Assuming the particle moves along the curve from (t=0) to (t=1):

[ \vec F=2xy,\vec i-3x,\vec j-5z,\vec k ]

Given curve:

[ \vec r(t)=t,\vec i+t^2,\vec j+t^3,\vec k ]

[ d\vec r=(1,\vec i+2t,\vec j+3t^2,\vec k),dt ]

Now,

[ \vec F(t)=2t(t^2)\vec i-3t,\vec j-5t^3\vec k ]

[ \vec F(t)=2t^3\vec i-3t\vec j-5t^3\vec k ]

Work done is

[ W=\int_C \vec F\cdot d\vec r ]

[ W=\int_0^1 (2t^3\vec i-3t\vec j-5t^3\vec k)\cdot(1\vec i+2t\vec j+3t^2\vec k),dt ]

[ W=\int_0^1 (2t^3-6t^2-15t^5),dt ]

[ W=\left[\frac{t^4}{2}-2t^3-\frac{5t^6}{2}\right]_0^1 ]

[ W=\frac12-2-\frac52=-4 ]

[ \boxed{W=-4\text{ units}} ]

So the total work done is (\boxed{-4}) units.
If the limits of (t) are not given, the general expression is

[ W=\frac{t^4}{2}-2t^3-\frac{5t^6}{2} ]

and the value depends on the interval of (t).

Given: [ \vec F=2xy,\vec i-3x,\vec j+5z,\vec k ] Curve: [ x=t,\quad y=t^2+1,\quad z=2t^2,\quad t=0\to 1 ]

Formula: [ W=\int_C \vec F\cdot d\vec r =\int_C (F_x,dx+F_y,dy+F_z,dz) ]

Here, [ dx=dt,\quad dy=2t,dt,\quad dz=4t,dt ]

So, [ W=\int_0^1 \left[2xy,dx-3x,dy+5z,dz\right] ]

Substitute (x=t,\ y=t^2+1,\ z=2t^2): [ W=\int_0^1 \left[2t(t^2+1),dt-3t(2t,dt)+5(2t^2)(4t,dt)\right] ]

[ W=\int_0^1 \left(2t^3+2t-6t^2+40t^3\right)dt ]

[ W=\int_0^1 (42t^3-6t^2+2t),dt ]

[ W=\left[\frac{42t^4}{4}-\frac{6t^3}{3}+\frac{2t^2}{2}\right]_0^1 ]

[ W=\left[\frac{21}{2}t^4-2t^3+t^2\right]_0^1 ]

[ W=\frac{21}{2}-2+1 ]

[ W=\frac{21}{2}-1=\frac{19}{2} ]

[ \boxed{W=\frac{19}{2}\ \text{units}} ]

So, the total work done is (\boxed{\frac{19}{2}}) units.

General definition of definite integral

Let $f(x)$ be defined on $[a,b]$. Divide $[a,b]$ into $n$ subintervals by
$a=x_0<x_1<x_2<\cdots<x_n=b.$ Let $\Delta x_i=x_i-x_{i-1}$ and choose any point $\xi_i\in[x_{i-1},x_i]$. Form the Riemann sum $S_n=\sum_{i=1}^{n} f(\xi_i)\Delta x_i.$ If $\lim_{\max \Delta x_i\to 0} S_n$ exists and is independent of the choice of partition and points $\xi_i$, then this limit is called the definite integral of $f$ from $a$ to $b$: $\int_a^b f(x)\,dx=\lim_{\max \Delta x_i\to 0}\sum_{i=1}^{n} f(\xi_i)\Delta x_i.$

For equal subintervals, $\Delta x=\dfrac{b-a}{n}$ and $\xi_i=a+i\Delta x$, so $\int_a^b f(x)\,dx=\lim_{n\to\infty}\frac{b-a}{n}\sum_{i=1}^{n} f\left(a+i\frac{b-a}{n}\right).$

Fundamental theorem of integral calculus

If $f$ is continuous on $[a,b]$ and $F$ is an antiderivative of $f$, i.e. $F'(x)=f(x)$, then $\int_a^b f(x)\,dx=F(b)-F(a).$ Also, $\frac{d}{dx}\int_a^x f(t)\,dt=f(x).$

Example: $\int_0^1 x^2\,dx=\left[\frac{x^3}{3}\right]_0^1=\frac{1}{3}-0=\frac{1}{3}.$

Evaluation by the method of summation

Let $I=\int_0^{\pi}\sin^2 x\,dx.$ Divide $[0,\pi]$ into $n$ equal parts, each of width $h=\frac{\pi}{n}.$ Taking right endpoints $x_i=ih$, we get $I=\lim_{n\to\infty} h\sum_{i=1}^{n}\sin^2(ih).$

Using $\sin^2\theta=\dfrac{1-\cos 2\theta}{2}$, $I=\lim_{n\to\infty} h\sum_{i=1}^{n}\frac{1-\cos(2ih)}{2}$ $=\lim_{n\to\infty}\frac{h}{2}\left[n-\sum_{i=1}^{n}\cos(2ih)\right].$

Now, $\sum_{i=1}^{n}\cos(2ih)=\sum_{i=1}^{n}\cos\left(\frac{2\pi i}{n}\right)=0.$ Hence,

$$I=\lim_{n\to\infty}\frac{h}{2}\cdot n =\lim_{n\to\infty}\frac{\pi/n}{2}\cdot n =\frac{\pi}{2}.$$

Therefore, $\boxed{\int_0^{\pi}\sin^2 x\,dx=\frac{\pi}{2}}$

Curvature. The curvature of a curve at a point is the rate at which the tangent direction turns per unit arc length: $K = \dfrac{d\psi}{ds}$, where $\psi$ is the angle the tangent makes with a fixed line and $s$ is the arc length.

Radius of curvature. It is the reciprocal of the curvature, $\rho = \dfrac{1}{K} = \dfrac{ds}{d\psi}$; it is the radius of the circle that touches the curve at the point with the same tangent and curvature.

For a polar curve $r = f(\theta)$,

$\rho = \frac{(r^2 + r_1^2)^{3/2}}{r^2 + 2r_1^2 - r\,r_2}, \qquad r_1 = \frac{dr}{d\theta},\; r_2 = \frac{d^2r}{d\theta^2}.$

For $r = a\cos\theta$: $r_1 = -a\sin\theta$, $r_2 = -a\cos\theta$.

$r^2 + r_1^2 = a^2, \qquad r^2 + 2r_1^2 - r r_2 = a^2\cos^2\theta + 2a^2\sin^2\theta + a^2\cos^2\theta = 2a^2.$

$\rho = \frac{a^3}{2a^2} = \frac{a}{2}.$

This is constant, as expected, because $r = a\cos\theta$ is a circle of diameter $a$ through the pole. The expression $\frac{2a}{3}\cos\theta$ in the question does not hold for this curve (at $\theta = 0$ the circle has $\rho = a/2$, not $2a/3$); it appears to be a misprint for a different curve.

Right angle at the pole. The circle $r = a\cos\theta$ has the diameter $OA = a$ along the initial line, with $O$ the pole. For any point $P$ on the curve, the angle in a semicircle gives $\angle OPA = 90^\circ$. Also, the angle $\phi$ between the radius vector and the tangent satisfies $\tan\phi = r/r_1 = -\cot\theta$, so $\phi = 90^\circ + \theta$. At the pole ($\theta = \pi/2$) the tangent is perpendicular to the initial line, which is the diameter $OA$.

Exact differential equation:
A differential equation of the form
$M(x,y)\,dx+N(x,y)\,dy=0$
is called exact if there exists a function $F(x,y)$ such that
$dF=M\,dx+N\,dy.$
Equivalently, it is exact when
$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}.$

Example:
$y\,dx+x\,dy=0$
is exact because
$\frac{\partial}{\partial y}(y)=1=\frac{\partial}{\partial x}(x).$
Its solution is $xy=C$.

Integrating factor:
If $M\,dx+N\,dy=0$ is not exact, then a multiplier $\mu(x,y)$ such that
$\mu M\,dx+\mu N\,dy=0$
becomes exact is called an integrating factor.

---

Solve:
$(1+xy)x\,dy+(1-xy)y\,dx=0$

Rewrite as
$y(1-xy)\,dx+x(1+xy)\,dy=0.$

Here
$M=y-xy^2,\qquad N=x+x^2y.$

Check exactness:
$\frac{\partial M}{\partial y}=1-2xy,\qquad \frac{\partial N}{\partial x}=1+2xy.$
Since they are not equal, the equation is not exact.

Take integrating factor
$\mu=\frac{1}{x^2y^2}.$

Then
$M'=\frac{y-xy^2}{x^2y^2}=\frac{1}{x^2y}-\frac{1}{x},$
$N'=\frac{x+x^2y}{x^2y^2}=\frac{1}{xy^2}+\frac{1}{y}.$

Now
$\frac{\partial M'}{\partial y}=-\frac{1}{x^2y^2}=\frac{\partial N'}{\partial x},$
so the equation becomes exact.

Let the solution be $F(x,y)=C$. Then
$F=\int M'\,dx=\int\left(\frac{1}{x^2y}-\frac{1}{x}\right)dx$
$F=-\frac{1}{xy}-\ln|x|+\phi(y).$

Differentiate with respect to $y$:
$\frac{\partial F}{\partial y}=\frac{1}{xy^2}+\phi'(y).$

But this must equal $N'$, so
$\frac{1}{xy^2}+\phi'(y)=\frac{1}{xy^2}+\frac{1}{y}.$
Hence
$\phi'(y)=\frac{1}{y}\quad\Rightarrow\quad \phi(y)=\ln|y|.$

Therefore,
$F=-\frac{1}{xy}-\ln|x|+\ln|y|=C.$

So the general solution is
$\boxed{\ln\left|\frac{y}{x}\right|-\frac{1}{xy}=C}$

Also, $x=0$ and $y=0$ are singular solutions.

For a polar curve (r=f(\theta)), the pedal equation is obtained from

[ \frac{1}{p^2}=\frac{1}{r^2}+\frac{1}{r^4}\left(\frac{dr}{d\theta}\right)^2 ]

where (p) is the perpendicular distance from the pole to the tangent.

Given,

[ r^2\cos 2\theta=a^2 ]

[ r^2=a^2\sec 2\theta ]

Differentiate with respect to (\theta):

[ 2r\frac{dr}{d\theta}=a^2\cdot 2\sec 2\theta \tan 2\theta ]

[ \frac{dr}{d\theta}=\frac{a^2\sec 2\theta \tan 2\theta}{r} ]

Since (a^2\sec 2\theta=r^2),

[ \frac{dr}{d\theta}=r\tan 2\theta ]

Now,

[ \frac{1}{p^2}=\frac{1}{r^2}+\frac{1}{r^4}\left(r\tan 2\theta\right)^2 ]

[ \frac{1}{p^2}=\frac{1}{r^2}+\frac{\tan^2 2\theta}{r^2} ]

[ \frac{1}{p^2}=\frac{1+\tan^2 2\theta}{r^2} ]

[ \frac{1}{p^2}=\frac{\sec^2 2\theta}{r^2} ]

[ p^2=r^2\cos^2 2\theta ]

But from the given curve,

[ r^2\cos 2\theta=a^2 ]

Hence,

[ p=\frac{a^2}{r} ]

Therefore, the pedal equation is

[ \boxed{rp=a^2} ]

or equivalently,

[ \boxed{p=\frac{a^2}{r}} ]

Let
[ L=\lim_{x\to 0}\left(\frac{\sin x}{x}\right)^{1/x}. ] Take logarithm: [ \ln L=\lim_{x\to 0}\frac{\ln\left(\frac{\sin x}{x}\right)}{x}. ] This is a (0/0) form. Using the expansion [ \frac{\sin x}{x}=1-\frac{x^2}{6}+\frac{x^4}{120}-\cdots, ] we get [ \ln\left(\frac{\sin x}{x}\right) =\ln\left(1-\frac{x^2}{6}+\cdots\right) =-\frac{x^2}{6}+O(x^4). ] Hence [ \ln L=\lim_{x\to 0}\frac{-\frac{x^2}{6}+O(x^4)}{x} =\lim_{x\to 0}\left(-\frac{x}{6}+O(x^3)\right)=0. ] Therefore, [ \ln L=0 \implies L=e^0=1. ]

[ \boxed{\lim_{x\to 0}\left(\frac{\sin x}{x}\right)^{1/x}=1} ]

The curve is
[ F(x,y)=(x-1)(x-2)(x+y)+x^2+x+1=0. ] Expanding, [ F(x,y)=x^3+x^2y-2x^2-3xy+3x+2y+1. ]

1. Asymptotes parallel to the (y)-axis

Write (F(x,y)) as a polynomial in (y): [ F(x,y)=(x^2-3x+2)y+(x^3-2x^2+3x+1). ] For a vertical asymptote (x=k), the coefficient of the highest power of (y) must vanish: [ k^2-3k+2=0 ] [ (k-1)(k-2)=0 \implies k=1,;2. ] Hence the vertical asymptotes are [ \boxed{x=1,\qquad x=2.} ]

2. Oblique asymptote

Let the asymptote be [ y=mx+c. ] Substitute into (F(x,y)=0): [ F(x,mx+c)=x^3+x^2(mx+c)-2x^2-3x(mx+c)+3x+2(mx+c)+1. ] Collecting powers of (x), [ F(x,mx+c)=(1+m)x^3+(c-2-3m)x^2+(3-3c+2m)x+(2c+1). ] For an asymptote, the coefficients of (x^3) and (x^2) must vanish: [ 1+m=0 \implies m=-1, ] [ c-2-3m=0 \implies c-2+3=0 \implies c=-1. ] Thus the oblique asymptote is [ y=-x-1, ] or [ \boxed{x+y+1=0.} ]

Therefore, the asymptotes of the given curve are [ \boxed{x=1,\quad x=2,\quad x+y+1=0.} ]

Euler’s Theorem on Homogeneous Functions of Two Variables

Statement:
If (z=f(x,y)) is a homogeneous function of degree (n), i.e. [ f(tx,ty)=t^n f(x,y), ] then [ x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=nz. ]

Proof:
Since (f) is homogeneous of degree (n), [ f(tx,ty)=t^n f(x,y). ] Differentiate both sides with respect to (t): [ \frac{d}{dt}f(tx,ty)=nt^{n-1}f(x,y). ] By the chain rule, [ x f_x(tx,ty)+y f_y(tx,ty)=nt^{n-1}f(x,y). ] Put (t=1): [ x f_x(x,y)+y f_y(x,y)=n f(x,y). ] Hence, [ \boxed{x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=nz}. ]

---

OR

Given [ u=\tan^{-1}\frac{x^3+y^3}{x-y},\qquad x\neq y. ] Let [ v=\tan u=\frac{x^3+y^3}{x-y}. ] Then [ u=\tan^{-1}v. ] Differentiating, [ \frac{\partial u}{\partial x}=\frac{1}{1+v^2}\frac{\partial v}{\partial x}, \qquad \frac{\partial u}{\partial y}=\frac{1}{1+v^2}\frac{\partial v}{\partial y}. ] Therefore, [ x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}

\frac{1}{1+v^2} \left( x\frac{\partial v}{\partial x}+y\frac{\partial v}{\partial y} \right). ]

Now, [ v=\frac{x^3+y^3}{x-y} ] is a homogeneous function of degree (2), because numerator is of degree (3) and denominator is of degree (1). Hence, by Euler’s theorem, [ x\frac{\partial v}{\partial x}+y\frac{\partial v}{\partial y}=2v. ] So, [ x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}

\frac{2v}{1+v^2}. ] Since (v=\tan u), [ \frac{2v}{1+v^2}

\frac{2\tan u}{1+\tan^2 u}

2\sin u\cos u

\sin 2u. ] Thus, [ \boxed{x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=\sin 2u}. ]

Hence proved.

The cycloid has one arch for
[ -\pi \le \theta \le \pi, ] since (y=a(1+\cos\theta)\ge 0), and (y=0) at (\theta=\pm\pi).

Surface area generated by revolving about the base (x)-axis is [ S=\int 2\pi y,ds. ]

Given [ x=a(\theta+\sin\theta),\qquad y=a(1+\cos\theta), ] we have [ \frac{dx}{d\theta}=a(1+\cos\theta),\qquad \frac{dy}{d\theta}=-a\sin\theta. ] Thus [ ds=\sqrt{\left(\frac{dx}{d\theta}\right)^2+\left(\frac{dy}{d\theta}\right)^2},d\theta ] [ =a\sqrt{(1+\cos\theta)^2+\sin^2\theta},d\theta ] [ =a\sqrt{2+2\cos\theta},d\theta =2a\cos\frac{\theta}{2},d\theta. ]

Also, [ y=a(1+\cos\theta)=2a\cos^2\frac{\theta}{2}. ]

Therefore, [ S=\int_{-\pi}^{\pi}2\pi\left(2a\cos^2\frac{\theta}{2}\right) \left(2a\cos\frac{\theta}{2}\right)d\theta ] [ =8\pi a^2\int_{-\pi}^{\pi}\cos^3\frac{\theta}{2},d\theta. ]

Put (t=\frac{\theta}{2}), so (d\theta=2dt). Limits become (-\frac{\pi}{2}) to (\frac{\pi}{2}).

[ S=16\pi a^2\int_{-\pi/2}^{\pi/2}\cos^3 t,dt ] [ =16\pi a^2\left[\sin t-\frac{\sin^3 t}{3}\right]_{-\pi/2}^{\pi/2} ] [ =16\pi a^2\left(\frac{2}{3}+\frac{2}{3}\right) =\frac{64\pi a^2}{3}. ]

[ \boxed{S=\frac{64\pi a^2}{3}\text{ square units}} ]

First alternative

Let the cylindrical surface be parametrized by
[ x=2\cos\theta,\qquad y=2\sin\theta,\qquad z=z ] where [ 0\le \theta\le \frac{\pi}{2},\qquad 0\le z\le 4. ]

The outward unit normal to the cylinder is [ \hat n=\cos\theta,\vec i+\sin\theta,\vec j, ] and [ dS=2,d\theta,dz. ]

Now [ \vec F=yz,\vec i+xz,\vec j-2x^2z,\vec k ] becomes [ \vec F=2z\sin\theta,\vec i+2z\cos\theta,\vec j-8z\cos^2\theta,\vec k. ]

Hence [ \vec F\cdot \hat n =2z\sin\theta\cos\theta+2z\cos\theta\sin\theta =4z\sin\theta\cos\theta. ]

Therefore, [ \iint_S \vec F\cdot \hat n,dS =\int_0^4\int_0^{\pi/2}4z\sin\theta\cos\theta\cdot 2,d\theta,dz ] [ =8\left(\int_0^4 z,dz\right)\left(\int_0^{\pi/2}\sin\theta\cos\theta,d\theta\right) ] [ =8\cdot 8\cdot \frac12=32. ]

[ \boxed{\iint_S \vec F\cdot \hat n,dS=32} ]

---

OR alternative

Given [ \vec F=(2x-y)\vec i-yz^2\vec j-y^2z\vec k. ]

Compute (\nabla\times \vec F): [ \nabla\times \vec F

\begin{vmatrix} \vec i&\vec j&\vec k\ \dfrac{\partial}{\partial x}&\dfrac{\partial}{\partial y}&\dfrac{\partial}{\partial z}\ 2x-y&-yz^2&-y^2z \end{vmatrix} =\vec k. ]

By Stokes’ theorem, [ \oint_C \vec F\cdot d\vec r

\iint_S (\nabla\times \vec F)\cdot \hat n,dS

\iint_S \vec k\cdot \hat n,dS. ]

For the sphere (x^2+y^2+z^2=a^2), the outward unit normal is [ \hat n=\frac{x\vec i+y\vec j+z\vec k}{a}. ] Thus [ \vec k\cdot \hat n=\frac{z}{a}. ]

Using spherical coordinates, [ z=a\cos\phi,\qquad dS=a^2\sin\phi,d\phi,d\theta, ] with [ 0\le \phi\le \frac{\pi}{2},\qquad 0\le \theta\le 2\pi. ]

So [ \oint_C \vec F\cdot d\vec r

\int_0^{2\pi}\int_0^{\pi/2} \frac{a\cos\phi}{a}\cdot a^2\sin\phi,d\phi,d\theta ] [

a^2\int_0^{2\pi}d\theta\int_0^{\pi/2}\cos\phi\sin\phi,d\phi ] [

a^2(2\pi)\left[\frac{\sin^2\phi}{2}\right]_0^{\pi/2}

\pi a^2. ]

[ \boxed{\oint_C \vec F\cdot d\vec r=\pi a^2} ]

If the boundary (C) is taken in the opposite direction, the sign becomes (-\pi a^2).

Assuming the intended integral is
[ \int_0^{\pi}\int_0^{\pi(1+\cos\theta)} r^3\sin\theta\cos\theta,dr,d\theta ] (the limits indicate the inner integration is with respect to (r)).

First integrate with respect to (r): [ \int_0^{\pi(1+\cos\theta)} r^3,dr =\left[\frac{r^4}{4}\right]_0^{\pi(1+\cos\theta)} =\frac{\pi^4}{4}(1+\cos\theta)^4. ]

So the integral becomes [ I=\frac{\pi^4}{4}\int_0^\pi (1+\cos\theta)^4\sin\theta\cos\theta,d\theta. ]

Let [ u=1+\cos\theta,\qquad du=-\sin\theta,d\theta,\qquad \cos\theta=u-1. ] When (\theta=0), (u=2); when (\theta=\pi), (u=0). Hence [ I=\frac{\pi^4}{4}\int_2^0 u^4(u-1)(-du) =\frac{\pi^4}{4}\int_0^2 (u^5-u^4),du. ]

Now [ I=\frac{\pi^4}{4}\left[\frac{u^6}{6}-\frac{u^5}{5}\right]_0^2 =\frac{\pi^4}{4}\left(\frac{64}{6}-\frac{32}{5}\right). ] [ \frac{64}{6}-\frac{32}{5} =\frac{32}{3}-\frac{32}{5} =\frac{160-96}{15} =\frac{64}{15}. ] Therefore [ I=\frac{\pi^4}{4}\cdot\frac{64}{15} =\frac{16\pi^4}{15}. ]

[ \boxed{\frac{16\pi^4}{15}} ]

The given equation is Clairaut’s equation of the form
[ y=px+f(p),\qquad f(p)=\frac{ap}{\sqrt{1+p^2}}. ]

General solution

Replace (p) by an arbitrary constant (c): [ \boxed{y=cx+\frac{ac}{\sqrt{1+c^2}}} ]

Singular solution

Differentiate the given equation partially with respect to (p): [ x+f'(p)=0 ] Now, [ f(p)=\frac{ap}{\sqrt{1+p^2}} ] [ f'(p)=a\frac{d}{dp}\left(\frac{p}{\sqrt{1+p^2}}\right) =a\cdot \frac{1}{(1+p^2)^{3/2}} ] Hence, [ x+\frac{a}{(1+p^2)^{3/2}}=0 ] [ x=-\frac{a}{(1+p^2)^{3/2}} ]

Put (p=\tan\theta). Then [ \sqrt{1+p^2}=\sec\theta ] So, [ x=-a\cos^3\theta ] Also, [ y=px+\frac{ap}{\sqrt{1+p^2}} ] [ y=\tan\theta(-a\cos^3\theta)+a\sin\theta ] [ y=-a\sin\theta\cos^2\theta+a\sin\theta =a\sin\theta(1-\cos^2\theta) =a\sin^3\theta ]

Thus the singular solution in parametric form is [ x=-a\cos^3\theta,\qquad y=a\sin^3\theta ] Eliminating (\theta), [ x^{2/3}+y^{2/3}=a^{2/3} ]

[ \boxed{\text{General solution: } y=cx+\frac{ac}{\sqrt{1+c^2}}} ] [ \boxed{\text{Singular solution: } x^{2/3}+y^{2/3}=a^{2/3}} ]

The singular solution is the envelope of the family of general solutions.

Solve (\frac{d^2y}{dx^2}-y=x^2\cos x)

Given: [ (D^2-1)y=x^2\cos x ]

Complementary function (C.F.): [ D^2-1=0 \Rightarrow D=\pm 1 ] [ \text{C.F.}=C_1e^x+C_2e^{-x} ]

Particular integral (P.I.): [ \text{P.I.}=\frac{1}{D^2-1}x^2\cos x ] Using (\cos x=\Re(e^{ix})), [ \text{P.I.}=\Re\left[\frac{1}{D^2-1}x^2e^{ix}\right] ] Now, [ \frac{1}{D^2-1}x^2e^{ix} =e^{ix}\frac{1}{(D+i)^2-1}x^2 ] [ =e^{ix}\frac{1}{D^2+2iD-2}x^2 ] [ =e^{ix}\left(-\frac12-\frac{i}{2}D+\frac14D^2\right)x^2 ] [ =e^{ix}\left(-\frac{x^2}{2}-ix+\frac12\right) ] Taking real part, [ \text{P.I.}=\left(\frac{1-x^2}{2}\right)\cos x+x\sin x ]

Hence the general solution is [ \boxed{y=C_1e^x+C_2e^{-x}+\left(\frac{1-x^2}{2}\right)\cos x+x\sin x} ]

---

OR

Complementary function:
The general solution of the homogeneous equation obtained by putting the right-hand side equal to zero is called the complementary function. It contains arbitrary constants equal to the order of the differential equation.

Particular integral:
Any solution of the non-homogeneous equation, containing no arbitrary constants, is called a particular integral.
General solution = C.F. + P.I.

Proof: [ \frac{1}{D-\alpha}Q=e^{\alpha x}\int Qe^{-\alpha x},dx ]

Let [ y=e^{\alpha x}\int Qe^{-\alpha x},dx ] Differentiate: [ \frac{dy}{dx}=\alpha e^{\alpha x}\int Qe^{-\alpha x},dx+e^{\alpha x}Qe^{-\alpha x} ] [ \frac{dy}{dx}=\alpha y+Q ] [ \frac{dy}{dx}-\alpha y=Q ] [ (D-\alpha)y=Q ] Therefore, [ y=\frac{1}{D-\alpha}Q ] Hence, [ \boxed{\frac{1}{D-\alpha}Q=e^{\alpha x}\int Qe^{-\alpha x},dx} ] Proved.

1. Definition of a first-order linear differential equation

A differential equation of the first order is called linear if it can be written in the form

[ \frac{dy}{dx}+P(x)y=Q(x), ]

where (P(x)) and (Q(x)) are functions of (x) only, or constants.
It is linear in (y) and (\frac{dy}{dx}), i.e. (y) and (\frac{dy}{dx}) occur only in the first degree and are not multiplied together.

If (Q(x)=0), it is called homogeneous; otherwise it is non-homogeneous.

---

2. Method of solving a first-order linear differential equation

Write the equation in standard form:

[ \frac{dy}{dx}+P(x)y=Q(x). ]

Find the integrating factor:

[ \text{I.F.}=e^{\int P(x),dx}. ]

Multiply both sides of the equation by the integrating factor:

[ e^{\int P,dx}\frac{dy}{dx}+P(x)e^{\int P,dx}y=Q(x)e^{\int P,dx}. ]

The left-hand side becomes the derivative of (y) times the integrating factor:

[ \frac{d}{dx}\left(y,e^{\int P,dx}\right)=Q(x)e^{\int P,dx}. ]

Integrating both sides,

[ y,e^{\int P,dx}=\int Q(x)e^{\int P,dx},dx+C. ]

Hence the general solution is

[ y=e^{-\int P,dx}\left[\int Q(x)e^{\int P,dx},dx+C\right]. ]

---

3. Reduction of (\frac{dy}{dx}+2xy=x^3y^3) to linear form and solution

Given:

[ \frac{dy}{dx}+2xy=x^3y^3. ]

This is a Bernoulli equation of the form

[ \frac{dy}{dx}+P(x)y=Q(x)y^n, ]

with (P(x)=2x,\ Q(x)=x^3,\ n=3).

Divide throughout by (y^3):

[ y^{-3}\frac{dy}{dx}+2xy^{-2}=x^3. ]

Let

[ v=y^{-2}. ]

Then

[ \frac{dv}{dx}=-2y^{-3}\frac{dy}{dx}, ]

so

[ y^{-3}\frac{dy}{dx}=-\frac12\frac{dv}{dx}. ]

Substituting in the equation,

[ -\frac12\frac{dv}{dx}+2xv=x^3. ]

Multiplying by (-2),

[ \frac{dv}{dx}-4xv=-2x^3. ]

This is a first-order linear differential equation in (v), with

[ P(x)=-4x,\qquad Q(x)=-2x^3. ]

Its integrating factor is

[ \text{I.F.}=e^{\int -4x,dx}=e^{-2x^2}. ]

Multiplying the linear equation by (e^{-2x^2}),

[ e^{-2x^2}\frac{dv}{dx}-4xe^{-2x^2}v=-2x^3e^{-2x^2}. ]

[ \frac{d}{dx}\left(v e^{-2x^2}\right)=-2x^3e^{-2x^2}. ]

Integrating both sides,

[ v e^{-2x^2}=\int -2x^3e^{-2x^2},dx+C. ]

Put (t=x^2), so that (dt=2x,dx). Then

[ \int -2x^3e^{-2x^2},dx

-\int x^2e^{-2x^2}(2x,dx)

-\int t e^{-2t},dt. ]

Now,

[ \int t e^{-2t},dt=e^{-2t}\left(-\frac{t}{2}-\frac14\right). ]

Therefore,

[ -\int t e^{-2t},dt

e^{-2t}\left(\frac{t}{2}+\frac14\right). ]

Hence,

[ v e^{-2x^2}

e^{-2x^2}\left(\frac{x^2}{2}+\frac14\right)+C. ]

Multiplying by (e^{2x^2}),

[ v=\frac{x^2}{2}+\frac14+C e^{2x^2}. ]

Since (v=y^{-2}),

[ \frac{1}{y^2}=\frac{x^2}{2}+\frac14+C e^{2x^2}. ]

Thus,

[ y^2=\frac{1}{\frac{x^2}{2}+\frac14+C e^{2x^2}}. ]

Therefore,

[ \boxed{ y=\pm \frac{1}{\sqrt{\frac{x^2}{2}+\frac14+C e^{2x^2}}} } ]

where (C) is an arbitrary constant.
Also, (y=0) is a solution, which is excluded during the substitution (v=y^{-2}).

Pedal equation. The pedal equation of a curve is the relation between $p$, the perpendicular distance from the pole (origin) to the tangent at a point $P$, and $r$, the distance of $P$ from the pole.

Deduction from the Cartesian equation $f(x,y)=0$.

Let the tangent at $P(x,y)$ make angle $\psi$ with the $x$-axis and let $OP$ make angle $\theta$, so $\tan\psi=\dfrac{dy}{dx}$, $\tan\theta=\dfrac{y}{x}$. The angle between $OP$ and the tangent is $\phi=\psi-\theta$, and $p=r\sin\phi$.

$\tan\phi=\frac{\tan\psi-\tan\theta}{1+\tan\psi\tan\theta}=\frac{x\,dy-y\,dx}{x\,dx+y\,dy}.$

Since $(x\,dx+y\,dy)^2+(x\,dy-y\,dx)^2=(x^2+y^2)(dx^2+dy^2)$, $\sin\phi=\frac{x\,dy-y\,dx}{\sqrt{(x^2+y^2)(dx^2+dy^2)}}\quad\Rightarrow\quad p=r\sin\phi=\frac{x\,dy-y\,dx}{\sqrt{dx^2+dy^2}}=\frac{x\frac{dy}{dx}-y}{\sqrt{1+\left(\frac{dy}{dx}\right)^2}} .$

Procedure. Substitute $\dfrac{dy}{dx}$ from $f(x,y)=0$ into this expression for $p$, then eliminate $x,y$ between $f(x,y)=0$, $r^2=x^2+y^2$ and the expression for $p$. The result is a relation $F(p,r)=0$, the pedal equation.

Seven indeterminate forms

The seven standard indeterminate forms are:

1. $\dfrac{0}{0}$
2. $\dfrac{\infty}{\infty}$
3. $0 \times \infty$
4. $\infty - \infty$
5. $0^0$
6. $1^\infty$
7. $\infty^0$

---

Prove that

[ \lim_{x \to 0} \frac{\tan x - \sin x}{x^3} = \frac{1}{2} ]

We have [ \tan x = \frac{\sin x}{\cos x} ] so [ \tan x - \sin x = \frac{\sin x}{\cos x} - \sin x = \sin x\left(\frac{1}{\cos x} - 1\right) = \frac{\sin x(1-\cos x)}{\cos x} ]

Therefore, [ \frac{\tan x - \sin x}{x^3}

\frac{\sin x(1-\cos x)}{x^3 \cos x} ]

Now write it as a product: [ \frac{\tan x - \sin x}{x^3}

\left(\frac{\sin x}{x}\right) \left(\frac{1-\cos x}{x^2}\right) \left(\frac{1}{\cos x}\right) ]

Taking the limit as $x \to 0$, [ \lim_{x \to 0} \frac{\sin x}{x} = 1 ] [ \lim_{x \to 0} \frac{1-\cos x}{x^2} = \frac{1}{2} ] [ \lim_{x \to 0} \frac{1}{\cos x} = 1 ]

Hence, [ \lim_{x \to 0} \frac{\tan x - \sin x}{x^3}

1 \cdot \frac{1}{2} \cdot 1

\frac{1}{2} ]

[ \boxed{\lim_{x \to 0} \frac{\tan x - \sin x}{x^3} = \frac{1}{2}} ]

Hence proved.

Trace of the curve
[ y^2=(x-2a)^3 ] Assume (a>0).

1. Symmetry
   The equation contains (y) only as (y^2), so the curve is symmetric about the (x)-axis.

2. Domain
   [ y^2\ge 0 \implies (x-2a)^3\ge 0 \implies x\ge 2a. ] Hence no part of the curve lies to the left of the line (x=2a).

3. Intercepts
   Put (y=0): [ (x-2a)^3=0 \implies x=2a. ] So the curve meets the (x)-axis only at ((2a,0)).
   For (x=0), (y^2=-8a^3<0), so there is no (y)-intercept.

4. Special point and tangent
   Near ((2a,0)), write (X=x-2a). Then [ y^2=X^3 \implies y=\pm X^{3/2}. ] Both branches meet at ((2a,0)), and the tangent there is the (x)-axis. Thus ((2a,0)) is a cusp.

5. Asymptotes
   As (x\to\infty), (y\sim x^{3/2}), so (y/x\to\infty). Therefore the curve has no asymptote.

6. Shape
   Parametrically, [ x=2a+t^2,\qquad y=t^3. ] For (t>0), we get the upper branch; for (t<0), the lower branch. The curve is symmetric about the (x)-axis and opens to the right.

Conclusion: The curve is a semicubical parabola with a cusp at ((2a,0)), symmetric about the (x)-axis, lying entirely to the right of (x=2a), and having no asymptote.

First option:

Given
[ z=\frac{\cos y}{x},\qquad x=u^2-v,\qquad y=e^u ]

Using the chain rule, [ \frac{\partial z}{\partial v}

\frac{\partial z}{\partial x}\frac{\partial x}{\partial v} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial v} ]

Now, [ \frac{\partial z}{\partial x}=-\frac{\cos y}{x^2},\qquad \frac{\partial x}{\partial v}=-1 ] [ \frac{\partial z}{\partial y}=-\frac{\sin y}{x},\qquad \frac{\partial y}{\partial v}=0 ]

Therefore, [ \frac{\partial z}{\partial v}

\left(-\frac{\cos y}{x^2}\right)(-1) + \left(-\frac{\sin y}{x}\right)(0)

\frac{\cos y}{x^2} ]

Substituting (x=u^2-v) and (y=e^u), [ \boxed{\frac{\partial z}{\partial v}=\frac{\cos(e^u)}{(u^2-v)^2}} ]

---

Second option:

Let
[ v=\frac{x+y}{\sqrt{x}+\sqrt{y}},\qquad u=\cos^{-1}v ]

The function (v) is homogeneous of degree (\frac12), because [ v(tx,ty)=\frac{tx+ty}{\sqrt{tx}+\sqrt{ty}} =\sqrt{t},\frac{x+y}{\sqrt{x}+\sqrt{y}} =\sqrt{t},v(x,y) ]

By Euler’s theorem for homogeneous functions, [ x\frac{\partial v}{\partial x}+y\frac{\partial v}{\partial y}

\frac12 v ]

Since (u=\cos^{-1}v), [ \frac{\partial u}{\partial x}

-\frac{1}{\sqrt{1-v^2}}\frac{\partial v}{\partial x}, \qquad \frac{\partial u}{\partial y}

-\frac{1}{\sqrt{1-v^2}}\frac{\partial v}{\partial y} ]

Hence, [ x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}

-\frac{1}{\sqrt{1-v^2}} \left( x\frac{\partial v}{\partial x}+y\frac{\partial v}{\partial y} \right) ] [

-\frac{1}{\sqrt{1-v^2}}\cdot \frac12 v

-\frac12 \frac{v}{\sqrt{1-v^2}} ]

But (v=\cos u), so [ \frac{v}{\sqrt{1-v^2}}=\frac{\cos u}{\sin u}=\cot u ]

Thus, [ x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}

-\frac12 \cot u ]

Therefore, [ \boxed{x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}+\frac12 \cot u=0} ]

The curve
[ y^2=x^2(a-x) ] has a loop between (x=0) and (x=a), since (y=0) at (x=0) and (x=a).

Volume generated by revolving the loop about the (x)-axis is [ V=\pi\int_0^a y^2,dx ] [ =\pi\int_0^a x^2(a-x),dx ] [ =\pi\int_0^a (ax^2-x^3),dx ] [ =\pi\left[\frac{ax^3}{3}-\frac{x^4}{4}\right]_0^a ] [ =\pi\left(\frac{a^4}{3}-\frac{a^4}{4}\right) ] [ =\pi\left(\frac{a^4}{12}\right) ]

[ \boxed{V=\frac{\pi a^4}{12}} ]

So, the required volume is (\displaystyle \frac{\pi a^4}{12}) cubic units.

Stokes' theorem. Here $\vec F=(y+y^2z)\,\vec i+(y+z^2)\,\vec j+(z+z^2)\,\vec k$, and $C$ is the triangle with vertices $(1,0,0),(0,1,0),(0,0,1)$, traversed in that order (anticlockwise seen from outside, normal $(1,1,1)$). The surface $S$ is the triangular part of the plane $x+y+z=1$ bounded by $C$.

$\operatorname{curl}\vec F=\begin{vmatrix}\vec i&\vec j&\vec k\\ \partial_x&\partial_y&\partial_z\\ y+y^2z & y+z^2 & z+z^2\end{vmatrix}=(-2z)\,\vec i+y^2\,\vec j-(1+2yz)\,\vec k.$

Unit normal $\hat n=\dfrac{1}{\sqrt3}(1,1,1)$; projecting onto the $xy$-plane, $dS=\sqrt3\,dx\,dy$, so $\hat n\,dS=(1,1,1)\,dx\,dy$ and, with $z=1-x-y$, $\operatorname{curl}\vec F\cdot(1,1,1)=-2z+y^2-1-2yz=-3+2x+3y^2+2xy.$

Over the triangle $x,y\ge0,\ x+y\le1$ (area $\tfrac12$, $\iint x=\tfrac16$, $\iint y^2=\tfrac1{12}$, $\iint xy=\tfrac1{24}$): $\oint_C\vec F\cdot d\vec r=\iint(-3+2x+3y^2+2xy)\,dx\,dy=-\frac32+\frac13+\frac14+\frac1{12}=-\frac56 .$

---

OR

Stokes' theorem: If $S$ is an oriented piecewise-smooth surface bounded by a simple closed piecewise-smooth curve $C$ (positively oriented) and $\vec F$ has continuous partial derivatives, then $\oint_C\vec F\cdot d\vec r=\iint_S\operatorname{curl}\vec F\cdot\hat n\,dS .$

Take $\vec F=\vec r=x\vec i+y\vec j+z\vec k$. Then $\operatorname{curl}\vec r=\begin{vmatrix}\vec i&\vec j&\vec k\\ \partial_x&\partial_y&\partial_z\\ x&y&z\end{vmatrix}=\vec 0,$ so $\displaystyle\oint_C\vec r\cdot d\vec r=\iint_S\vec 0\cdot\hat n\,dS=0$. $\blacksquare$

First part: P.I. of ((D^2+P_1D+P_2)y=e^{ax})

Let
[ \phi(D)=D^2+P_1D+P_2,\qquad D\equiv \frac{d}{dx}. ] The auxiliary equation is
[ \phi(m)=m^2+P_1m+P_2=0. ]

The particular integral depends on whether (a) is a root of the auxiliary equation.

1. If (\phi(a)\neq 0): [ \text{P.I.}=\frac{1}{\phi(D)}e^{ax} =\frac{e^{ax}}{\phi(a)} =\frac{e^{ax}}{a^2+P_1a+P_2}. ]

2. If (\phi(a)=0) but (\phi'(a)\neq 0) (simple root): [ \text{P.I.}=\frac{xe^{ax}}{\phi'(a)} =\frac{xe^{ax}}{2a+P_1}. ]

3. If (\phi(a)=0) and (\phi'(a)=0) (double root): [ \text{P.I.}=\frac{x^2e^{ax}}{\phi''(a)} =\frac{x^2e^{ax}}{2}. ]

---

P.I. of ((D-2)^2y=6e^{2x})

Here
[ \phi(D)=(D-2)^2. ] For (e^{2x}), we have (a=2).
[ \phi(2)=0,\qquad \phi'(2)=0,\qquad \phi''(2)=2. ] So (a=2) is a double root. Hence [ \text{P.I.}=\frac{6x^2e^{2x}}{\phi''(2)} =\frac{6x^2e^{2x}}{2} =3x^2e^{2x}. ]

[ \boxed{\text{P.I.}=3x^2e^{2x}} ]

---

OR: Solve ((D^2+3D+2)y=e^{2x}), (y(0)=1,\ y'(0)=0)

Auxiliary equation: [ m^2+3m+2=0 ] [ (m+1)(m+2)=0 ] [ m=-1,\ -2. ] So [ \text{C.F.}=C_1e^{-x}+C_2e^{-2x}. ]

For (e^{2x}), (a=2): [ \phi(2)=2^2+3(2)+2=4+6+2=12\neq 0. ] Thus [ \text{P.I.}=\frac{e^{2x}}{12}. ]

General solution: [ y=C_1e^{-x}+C_2e^{-2x}+\frac{e^{2x}}{12}. ]

Now apply initial conditions.

[ y(0)=C_1+C_2+\frac{1}{12}=1 ] [ C_1+C_2=\frac{11}{12}. \quad (1) ]

Differentiate: [ y'=-C_1e^{-x}-2C_2e^{-2x}+\frac{2e^{2x}}{12}. ] [ y'=-C_1e^{-x}-2C_2e^{-2x}+\frac{e^{2x}}{6}. ] At (x=0): [ -C_1-2C_2+\frac{1}{6}=0 ] [ C_1+2C_2=\frac{1}{6}. \quad (2) ]

Subtract (1) from (2): [ C_2=\frac{1}{6}-\frac{11}{12} =\frac{2}{12}-\frac{11}{12} =-\frac{9}{12} =-\frac{3}{4}. ] Then from (1): [ C_1=\frac{11}{12}+\frac{3}{4} =\frac{11}{12}+\frac{9}{12} =\frac{20}{12} =\frac{5}{3}. ]

Therefore, [ \boxed{y=\frac{5}{3}e^{-x}-\frac{3}{4}e^{-2x}+\frac{1}{12}e^{2x}} ]

This satisfies (y(0)=1) and (y'(0)=0).

[ \int_0^1 \int_1^2 x^2 y , dy , dx ]

First integrate with respect to (y):

[ \int_1^2 x^2 y , dy = x^2 \left[\frac{y^2}{2}\right]_1^2 = x^2\left(\frac{4}{2}-\frac{1}{2}\right) = \frac{3}{2}x^2 ]

Now integrate with respect to (x):

[ \int_0^1 \frac{3}{2}x^2 , dx = \frac{3}{2}\left[\frac{x^3}{3}\right]_0^1 = \frac{3}{2}\cdot \frac{1}{3} = \frac{1}{2} ]

[ \boxed{\frac{1}{2}} ]

(i)
[ \left(\frac{d^2y}{dx^2}\right)^2+3\left(\frac{dy}{dx}\right)^4+y=0 ]

• Highest derivative present is (\dfrac{d^2y}{dx^2}), so the order is 2.
• The equation is polynomial in derivatives. The highest-order derivative (\dfrac{d^2y}{dx^2}) occurs with power (2), so the degree is 2.

[ \boxed{\text{Order }=2,\quad \text{Degree }=2} ]

---

(ii)
[ \frac{dy}{dx}+\cos\left(\frac{dy}{dx}\right)=0 ]

• Highest derivative present is (\dfrac{dy}{dx}), so the order is 1.
• The equation contains (\cos\left(\dfrac{dy}{dx}\right)), so it is not a polynomial equation in derivatives. Hence the degree is not defined.

[ \boxed{\text{Order }=1,\quad \text{Degree does not exist}} ]