not absolutely convergent; the classical integral \int_1^\infty |\sin x|/x\,dx diverges. Hence for a=0, convergence is conditional.

Final Answer:

\boxed{ \begin{cases} \text{absolutely convergent}, & a>0,\\ \text{conditionally convergent}, & a=0. \end{cases}}

Thus the improper integral converges for every a\ge0.

Solution:

Given

(A,d) is a metric space. We have to prove

|d(x_1,y_1)-d(x_2,y_2)| \le d(x_1,x_2)+d(y_1,y_2).

Step 1: Use the triangle inequality to bound d(x_1,y_1)

By travelling from x_1 to x_2, then to y_2, and finally to y_1, the triangle inequality gives

d(x_1,y_1) \le d(x_1,x_2)+d(x_2,y_2)+d(y_2,y_1).

Because a metric is symmetric,

d(y_2,y_1)=d(y_1,y_2).

Therefore

d(x_1,y_1)-d(x_2,y_2) \le d(x_1,x_2)+d(y_1,y_2). \tag{1}

Step 2: Reverse the roles of the two pairs

Again by the triangle inequality,

d(x_2,y_2) \le d(x_2,x_1)+d(x_1,y_1)+d(y_1,y_2).

Since d(x_2,x_1)=d(x_1,x_2),

d(x_2,y_2)-d(x_1,y_1) \le d(x_1,x_2)+d(y_1,y_2). \tag{2}

Step 3: Combine the two estimates

Equations (1) and (2) say that both a real number and its negative are bounded above by the same non-negative quantity. Hence

-\left[d(x_1,x_2)+d(y_1,y_2)\right] \le d(x_1,y_1)-d(x_2,y_2) \le d(x_1,x_2)+d(y_1,y_2).

Therefore

|d(x_1,y_1)-d(x_2,y_2)| \le d(x_1,x_2)+d(y_1,y_2).

Final Answer:

\boxed{|d(x_1,y_1)-d(x_2,y_2)| \le d(x_1,x_2)+d(y_1,y_2).}

Solution:

Given series

x^2(\log2)^q+x^3(\log3)^q+x^4(\log4)^q+\cdots =\sum_{n=2}^{\infty}x^n(\log n)^q.

We take x,q\in\mathbb R and test both the interior and the end points.

Step 1: Apply the root test

Let

a_n=x^n(\log n)^q.

Then

|a_n|^{1/n} =|x|(\log n)^{q/n}.

Now

(\log n)^{q/n} =\exp\left(\frac{q\log\log n}{n}\right).

Since

\frac{\log\log n}{n}\longrightarrow0,

we obtain

\lim_{n\to\infty}(\log n)^{q/n}=1.

Thus

\lim_{n\to\infty}|a_n|^{1/n}=|x|.

Therefore, by the root test:

|x|<1\Rightarrow\text{absolute convergence}, [/katex] <p>and</p> [katex display="true"] |x|>1\Rightarrow\text{divergence}.

It remains to test x=\pm1.

Step 2: End point x=1

The series becomes

\sum_{n=2}^{\infty}(\log n)^q.

If q\ge0, then (\log n)^q does not tend to zero (for q=0 it equals 1, and for q>0 it tends to infinity). Hence the series diverges.

If q<0[/katex], write [katex]q=-p[/katex] with [katex]p>0. Then

(\log n)^q=\frac1{(\log n)^p}.

For sufficiently large n,

n>(\log n)^p,

hence

\frac1{(\log n)^p}>\frac1n.

Since \sum1/n diverges, comparison gives divergence here also.

Thus x=1 gives divergence for every real q.

Step 3: End point x=-1

We obtain

\sum_{n=2}^{\infty}(-1)^n(\log n)^q.

If q\ge0, the terms do not tend to zero, so the series diverges.

If q<0[/katex], put [katex]q=-p[/katex], [katex]p>0. Then

\sum_{n=2}^{\infty}\frac{(-1)^n}{(\log n)^p}.

The positive terms 1/(\log n)^p decrease eventually and tend to zero. Hence the alternating-series test gives convergence.

However, the absolute series \sum1/(\log n)^p diverges as shown above. Thus convergence at x=-1 is conditional when q<0[/katex].</p><p><strong>Final Answer:</strong></p> [katex display="true"] \boxed{ \begin{array}{ll} |x|<1:&\text{absolutely convergent for every real }q,\\ |x|>1:&\text{divergent},\\ x=1:&\text{divergent for every real }q,\\ x=-1:&\text{conditionally convergent if }q<0,\ \text{divergent if }q\ge0. \end{array}} [/katex] </div> </div><div id="q5a" class="math-qna-wrapper"> <div class="question-box"> <h2>HPAS 2017 Maths Optional Paper-2 Question 5(a)</h2> <p>Prove that the function [katex]u(x,y) = x^3 - 3xy^2 is harmonic and obtain its conjugate.

Solution:

Given

u(x,y)=x^3-3xy^2.

Condition for harmonicity

A twice continuously differentiable real-valued function u(x,y) is harmonic if

u_{xx}+u_{yy}=0.

Step 1: Compute the second partial derivatives

u_x=3x^2-3y^2,

so

u_{xx}=6x.

Also,

u_y=-6xy,

and therefore

u_{yy}=-6x.

Hence

u_{xx}+u_{yy}=6x-6x=0.

Thus u is harmonic.

Step 2: Use the Cauchy-Riemann equations to find the harmonic conjugate

If f(z)=u+iv is analytic, then

u_x=v_y,\qquad u_y=-v_x.

From v_y=u_x,

v_y=3x^2-3y^2.

Integrating with respect to y, treating x as constant,

v=3x^2y-y^3+\phi(x),

where \phi(x) is an arbitrary function of x.

Differentiate this expression with respect to x:

v_x=6xy+\phi'(x).

But the second Cauchy-Riemann equation gives

v_x=-u_y=6xy.

Therefore

6xy+\phi'(x)=6xy,

so

\phi'(x)=0.

Hence \phi(x)=C, a real constant.

Therefore a harmonic conjugate is

v(x,y)=3x^2y-y^3+C.

Final Answer:

\boxed{v(x,y)=3x^2y-y^3+C.}

Consequently, apart from an additive imaginary constant,

f(z)=z^3.

Solution:

Integral

I=\int_C\frac{e^{3z}}{z-\pi i}\,dz,

where C is the circle |z-1|=4.

Cauchy's integral formula

If f is analytic on and inside a positively oriented simple closed contour C and a lies inside C, then

\int_C\frac{f(z)}{z-a}\,dz=2\pi i\,f(a).

Step 1: Check whether the pole lies inside the contour

The only singularity is

z=\pi i.

Its distance from the centre 1 of the circle is

|\pi i-1| =\sqrt{(-1)^2+\pi^2} =\sqrt{1+\pi^2}.

Since

1+\pi^2<16, [/katex] <p>we have</p> [katex display="true"] \sqrt{1+\pi^2}<4. [/katex] <p>Hence [katex]z=\pi i lies inside C.

Step 2: Apply Cauchy's integral formula

Take

f(z)=e^{3z},\qquad a=\pi i.

Then

I=2\pi i\,e^{3\pi i}.

Using Euler's formula,

e^{3\pi i}=\cos3\pi+i\sin3\pi=-1.

Therefore

I=-2\pi i.

Final Answer:

\boxed{\displaystyle \int_C\frac{e^{3z}}{z-\pi i}\,dz=-2\pi i}.

Solution:

Given PDE

y(p^2+q^2)=qz,

where

p=\frac{\partial z}{\partial x},\qquad q=\frac{\partial z}{\partial y}.

Write

F(x,y,z,p,q)=y(p^2+q^2)-qz=0.

Charpit's equations

For F=0, Charpit's auxiliary equations are

\frac{dx}{F_p} = \frac{dy}{F_q} = \frac{dz}{pF_p+qF_q} = \frac{-dp}{F_x+pF_z} = \frac{-dq}{F_y+qF_z}.

Step 1: Calculate the partial derivatives of F

F_x=0,\qquad F_y=p^2+q^2,\qquad F_z=-q, F_p=2yp,\qquad F_q=2yq-z.

Also,

pF_p+qF_q =2yp^2+2yq^2-qz.

Since the original PDE gives y(p^2+q^2)=qz,

pF_p+qF_q =2qz-qz =qz.

Thus the relevant Charpit relations include

\frac{-dp}{-pq} = \frac{-dq}{p^2}.

Therefore

\frac{dp}{pq}=-\frac{dq}{p^2}.

Step 2: Obtain a first integral in p,q

Cross-multiplying,

p^2\,dp=-pq\,dq.

For the non-degenerate case p\ne0, divide by p:

p\,dp+q\,dq=0.

Integrating,

\frac12p^2+\frac12q^2=C.

Write 2C=a^2. Then

p^2+q^2=a^2. \tag{1}

Step 3: Use the original PDE

Substituting (1) into y(p^2+q^2)=qz,

a^2y=qz.

Hence

q=\frac{a^2y}{z}.

Since q=\partial z/\partial y,

\frac{\partial z}{\partial y} =\frac{a^2y}{z}.

Multiply by z:

z\frac{\partial z}{\partial y}=a^2y.

Integrating with respect to y,

\frac{z^2}{2} =\frac{a^2y^2}{2}+\Phi(x),

or, after writing 2\Phi(x)=\phi(x),

z^2=a^2y^2+\phi(x). \tag{2}

Step 4: Determine \phi(x)

Differentiate (2) with respect to x:

2zp=\phi'(x). \tag{3}

From (1),

p^2=a^2-q^2.

Since q=a^2y/z,

p^2 =a^2-\frac{a^4y^2}{z^2} =\frac{a^2(z^2-a^2y^2)}{z^2}.

By (2), z^2-a^2y^2=\phi(x). Thus

p^2=\frac{a^2\phi(x)}{z^2}.

Squaring (3),

[\phi'(x)]^2 =4z^2p^2 =4a^2\phi(x).

A convenient general solution of this equation is

\phi(x)=(ax+b)^2,

where b is a second arbitrary constant. Indeed,

\phi'(x)=2a(ax+b)

and therefore

[\phi'(x)]^2=4a^2(ax+b)^2=4a^2\phi(x).

Substituting this into (2),

z^2=a^2y^2+(ax+b)^2.

Verification

From the proposed solution,

p=\frac{a(ax+b)}{z},\qquad q=\frac{a^2y}{z}.

Then

p^2+q^2 = \frac{a^2(ax+b)^2+a^4y^2}{z^2} =a^2.

Hence

y(p^2+q^2)=a^2y=qz.

Thus the PDE is satisfied.

Final Answer:

\boxed{z^2=a^2y^2+(ax+b)^2},

where a and b are arbitrary constants.

Solution:

Given PDE

x^2r+2xys+y^2t=0,

where

r=z_{xx},\qquad s=z_{xy},\qquad t=z_{yy}.

Step 1: Introduce the Euler differential operator

Define

E=x\frac{\partial}{\partial x}+y\frac{\partial}{\partial y}.

Then

Ez=xz_x+yz_y=xp+yq.

Apply E once more:

E^2z =x\frac{\partial}{\partial x}(xp+yq) +y\frac{\partial}{\partial y}(xp+yq).

Now

\frac{\partial}{\partial x}(xp+yq) =p+xr+ys,

and

\frac{\partial}{\partial y}(xp+yq) =xs+q+yt.

Therefore

E^2z =x(p+xr+ys)+y(xs+q+yt).

Expanding,

E^2z =xp+yq+x^2r+2xys+y^2t.

Since Ez=xp+yq,

x^2r+2xys+y^2t=E^2z-Ez.

Hence the PDE becomes

E^2z-Ez=0. \tag{1}

Step 2: Choose variables adapted to the operator E

For x\ne0, put

\xi=\frac{y}{x},\qquad \eta=\log|x|.

Along a ray y/x=\xi=\text{constant}, scaling (x,y) changes only \eta. In these variables,

E=\frac{\partial}{\partial\eta}.

Therefore (1) becomes

\frac{\partial^2z}{\partial\eta^2} -\frac{\partial z}{\partial\eta}=0.

Step 3: Solve the resulting ordinary differential equation

For each fixed \xi, let Z(\eta)=z(\xi,\eta). Then

Z''-Z'=0.

The auxiliary equation is

m^2-m=0,

so

m(m-1)=0,

giving m=0,1. Hence

Z=A(\xi)+B(\xi)e^\eta,

where A and B are arbitrary functions because \xi is merely a parameter in the ODE.

Since e^\eta=|x|, on any region where x>0 this is x. Absorbing the sign locally into the arbitrary function gives the standard form

z=\phi\left(\frac{y}{x}\right) +x\,\psi\left(\frac{y}{x}\right).

Final Answer:

\boxed{ z=\phi\left(\frac{y}{x}\right) +x\,\psi\left(\frac{y}{x}\right) },

where \phi and \psi are arbitrary functions (locally for x\ne0).

Solution:

Required result

L\left\{\int_0^x\frac{1-e^{-u}}{u}\,du\right\}(p) = \frac1p\log\left(1+\frac1p\right).

Step 1: Use the Laplace transform of an integral

If

F(x)=\int_0^x f(u)\,du,

then

L\{F(x)\}(p)=\frac1pL\{f(x)\}(p).

Here

f(x)=\frac{1-e^{-x}}{x}.

Therefore

L\left\{\int_0^x\frac{1-e^{-u}}{u}\,du\right\}(p) = \frac1p L\left\{\frac{1-e^{-x}}{x}\right\}(p). \tag{1}

Step 2: Evaluate L\{(1-e^{-x})/x\}

Observe that

\int_0^1e^{-tx}\,dt = \left[-\frac1x e^{-tx}\right]_{0}^{1} = \frac{1-e^{-x}}{x}.

Hence

\frac{1-e^{-x}}{x} = \int_0^1e^{-tx}\,dt.

Therefore, for p>0,

L\left\{\frac{1-e^{-x}}{x}\right\}(p) = \int_0^\infty e^{-px} \left(\int_0^1e^{-tx}\,dt\right)dx.

Interchanging the order of integration,

= \int_0^1\left(\int_0^\infty e^{-(p+t)x}\,dx\right)dt.

The inner integral is

\int_0^\infty e^{-(p+t)x}\,dx = \frac1{p+t}.

Thus

L\left\{\frac{1-e^{-x}}{x}\right\}(p) = \int_0^1\frac{dt}{p+t}.

Integrating,

= [\log(p+t)]_0^1 = \log(p+1)-\log p.

Hence

L\left\{\frac{1-e^{-x}}{x}\right\}(p) = \log\left(\frac{p+1}{p}\right) = \log\left(1+\frac1p\right). \tag{2}

Step 3: Substitute into (1)

L\left\{\int_0^x\frac{1-e^{-u}}{u}\,du\right\}(p) = \frac1p\log\left(1+\frac1p\right).

Final Answer:

\boxed{ L\left\{\int_0^x\frac{1-e^{-u}}{u}\,du\right\}(p) = \frac1p\log\left(1+\frac1p\right) }.

Solution:

Required inverse transform

L^{-1}\left\{\frac{p^2}{(p^2+a^2)^2}\right\}(x).

Convolution theorem

If L\{f\}=F and L\{g\}=G, then

L^{-1}\{F(p)G(p)\} = (f*g)(x) = \int_0^x f(t)g(x-t)\,dt.

Step 1: Factor the transform

\frac{p^2}{(p^2+a^2)^2} = \frac{p}{p^2+a^2}\cdot \frac{p}{p^2+a^2}.

But

L\{\cos ax\} = \frac{p}{p^2+a^2}.

Therefore, by convolution,

L^{-1}\left\{\frac{p^2}{(p^2+a^2)^2}\right\} = \int_0^x \cos(at)\cos(a(x-t))\,dt.

Step 2: Evaluate the convolution integral

Use

\cos A\cos B = \frac12[\cos(A-B)+\cos(A+B)].

With A=at and B=a(x-t),

A-B=2at-ax,\qquad A+B=ax.

Hence

\cos(at)\cos(a(x-t)) = \frac12[\cos(2at-ax)+\cos(ax)].

Therefore

\int_0^x\cos(at)\cos(a(x-t))\,dt = \frac12\int_0^x\cos(2at-ax)\,dt + \frac12\int_0^x\cos(ax)\,dt.

The second integral is immediate:

\frac12\int_0^x\cos(ax)\,dt = \frac{x}{2}\cos(ax).

For the first integral,

\frac12\int_0^x\cos(2at-ax)\,dt = \frac12\left[ \frac{\sin(2at-ax)}{2a} \right]_0^x.

Thus

= \frac1{4a} [\sin(ax)-\sin(-ax)].

Since \sin(-ax)=-\sin(ax),

= \frac1{4a}[2\sin(ax)] = \frac{\sin(ax)}{2a}.

Combining the two parts,

L^{-1}\left\{\frac{p^2}{(p^2+a^2)^2}\right\} = \frac{x}{2}\cos(ax)+\frac{\sin(ax)}{2a}.

Final Answer:

\boxed{ L^{-1}\left\{\frac{p^2}{(p^2+a^2)^2}\right\}(x) = \frac12\left[x\cos(ax)+\frac{\sin(ax)}{a}\right] }.

Solution:

Problem setup

This is the classical brachistochrone problem. We seek the curve along which a particle slides, without friction, between two fixed points in the least time under uniform gravity.

Take the initial point as the origin. Let the x-axis be horizontal and let y be measured vertically downward. Then gravity acts in the positive y-direction.

Step 1: Express the speed using conservation of energy

If the particle starts from rest at y=0, then after descending a vertical distance y, the loss of potential energy equals the gain of kinetic energy:

mgy=\frac12mv^2.

Hence

v=\sqrt{2gy}.

Step 2: Write the time functional

For an element of arc length ds,

dt=\frac{ds}{v}.

Since

ds=\sqrt{dx^2+dy^2} =\sqrt{1+(y')^2}\,dx,

where y'=dy/dx, the total time is

T=\int dt = \frac1{\sqrt{2g}} \int \sqrt{\frac{1+(y')^2}{y}}\,dx.

Thus, apart from the constant factor 1/\sqrt{2g}, we have to extremize

\int F(y,y')\,dx, \qquad F(y,y')=\sqrt{\frac{1+(y')^2}{y}}.

Step 3: Apply Beltrami's identity

Because F does not contain x explicitly, the Euler-Lagrange equation has the first integral

F-y'\frac{\partial F}{\partial y'}=C.

Now

F=\frac{\sqrt{1+(y')^2}}{\sqrt y}.

Also,

\frac{\partial F}{\partial y'} = \frac{y'}{\sqrt y\sqrt{1+(y')^2}}.

Therefore

F-y'F_{y'} = \frac{\sqrt{1+(y')^2}}{\sqrt y} - \frac{(y')^2}{\sqrt y\sqrt{1+(y')^2}}.

Taking the common denominator,

F-y'F_{y'} = \frac{1+(y')^2-(y')^2} {\sqrt y\sqrt{1+(y')^2}} = \frac1{\sqrt{y(1+(y')^2)}}.

Hence

\frac1{\sqrt{y(1+(y')^2)}}=C.

Choose the positive constant in the form

C=\frac1{\sqrt{2a}},

where a>0. Then

y(1+(y')^2)=2a.

Thus

(y')^2=\frac{2a-y}{y}. \tag{1}

Step 4: Introduce a cycloidal parameter

Set

y=a(1-\cos\theta). \tag{2}

Then

2a-y = 2a-a(1-\cos\theta) = a(1+\cos\theta).

From (1),

\left(\frac{dy}{dx}\right)^2 = \frac{1+\cos\theta}{1-\cos\theta}.

Using

1+\cos\theta=2\cos^2\frac\theta2,\qquad 1-\cos\theta=2\sin^2\frac\theta2,

we get

\frac{dy}{dx}=\cot\frac\theta2

for the descending branch. Hence

\frac{dx}{dy}=\tan\frac\theta2.

Differentiate (2):

dy=a\sin\theta\,d\theta.

Therefore

dx = \tan\frac\theta2\,dy = a\sin\theta\tan\frac\theta2\,d\theta.

Now

\sin\theta\tan\frac\theta2 = 2\sin\frac\theta2\cos\frac\theta2 \cdot \frac{\sin(\theta/2)}{\cos(\theta/2)} = 2\sin^2\frac\theta2 = 1-\cos\theta.

Hence

dx=a(1-\cos\theta)\,d\theta.

Integrating,

x=a(\theta-\sin\theta)+C_1.

Taking the initial point at (0,0) and \theta=0 gives C_1=0.

Thus the required curve is represented parametrically by

x=a(\theta-\sin\theta),\qquad y=a(1-\cos\theta).

These are the parametric equations of a cycloid. The constant a and the terminal value of \theta are fixed by the coordinates of the second prescribed point.

Final Answer:

The curve of quickest descent is an arc of an inverted cycloid:

\boxed{ x=a(\theta-\sin\theta),\qquad y=a(1-\cos\theta) }.

Solution:

Equation

x\log_{10}x-1.2=0.

We use the Newton-Raphson method to obtain the root correct to four decimal places.

Newton-Raphson formula

For f(x)=0,

x_{n+1} = x_n-\frac{f(x_n)}{f'(x_n)}.

Here

f(x)=x\log_{10}x-1.2.

Since

\log_{10}x=\frac{\ln x}{\ln10},

we get

\frac{d}{dx}\left[x\log_{10}x\right] = \log_{10}x+\frac1{\ln10}.

Therefore

f'(x) = \log_{10}x+\frac1{\ln10}.

Step 1: Locate the positive root approximately

At x=2,

f(2)=2\log_{10}2-1.2 \approx2(0.30103)-1.2 =-0.59794<0.

At x=3,

f(3)=3\log_{10}3-1.2 \approx3(0.477121)-1.2 =0.231363>0.

Thus a root lies between 2 and 3. We choose x_0=2.7.

Step 2: First Newton iteration

For x_0=2.7,

f(2.7)\approx-0.0353178368,

and

f'(2.7) = \log_{10}(2.7)+\frac1{\ln10} \approx0.8656582461.

Hence

x_1 = 2.7-\frac{-0.0353178368}{0.8656582461} \approx2.7407988221.

Step 3: Second Newton iteration

At x_1=2.7407988221,

f(x_1)\approx0.0001332015,

and

f'(x_1)\approx0.8721716409.

Therefore

x_2 = 2.7407988221 - \frac{0.0001332015}{0.8721716409} \approx2.7406460981.

Step 4: Third Newton iteration

At x_2=2.7406460981,

f(x_2)\approx1.85\times10^{-9},

and

f'(x_2)\approx0.8721474403.

Thus

x_3 = 2.7406460981 - \frac{1.85\times10^{-9}}{0.8721474403} \approx2.7406460960.

The successive values now agree to considerably more than four decimal places.

Final Answer:

\boxed{x\approx2.7406}

correct to four decimal places.