Given
[ \vec F=2xy,\vec i-3x,\vec j-5z,\vec k ] and the curve
[ \vec r(t)=t,\vec i+t^2,\vec j+t^3,\vec k,\qquad 0\le t\le 1. ]

Then [ x=t,\quad y=t^2,\quad z=t^3 ] so [ \vec F=2t^3,\vec i-3t,\vec j-5t^3,\vec k. ]

Also, [ d\vec r=\frac{d\vec r}{dt},dt=(1,,2t,,3t^2),dt. ]

Work done is [ W=\int_C \vec F\cdot d\vec r =\int_0^1 (2t^3,,-3t,,-5t^3)\cdot(1,,2t,,3t^2),dt. ]

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

[ =\left[\frac12 t^4-2t^3-\frac52 t^6\right]_0^1 ]

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

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

So the total work done is (\boxed{-4}) units.

Given
[ \vec F=2xy,\hat i-3x,\hat j+5z,\hat k ] and the curve [ x=t,\quad y=t^2+1,\quad z=2t^2,\quad t:0\to1. ]

Work done is [ W=\int_C \vec F\cdot d\vec r =\int_C (F_x,dx+F_y,dy+F_z,dz). ]

Now, [ dx=dt,\qquad dy=2t,dt,\qquad dz=4t,dt. ]

Also, [ F_x=2xy=2t(t^2+1)=2t^3+2t, ] [ F_y=-3x=-3t, ] [ F_z=5z=10t^2. ]

So, [ \vec F\cdot d\vec r =(2t^3+2t)dt+(-3t)(2t,dt)+(10t^2)(4t,dt) ] [ =(2t^3+2t-6t^2+40t^3),dt =(42t^3-6t^2+2t),dt. ]

Therefore, [ W=\int_0^1 (42t^3-6t^2+2t),dt ] [ =\left[\frac{42t^4}{4}-\frac{6t^3}{3}+\frac{2t^2}{2}\right]_0^1 ] [ =\left[\frac{21}{2}t^4-2t^3+t^2\right]_0^1 ] [ =\frac{21}{2}-2+1 =\frac{19}{2}. ]

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

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

General definition of definite integral

Let (f(x)) be bounded on ([a,b]). Divide ([a,b]) into (n) equal subintervals of width
[ h=\frac{b-a}{n}. ] Let the points of division be [ a=x_0<x_1<x_2<\cdots<x_n=b. ] Choose any point (\xi_i\in[x_{i-1},x_i]). Form the Riemann sum [ S_n=\sum_{i=1}^n f(\xi_i)h. ] If (\lim_{n\to\infty}S_n) exists and is independent of the choice of (\xi_i), then this limit is called the definite integral of (f(x)) from (a) to (b), written [ \int_a^b f(x),dx=\lim_{n\to\infty}\sum_{i=1}^n f(\xi_i)h. ]

Fundamental theorem of integral calculus

If (f(x)) is continuous on ([a,b]) and (F(x)) is an antiderivative of (f(x)), i.e. (F'(x)=f(x)), then [ \int_a^b f(x),dx=F(b)-F(a). ]

Example: [ \int_0^1 x^2,dx=\left[\frac{x^3}{3}\right]_0^1=\frac13-0=\frac13. ]

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Evaluate by the method of summation: [ \int_0^2 \sin^2 x,dx ]

Let [ h=\frac{2}{n},\qquad x_i=0+ih=\frac{2i}{n}. ] Then [ \int_0^2 \sin^2 x,dx =\lim_{n\to\infty} h\sum_{i=1}^n \sin^2\left(\frac{2i}{n}\right). ] Using (\sin^2\theta=\dfrac{1-\cos 2\theta}{2}), [ \sin^2\left(\frac{2i}{n}\right) =\frac{1-\cos\left(\frac{4i}{n}\right)}{2}. ] So [ I=\lim_{n\to\infty}\frac{2}{n}\sum_{i=1}^n \frac{1-\cos\left(\frac{4i}{n}\right)}{2} ] [ I=\lim_{n\to\infty}\frac{1}{n}\left[n-\sum_{i=1}^n\cos\left(\frac{4i}{n}\right)\right] ] [ I=1-\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n\cos\left(\frac{4i}{n}\right). ]

Now use the formula [ \sum_{i=1}^n \cos(i\alpha)

\frac{\sin\left(\frac{n\alpha}{2}\right)\cos\left(\frac{(n+1)\alpha}{2}\right)} {\sin\left(\frac{\alpha}{2}\right)}. ] Here (\alpha=\dfrac{4}{n}). Hence [ \sum_{i=1}^n \cos\left(\frac{4i}{n}\right)

\frac{\sin 2\cos\left(2+\frac{2}{n}\right)} {\sin\left(\frac{2}{n}\right)}. ] Therefore, [ \frac{1}{n}\sum_{i=1}^n \cos\left(\frac{4i}{n}\right)

\frac{\sin 2\cos\left(2+\frac{2}{n}\right)} {n\sin\left(\frac{2}{n}\right)}. ] Taking (n\to\infty), [ n\sin\left(\frac{2}{n}\right)\to 2,\qquad \cos\left(2+\frac{2}{n}\right)\to \cos 2. ] Thus [ \lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n \cos\left(\frac{4i}{n}\right)

\frac{\sin 2\cos 2}{2}

\frac{\sin 4}{4}. ] Hence [ I=1-\frac{\sin 4}{4}. ]

[ \boxed{\int_0^2 \sin^2 x,dx=1-\frac{\sin 4}{4}} ]

This agrees with the antiderivative method: [ \int \sin^2 x,dx=\frac{x}{2}-\frac{\sin 2x}{4}, ] so [ \int_0^2 \sin^2 x,dx

\left[\frac{x}{2}-\frac{\sin 2x}{4}\right]_0^2

1-\frac{\sin 4}{4}. ]

Curvature. If $\psi$ is the angle the tangent makes with a fixed line and $s$ is the arc length measured along the curve, the curvature at a point is the rate of turning of the tangent:

$K=\frac{d\psi}{ds}$

Radius of curvature. It is the reciprocal of the curvature:

$\rho=\frac1K=\frac{ds}{d\psi}=\frac{\left[1+\left(\frac{dy}{dx}\right)^2\right]^{3/2}}{\left|\frac{d^2y}{dx^2}\right|}$

It is the radius of the circle of curvature, which has the same tangent and curvature as the curve at that point.

For $\sqrt x+\sqrt y=1$. Differentiating,

$\frac{1}{2\sqrt x}+\frac{1}{2\sqrt y}\frac{dy}{dx}=0\ \Rightarrow\ \frac{dy}{dx}=-\sqrt{\frac yx}$

$\frac{d^2y}{dx^2}=-\frac{d}{dx}\sqrt{\frac yx}=\frac{\sqrt x+\sqrt y}{2x\sqrt x}=\frac{1}{2x^{3/2}}$

Also $1+\left(\frac{dy}{dx}\right)^2=\dfrac{x+y}{x}$, so

$\rho=\frac{\left(\frac{x+y}{x}\right)^{3/2}}{\frac{1}{2x^{3/2}}}=2\,(x+y)^{3/2}$

This depends on the point (for example $\rho=\tfrac{1}{\sqrt2}$ at $\left(\tfrac14,\tfrac14\right)$ and $\rho=2$ at $(1,0)$), so $\rho$ is not a constant for this curve.

Right angle at the pole. Squaring $\sqrt x+\sqrt y=1$ twice removes the roots:

$(x-y)^2-2(x+y)+1=0$

This is a parabola with axis $y=x$, vertex $\left(\tfrac14,\tfrac14\right)$, focus $\left(\tfrac12,\tfrac12\right)$ and directrix $x+y=0$. The pole (origin) lies on the directrix.

The curve touches the coordinate axes at $A(1,0)$ and $B(0,1)$, since $y=0$ gives $x=1$ with $\frac{dy}{dx}=0$ there, and $x=0$ gives $y=1$ with the tangent vertical. The tangents from the pole $O$ are therefore the two axes $OA$ and $OB$, and these meet at $90^\circ$. This is the directrix property of a parabola: the two tangents drawn from any point of the directrix are at right angles, and their chord of contact $AB$ ($x+y=1$) passes through the focus $\left(\tfrac12,\tfrac12\right)$.

Hence the curve is a parabola whose tangents from the pole are perpendicular, i.e. the chord of contact $AB$ subtends a right angle at the pole, and its radius of curvature is $\rho=2(x+y)^{3/2}$.

Exact differential equation:
A first-order differential equation
[ 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:
[ (2xy+y^2),dx+(x^2+2xy),dy=0 ] is exact because
[ \frac{\partial}{\partial y}(2xy+y^2)=2x+2y=\frac{\partial}{\partial x}(x^2+2xy). ] Its solution is
[ x^2y+xy^2=C. ]

Integrating factor:
An integrating factor is a function (\mu(x,y)) such that after multiplying a non-exact equation
[ M,dx+N,dy=0 ] by (\mu), the new equation
[ \mu M,dx+\mu N,dy=0 ] becomes exact, i.e.
[ \frac{\partial(\mu M)}{\partial y}=\frac{\partial(\mu N)}{\partial x}. ]

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Solve:

[ (1+xy)x,dy+(1-xy)y,dx=0 ]

Write in the form (M,dx+N,dy=0): [ M=y(1-xy)=y-xy^2,\qquad N=x(1+xy)=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 [ \mu M=\frac{1}{x^2y}-\frac1x,\qquad \mu N=\frac{1}{xy^2}+\frac1y. ]

Now, [ \frac{\partial(\mu M)}{\partial y}=-\frac{1}{x^2y^2}

\frac{\partial(\mu N)}{\partial x}. ] So the equation becomes exact.

Let the solution be (f(x,y)=C). Then [ f_x=\frac{1}{x^2y}-\frac1x. ] Integrating with respect to (x): [ f=-\frac{1}{xy}-\ln|x|+g(y). ] Differentiate with respect to (y): [ f_y=\frac{1}{xy^2}+g'(y). ] But [ f_y=\frac{1}{xy^2}+\frac1y. ] Hence [ g'(y)=\frac1y \implies g(y)=\ln|y|. ]

Therefore, [ f=\ln|y|-\ln|x|-\frac{1}{xy} =\ln\left|\frac yx\right|-\frac{1}{xy}. ]

So the general solution is [ \boxed{\ln\left|\frac yx\right|-\frac{1}{xy}=C} ] where (C) is an arbitrary constant.
For (xy\neq 0); the axes (x=0) and (y=0) are singular solutions.

For a polar curve (r=f(\theta)), the pedal equation is obtained from
[ p=r\sin\phi, ] where (\phi) is the angle between the radius vector and the tangent. Also, [ \tan\phi=\frac{r}{dr/d\theta}. ]

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

Differentiate with respect to (\theta): [ 2r\frac{dr}{d\theta}\cos 2\theta+r^2(-2\sin 2\theta)=0 ] [ r\frac{dr}{d\theta}\cos 2\theta=r^2\sin 2\theta ] [ \frac{dr}{d\theta}=r\tan 2\theta. ]

Now, [ p=\frac{r^2}{\sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2}} =\frac{r^2}{\sqrt{r^2+r^2\tan^2 2\theta}} ] [ p=\frac{r^2}{r\sec 2\theta}=r\cos 2\theta. ]

But from the curve, [ \cos 2\theta=\frac{a^2}{r^2}. ]

Hence, [ p=r\cdot \frac{a^2}{r^2}=\frac{a^2}{r}. ]

Therefore, the pedal equation is
[ \boxed{pr=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{1}{x}\ln\left(\frac{\sin x}{x}\right). ] Now, [ \ln\left(\frac{\sin x}{x}\right)=\ln(\sin x)-\ln x. ] So [ \ln L=\lim_{x\to 0}\frac{\ln(\sin x)-\ln x}{x}. ] This is of the form (\frac00), so applying L’Hospital’s rule: [ \ln L=\lim_{x\to 0}\left(\cot x-\frac1x\right). ] Now, [ \cot x-\frac1x=\frac{x\cos x-\sin x}{x\sin x}. ] Using expansions, [ x\cos x-\sin x=-\frac{x^3}{3}+O(x^5),\qquad x\sin x=x^2+O(x^4). ] Hence [ \ln L=\lim_{x\to 0}\frac{-\frac{x^3}{3}+O(x^5)}{x^2+O(x^4)} =\lim_{x\to 0}\left(-\frac{x}{3}+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 given curve is
[ (x-1)(x-2)(x+y)+x^2+x+1=0. ]

Expanding: [ (x^2-3x+2)(x+y)+x^2+x+1=0 ] [ x^3+x^2y-3x^2-3xy+2x+x^2+x+1=0 ] [ x^3+x^2y-2x^2-3xy+3x+2y+1=0. ]

1. Oblique asymptote

Let the asymptote be (y=mx+c).

Highest-degree terms are [ x^3+x^2y=x^2(x+y). ] For the direction, put (y=mx): [ x^2(x+mx)=(1+m)x^3. ] So, [ 1+m=0 \implies m=-1. ]

Now substitute (y=mx+c) into the curve and equate the coefficient of (x^2) to zero.

Coefficient of (x^2) is [ c-3m-2. ] Putting (m=-1): [ c-3(-1)-2=0 ] [ c+3-2=0 \implies c=-1. ]

Hence the oblique asymptote is [ y=-x-1 ] or [ x+y+1=0. ]

2. Vertical asymptotes

Put (x=a) in the curve: [ (a-1)(a-2)(a+y)+a^2+a+1=0. ] For a vertical asymptote, the coefficient of (y) must vanish: [ (a-1)(a-2)=0. ] Thus, [ a=1,\quad a=2. ] At (x=1), the remaining term is (3\ne0), and at (x=2), it is (7\ne0). So both give vertical asymptotes.

Therefore, [ x=1,\qquad x=2. ]

Final answer

The asymptotes of the curve are [ \boxed{x=1,\qquad x=2,\qquad 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) in (x) and (y), i.e. [ f(tx,ty)=t^n f(x,y) ] for all (t>0), and (f) is differentiable, then [ x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=nz. ]

Proof:
Let [ \phi(t)=f(tx,ty). ] By homogeneity, [ \phi(t)=t^n f(x,y). ] Differentiate both sides with respect to (t): [ \frac{d\phi}{dt}=x f_x(tx,ty)+y f_y(tx,ty). ] Also, [ \frac{d\phi}{dt}=n t^{n-1} f(x,y). ] Hence, [ x f_x(tx,ty)+y f_y(tx,ty)=n t^{n-1} f(x,y). ] Putting (t=1), we get [ x f_x(x,y)+y f_y(x,y)=n f(x,y), ] i.e. [ x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=nz. ] Hence proved.

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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}. ] The numerator (x^3+y^3) is homogeneous of degree (3), and the denominator (x-y) is homogeneous of degree (1). Therefore (v) is homogeneous of degree [ 3-1=2. ] By Euler’s theorem, [ x\frac{\partial v}{\partial x}+y\frac{\partial v}{\partial y}=2v. ] Now, since (u=\tan^{-1}v), [ \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}. ] Thus, [ 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)

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

For the cycloid $x=a(\theta+\sin\theta),\ y=a(1+\cos\theta)$, one arch is traced for $-\pi\le\theta\le\pi$ (since $y=0$ at $\theta=\pm\pi$), and its base is the $x$-axis. The surface area of revolution about the base is

$S=2\pi\int y\,ds,\qquad ds=\sqrt{\left(\frac{dx}{d\theta}\right)^2+\left(\frac{dy}{d\theta}\right)^2}\,d\theta$

Here $\dfrac{dx}{d\theta}=a(1+\cos\theta)$ and $\dfrac{dy}{d\theta}=-a\sin\theta$, so

$ds=a\sqrt{(1+\cos\theta)^2+\sin^2\theta}\,d\theta=a\sqrt{2+2\cos\theta}\,d\theta=2a\cos\frac\theta2\,d\theta$

Also $y=a(1+\cos\theta)=2a\cos^2\dfrac\theta2$. Therefore

$S=2\pi\int_{-\pi}^{\pi}2a\cos^2\frac\theta2\cdot2a\cos\frac\theta2\,d\theta=8\pi a^2\int_{-\pi}^{\pi}\cos^3\frac\theta2\,d\theta$

Put $u=\theta/2$, so $d\theta=2\,du$ and the limits become $-\pi/2$ to $\pi/2$. As $\cos^3u$ is even,

$S=8\pi a^2\cdot2\cdot2\int_0^{\pi/2}\cos^3u\,du=32\pi a^2\cdot\frac23$

since $\displaystyle\int_0^{\pi/2}\cos^3u\,du=\frac23$. Hence

$S=\frac{64\pi a^2}{3}$

First alternative

Take the outward normal to the cylinder $x^2+y^2=4$. Parametrize the surface in the first octant as

$x=2\cos\theta,\quad y=2\sin\theta,\quad z=z,\qquad 0\le\theta\le\frac\pi2,\ 0\le z\le4$

The outward unit normal is $\hat n=\cos\theta\,\vec i+\sin\theta\,\vec j$ and $dS=2\,d\theta\,dz$. On the surface

$\vec F=2z\sin\theta\,\vec i+2\cos\theta\,\vec j-8z\cos^2\theta\,\vec k$

so $\vec F\cdot\hat n=2z\sin\theta\cos\theta+2\sin\theta\cos\theta=2\sin\theta\cos\theta\,(z+1)$. Therefore

$\iint_S\vec F\cdot\hat n\,dS=\int_0^{\pi/2}\!\!\int_0^4 4\sin\theta\cos\theta\,(z+1)\,dz\,d\theta=4\cdot\frac12\cdot\left[\frac{z^2}{2}+z\right]_0^4=2(8+4)=24$

OR

By Stokes' theorem, $\displaystyle\oint_C\vec F\cdot d\vec r=\iint_S(\nabla\times\vec F)\cdot\hat n\,dS$, where $\vec F=(2x-y)\vec i-yz^2\vec j-y^2z\vec k$.

$\nabla\times\vec F=\begin{vmatrix}\vec i&\vec j&\vec k\\ \partial_x&\partial_y&\partial_z\\ 2x-y&-yz^2&-y^2z\end{vmatrix}=\vec i(-2yz+2yz)-\vec j(0-0)+\vec k(0+1)=\vec k$

So $(\nabla\times\vec F)\cdot\hat n\,dS=\vec k\cdot\hat n\,dS=dx\,dy$, the projection of $dS$ on the $xy$-plane. The projection of the upper hemisphere is the disc $x^2+y^2\le a^2$, hence

$\iint_S(\nabla\times\vec F)\cdot\hat n\,dS=\iint_{x^2+y^2\le a^2}dx\,dy=\pi a^2$

Check by the line integral: the boundary $C$ is the circle $x=a\cos t,\ y=a\sin t,\ z=0$, on which $\vec F\cdot d\vec r=(2x-y)\,dx$:

$\oint_C(2x-y)\,dx=\int_0^{2\pi}(2a\cos t-a\sin t)(-a\sin t)\,dt=a^2\int_0^{2\pi}\sin^2t\,dt=\pi a^2$

$\oint_C\vec F\cdot d\vec r=\pi a^2$

Assuming the intended double integral is
[ I=\int_0^\pi \int_0^{\pi(1+\cos\theta)} r^3\sin\theta\cos\theta,dr,d\theta, ] we evaluate as follows.

Inner integral: [ \int_0^{\pi(1+\cos\theta)} r^3,dr =\left[\frac{r^4}{4}\right]_0^{\pi(1+\cos\theta)} =\frac{\pi^4(1+\cos\theta)^4}{4}. ] So [ I=\frac{\pi^4}{4}\int_0^\pi (1+\cos\theta)^4\sin\theta\cos\theta,d\theta. ]

Put [ u=\cos\theta,\qquad du=-\sin\theta,d\theta. ] When (\theta=0), (u=1); when (\theta=\pi), (u=-1). Hence [ \int_0^\pi (1+\cos\theta)^4\sin\theta\cos\theta,d\theta =\int_{-1}^{1} u(1+u)^4,du. ] Now put (v=1+u), so (u=v-1): [ \int_{-1}^{1} u(1+u)^4,du =\int_0^2 (v-1)v^4,dv =\int_0^2 (v^5-v^4),dv. ] Thus [ \int_0^2 (v^5-v^4),dv =\left[\frac{v^6}{6}-\frac{v^5}{5}\right]_0^2 =\frac{64}{6}-\frac{32}{5} =\frac{64}{15}. ] Therefore, [ I=\frac{\pi^4}{4}\cdot \frac{64}{15} =\boxed{\frac{16\pi^4}{15}}. ]

So the value of the integral is
[ \boxed{\frac{16\pi^4}{15}}. ]

The given equation is Clairaut’s equation: [ y=px+\frac{ap}{\sqrt{1+p^2}},\qquad p=\frac{dy}{dx}. ]

General solution

For Clairaut’s equation (y=px+f(p)), the general solution is obtained by replacing (p) by an arbitrary constant (c): [ \boxed{y=cx+\frac{ac}{\sqrt{1+c^2}}} ] where (c) is an arbitrary constant.

Singular solution

The singular solution is the envelope of the family [ y=cx+\frac{ac}{\sqrt{1+c^2}}. ] Differentiate with respect to (c): [ x+\frac{a}{(1+c^2)^{3/2}}=0. ] Let (c=\tan\theta). Then [ x=-a\cos^3\theta,\qquad y=a\sin^3\theta. ] Eliminating (\theta), [ x^{2/3}+y^{2/3}=a^{2/3}. ]

Hence the singular solution is [ \boxed{x^{2/3}+y^{2/3}=a^{2/3}} ] (for (a>0); otherwise use (|a|^{2/3})).

Solve

[ \frac{d^2y}{dx^2}-y=x^2\cos x ]

In operator form, [ (D^2-1)y=x^2\cos x. ]

Auxiliary equation: [ m^2-1=0 \Rightarrow m=\pm 1. ]

So the complementary function is [ y_c=C_1e^x+C_2e^{-x}. ]

Now, [ y_p=\frac{1}{D^2-1}x^2\cos x. ] Using (\cos x=\Re(e^{ix})), [ y_p=\Re\left[\frac{1}{D^2-1}x^2e^{ix}\right]. ] Since (D(e^{ix}v)=e^{ix}(D+i)v), [ \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. ] Now, [ \frac{1}{D^2+2iD-2} =-\frac12\left(1-iD-\frac{D^2}{2}\right)^{-1} =-\frac12\left(1+iD-\frac{D^2}{2}+\cdots\right). ] Hence, [ \frac{1}{D^2+2iD-2}x^2 =-\frac12(x^2+2ix-1) =-\frac12(x+i)^2. ] Therefore, [ y_p=\Re\left[-\frac12e^{ix}(x+i)^2\right] =x\sin x+\frac{1-x^2}{2}\cos x. ]

Thus the general solution is [ \boxed{y=C_1e^x+C_2e^{-x}+x\sin x+\frac{1-x^2}{2}\cos x} ]

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OR

Complementary function (CF):
The general solution of the homogeneous linear differential equation (f(D)y=0) is called the complementary function. It contains as many arbitrary constants as the order of the differential equation.

Particular integral (PI):
A solution of the non-homogeneous equation (f(D)y=Q(x)), obtained without arbitrary constants, is called the particular integral.
General solution (=) CF (+) PI.

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

Let [ y=e^{\alpha x}\int Qe^{-\alpha x},dx. ] Differentiating, [ Dy=\alpha e^{\alpha x}\int Qe^{-\alpha x},dx+e^{\alpha x}Qe^{-\alpha x}. ] But [ e^{\alpha x}\int Qe^{-\alpha x},dx=y. ] So, [ Dy=\alpha y+Q. ] Thus, [ (D-\alpha)y=Q. ] Operating by (\frac{1}{D-\alpha}), [ 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 linear differential equation of first order

A first-order differential equation is said to be linear if the dependent variable $y$ and its derivative $\dfrac{dy}{dx}$ occur only in the first degree and are not multiplied together. Its standard form is

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

where $P(x)$ and $Q(x)$ are functions of $x$ alone (or constants). For example, $\dfrac{dy}{dx} + 2xy = x$ is linear, whereas $\dfrac{dy}{dx} + 2xy = x^3y^3$ is not linear, because $y$ appears in the third degree.

2. Method of solving

Step 1. Write the equation in the standard form $\dfrac{dy}{dx} + P y = Q$.

Step 2. Find the integrating factor (I.F.)

$\text{I.F.} = e^{\int P\,dx}.$

Step 3. Multiply both sides of the equation by the I.F. The left-hand side then becomes the exact derivative of $y\cdot\text{I.F.}$:

$\frac{d}{dx}\bigl(y\cdot e^{\int P\,dx}\bigr) = Q\,e^{\int P\,dx}.$

Step 4. Integrate both sides with respect to $x$:

$y\cdot e^{\int P\,dx} = \int Q\,e^{\int P\,dx}\,dx + c.$

Step 5. Solve for $y$ by dividing by the I.F. This gives the general solution containing one arbitrary constant $c$.

3. Reduction of $\dfrac{dy}{dx} + 2xy = x^3y^3$ to linear form and its solution

The given equation is of the Bernoulli type $\dfrac{dy}{dx} + P y = Q y^n$ with $P = 2x$, $Q = x^3$, $n = 3$.

Reduction to linear form. Divide throughout by $y^3$:

$y^{-3}\frac{dy}{dx} + 2x\,y^{-2} = x^3. \qquad \dots (1)$

Put

$v = y^{-2} \quad\Longrightarrow\quad \frac{dv}{dx} = -2y^{-3}\frac{dy}{dx} \quad\Longrightarrow\quad y^{-3}\frac{dy}{dx} = -\frac{1}{2}\frac{dv}{dx}.$

Substituting in (1):

$-\frac{1}{2}\frac{dv}{dx} + 2x\,v = x^3$

$\boxed{\frac{dv}{dx} - 4x\,v = -2x^3}$

This is a linear equation in $v$ and $\dfrac{dv}{dx}$, with $P = -4x$ and $Q = -2x^3$.

Solving. The integrating factor is

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

Multiplying the linear equation by $e^{-2x^2}$:

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

Integrating both sides:

$v\,e^{-2x^2} = \int -2x^3 e^{-2x^2}\,dx + c.$

To evaluate the integral, put $t = -2x^2$, so that $dt = -4x\,dx$, i.e. $x\,dx = -\dfrac{dt}{4}$. Then

$\int -2x^3 e^{-2x^2}dx = \int (-2x^2)\,e^{-2x^2}\,(x\,dx) = \int t\,e^{t}\left(-\frac{dt}{4}\right) = -\frac{1}{4}\int t\,e^{t}\,dt.$

Now $\int t\,e^{t}dt = e^{t}(t-1)$, hence

$-\frac{1}{4}\,e^{t}(t-1) = -\frac{1}{4}e^{-2x^2}\left(-2x^2-1\right) = \frac{1}{4}e^{-2x^2}\left(2x^2+1\right).$

Therefore

$v\,e^{-2x^2} = \frac{1}{4}\left(2x^2+1\right)e^{-2x^2} + c$

$v = \frac{2x^2+1}{4} + c\,e^{2x^2}.$

Replacing $v$ by $y^{-2}$:

$\frac{1}{y^2} = \frac{2x^2+1}{4} + c\,e^{2x^2}$

$\boxed{y^2 = \frac{4}{2x^2+1+4c\,e^{2x^2}}}$

This is the required general solution of the given equation.

Pedal equation. The pedal equation of a curve is a 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. It has the form $f(r, p) = 0$.

Deduction from the Cartesian equation

Let the curve be $f(x, y) = 0$, with $P(x, y)$ on it, the pole at the origin and the initial line along the $x$-axis. Then

$r^2 = x^2 + y^2, \qquad \tan\theta = \frac{y}{x}.$

Let the tangent at $P$ make an angle $\psi$ with the $x$-axis, so $\tan\psi = y' = \dfrac{dy}{dx}$, and let $\phi$ be the angle between the tangent and the radius vector $OP$. Then $\phi = \psi - \theta$ and

$\tan\phi = \frac{\tan\psi - \tan\theta}{1 + \tan\psi\tan\theta} = \frac{y' - y/x}{1 + y y'/x} = \frac{xy' - y}{x + yy'}.$

The perpendicular from the pole to the tangent is $p = r\sin\phi$. Now

$(xy' - y)^2 + (x + yy')^2 = (x^2 + y^2)(1 + y'^2) = r^2(1 + y'^2),$

so

$\sin\phi = \frac{xy' - y}{r\sqrt{1 + y'^2}} \quad\Rightarrow\quad p = \frac{xy' - y}{\sqrt{1 + y'^2}}, \qquad p^2 = \frac{(xy' - y)^2}{1 + y'^2}.$

To get the pedal equation of a given curve, find $y'$ from its Cartesian equation, substitute it in this relation, and eliminate $x$ and $y$ using $r^2 = x^2 + y^2$ and the curve equation, leaving a relation between $r$ and $p$ only.

Check: for the circle $x^2 + y^2 = a^2$, $y' = -x/y$ gives $p^2 = \dfrac{(x^2+y^2)^2/y^2}{(x^2+y^2)/y^2} = a^2$, i.e. $p = a$.

Seven indeterminate forms: $\frac{0}{0},\quad \frac{\infty}{\infty},\quad 0\cdot\infty,\quad \infty-\infty,\quad 0^0,\quad 1^\infty,\quad \infty^0.$

Proof:

$$\lim_{x\to 0}\frac{\tan x-\sin x}{x^3} =\lim_{x\to 0}\frac{\frac{\sin x}{\cos x}-\sin x}{x^3} =\lim_{x\to 0}\frac{\sin x(1-\cos x)}{x^3\cos x}.$$

Using (1-\cos x=2\sin^2\frac{x}{2}),

$$\lim_{x\to 0}\frac{\tan x-\sin x}{x^3} =\lim_{x\to 0}\frac{\sin x\cdot 2\sin^2\frac{x}{2}}{x^3\cos x}.$$

Now,

$$\sin^2\frac{x}{2}=\frac{x^2}{4}\left(\frac{\sin(x/2)}{x/2}\right)^2.$$

So,

$$\lim_{x\to 0}\frac{\tan x-\sin x}{x^3} = \lim_{x\to 0}\frac{\sin x}{x}\cdot \left(\frac{\sin(x/2)}{x/2}\right)^2\cdot \frac{1}{2\cos x}.$$

Since

$$\lim_{x\to 0}\frac{\sin x}{x}=1,\qquad \lim_{x\to 0}\frac{\sin(x/2)}{x/2}=1,\qquad \lim_{x\to 0}\cos x=1,$$

we get

$$\lim_{x\to 0}\frac{\tan x-\sin x}{x^3} =1\cdot 1^2\cdot \frac{1}{2\cdot 1} =\frac{1}{2}.$$

Hence proved.

Tracing $y^2 = (x - 2a)^3$

1. Symmetry: only even powers of $y$ occur, so the curve is symmetric about the $x$-axis.
2. Origin: $(0,0)$ does not satisfy the equation (it gives $0=-8a^3$), so the curve does not pass through the origin.
3. Intercepts: with $y=0$, $x=2a$; the curve meets the $x$-axis at $(2a, 0)$. With $x=0$, $y^2=-8a^3<0$, so there is no real intersection with the $y$-axis (taking $a>0$).
4. Region: $y^2\ge 0$ requires $(x-2a)^3\ge 0$, so $x\ge 2a$. No part of the curve lies to the left of $x=2a$.
5. Tangent at $(2a,0)$: shift the origin to $(2a,0)$ to get $y^2=x^3$. The lowest-degree term is $y^2=0$, so the tangent is $y=0$ (the $x$-axis), counted twice. This is a cusp at $(2a,0)$ with the $x$-axis as cuspidal tangent.
6. Asymptotes: none. For large $x$, $y=\pm(x-2a)^{3/2}$ and $y/x\to\infty$, so the branches go off steeply with no asymptote.
7. Shape: $\dfrac{dy}{dx}=\pm\dfrac32(x-2a)^{1/2}\ge 0$ for the upper branch, so $y$ increases as $x$ increases. Both branches go to infinity as $x\to\infty$.

Conclusion: the curve is a semi-cubical parabola, with a cusp at $(2a,0)$, symmetric about the $x$-axis, extending to the right of $x=2a$ with the upper branch rising and the lower branch falling.

OR: Asymptotes of $x^3 + x^2y - xy^2 - y^3 + x - y + 1 = 0$

The highest-degree terms are $\phi_3(x,y)=x^3+x^2y-xy^2-y^3$. Put $x=1,\ y=m$:

$\phi_3(m) = 1 + m - m^2 - m^3 = (1+m)(1-m^2) = (1+m)^2(1-m)$

so the slopes are $m=-1$ (twice) and $m=1$. The degree-1 terms give $\phi_2 = 0$ and the next group is $\phi_1 = x - y$, with $\phi_0=1$.

Asymptote for the simple root $m=1$: $c = -\dfrac{\phi_2(1)}{\phi_3'(1)}=0$ (since $\phi_2\equiv0$). So $y = x$ is an asymptote.

Asymptotes for the repeated root $m=-1$: $c$ satisfies

$\frac{c^2}{2!}\phi_3''(-1) + c\,\phi_2'(-1) + \phi_1(-1) = 0.$

Here $\phi_3(m) = (1+m)^2(1-m)$, so $\phi_3''(-1) = 2(1-(-1)) = 4$; $\phi_2'=0$; and $\phi_1(-1) = 1-(-1) = 2$. Hence

$2c^2 + 2 = 0 \Rightarrow c^2 = -1,$

which has no real value. So there are no real asymptotes for $m=-1$.

Result: the only real asymptote is $y = x$, that is, $x - y = 0$.

First alternative

Given
[ z=\frac{\cos y}{x},\qquad x=u^2-v,\qquad y=e^v ] we need (\frac{\partial z}{\partial v}), treating (u) as constant.

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

\frac{\partial z}{\partial x}\frac{dx}{dv} + \frac{\partial z}{\partial y}\frac{dy}{dv}. ]

Now, [ \frac{\partial z}{\partial x}=-\frac{\cos y}{x^2},\qquad \frac{\partial z}{\partial y}=-\frac{\sin y}{x}, ] and [ \frac{dx}{dv}=-1,\qquad \frac{dy}{dv}=e^v. ]

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

\left(-\frac{\cos y}{x^2}\right)(-1) + \left(-\frac{\sin y}{x}\right)e^v ] [

\frac{\cos y}{x^2}-\frac{e^v\sin y}{x}. ]

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

\frac{\cos(e^v)}{(u^2-v)^2}

\frac{e^v\sin(e^v)}{u^2-v} } ]

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Second alternative

Given [ u=\cos^{-1}\left(\frac{x+y}{\sqrt{x}+\sqrt{y}}\right), ] show that [ x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}+\frac12\cot u=0. ]

Let [ w=\cos u=\frac{x+y}{\sqrt{x}+\sqrt{y}}. ] Put [ p=\sqrt{x},\qquad q=\sqrt{y}. ] Then [ w=\frac{p^2+q^2}{p+q}. ] This is homogeneous of degree (1) in (p,q). Hence, by Euler’s theorem, [ p\frac{\partial w}{\partial p}+q\frac{\partial w}{\partial q}=w. ]

Also, [ \frac{\partial w}{\partial x}

\frac{\partial w}{\partial p}\frac{dp}{dx}

\frac{1}{2p}\frac{\partial w}{\partial p}, ] and [ \frac{\partial w}{\partial y}

\frac{\partial w}{\partial q}\frac{dq}{dy}

\frac{1}{2q}\frac{\partial w}{\partial q}. ]

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

p^2\cdot \frac{1}{2p}\frac{\partial w}{\partial p} + q^2\cdot \frac{1}{2q}\frac{\partial w}{\partial q} ] [

\frac12\left(p\frac{\partial w}{\partial p}+q\frac{\partial w}{\partial q}\right)

\frac12 w. ]

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

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

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

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

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

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

-\frac12\frac{w}{\sqrt{1-w^2}}. ]

But (w=\cos u), so [ \sqrt{1-w^2}=\sin u. ] Hence, [ x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}

-\frac12\frac{\cos u}{\sin u}

-\frac12\cot u. ]

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

Given: The curve is
[ y^2 = x^2(a-x) ]

To find: Volume of the solid formed by revolving the loop of the curve about the (x)-axis.

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Step 1: Determine the limits of integration

The curve is symmetric about the (x)-axis. It meets the (x)-axis where (y=0): [ x^2(a-x)=0 \implies x=0,; x=a ] So the loop lies between (x=0) and (x=a).

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Step 2: Use the volume formula

For revolution about the (x)-axis, [ V = \pi \int_{0}^{a} y^2 , dx ]

Substitute (y^2 = x^2(a-x)): [ V = \pi \int_{0}^{a} x^2(a-x) , dx ]

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Step 3: Evaluate the integral

[ V = \pi \int_{0}^{a} (a x^2 - x^3) , dx ] [ = \pi \left[ \frac{a x^3}{3} - \frac{x^4}{4} \right]_{0}^{a} ] [ = \pi \left( \frac{a^4}{3} - \frac{a^4}{4} \right) ] [ = \pi a^4 \left( \frac{1}{3} - \frac{1}{4} \right) ] [ = \pi a^4 \left( \frac{4-3}{12} \right) ] [ = \frac{\pi a^4}{12} ]

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Result: [ \boxed{V = \frac{\pi a^4}{12} \text{ cubic units}} ]

The volume of the solid generated by revolving the loop of the curve (y^2 = x^2(a-x)) about the (x)-axis is (\frac{\pi a^4}{12}) cubic units.

Solution (using Stoke’s theorem):

Let
[ \vec F=(y+y^2)\vec i+(y+z^2)\vec j+(z+x^2)\vec k. ] The curve (C) is the triangle with vertices
[ A(1,0,0),\quad B(0,1,0),\quad C(0,0,1), ] oriented as (A\to B\to C\to A). The surface (S) is the plane [ x+y+z=1 ] in the first octant. Its unit normal is [ \hat n=\frac{\vec i+\vec j+\vec k}{\sqrt3}. ] Also, [ d\vec S=\hat n,dS=(\vec i+\vec j+\vec k),dx,dy. ]

Now 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}\ y+y^2&y+z^2&z+x^2 \end{vmatrix} ] [ =\left(\frac{\partial}{\partial y}(z+x^2)-\frac{\partial}{\partial z}(y+z^2)\right)\vec i -\left(\frac{\partial}{\partial x}(z+x^2)-\frac{\partial}{\partial z}(y+y^2)\right)\vec j +\left(\frac{\partial}{\partial x}(y+z^2)-\frac{\partial}{\partial y}(y+y^2)\right)\vec k ] [ =(-2z)\vec i+(-2x)\vec j-(1+2y)\vec k. ]

Thus, [ (\nabla\times\vec F)\cdot(\vec i+\vec j+\vec k) =-2z-2x-1-2y =-1-2(x+y+z). ] On the plane (S), (x+y+z=1), so [ (\nabla\times\vec F)\cdot(\vec i+\vec j+\vec k)=-1-2(1)=-3. ]

By Stoke’s theorem, [ \int_C \vec F\cdot d\vec r =\iint_S(\nabla\times\vec F)\cdot d\vec S =\iint_D (-3),dx,dy, ] where (D) is the projection of (S) on the (xy)-plane: [ D:\ x\ge0,\ y\ge0,\ x+y\le1. ] The area of (D) is (\frac12). Hence [ \int_C \vec F\cdot d\vec r=-3\cdot \frac12=-\frac32. ]

[ \boxed{\int_C \vec F\cdot d\vec r=-\frac32} ]

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OR

Stoke’s theorem:
Let (S) be an open surface bounded by a simple closed curve (C). If (\vec F) is a continuously differentiable vector field, then [ \oint_C \vec F\cdot d\vec r =\iint_S(\nabla\times\vec F)\cdot \hat n,dS, ] where (C) is traversed in the positive direction relative to the normal (\hat n) of (S).

To deduce [ \oint_C d\vec r=0, ] let (\vec a) be an arbitrary constant vector. Put (\vec F=\vec a). Then [ \nabla\times\vec a=0. ] By Stoke’s theorem, [ \vec a\cdot\oint_C d\vec r =\oint_C \vec a\cdot d\vec r =\iint_S(\nabla\times\vec a)\cdot \hat n,dS =\iint_S 0,dS=0. ] Since (\vec a) is arbitrary, we must have [ \oint_C d\vec r=0. ]

[ \boxed{\oint_C d\vec r=0} ]

First alternative

For the differential equation
$(D^2+P_1D+P_2)y=e^{ax},$ where $D\equiv \frac{d}{dx}$, let
$f(D)=D^2+P_1D+P_2.$

The particular integral is found as follows:

1. If $f(a)\ne 0$,
   $\text{P.I.}=\frac{e^{ax}}{f(a)}.$

2. If $f(a)=0$ but $f'(a)\ne 0$,
   $\text{P.I.}=\frac{x e^{ax}}{f'(a)}.$

3. If $f(a)=f'(a)=0$ but $f''(a)\ne 0$,
   $\text{P.I.}=\frac{x^2 e^{ax}}{f''(a)}.$

In general, for a repeated factor $(D-a)^n$,
$\text{P.I.}=\frac{x^n}{n!}e^{ax}.$

Now for
$(D-2)^2y=6e^{2x},$ here $f(D)=(D-2)^2$, so $f(2)=0$ and $f'(2)=0$, but
$f''(2)=2.$ Hence

$$\text{P.I.}=6\cdot \frac{x^2 e^{2x}}{f''(2)} =6\cdot \frac{x^2 e^{2x}}{2} =3x^2e^{2x}.$$

So,
$\boxed{\text{P.I.}=3x^2e^{2x}}$

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OR alternative

Solve
$(D^2+3D+2)y=e^{2x},$ given $y=1,\ \frac{dy}{dx}=0$ when $x=0$.

Auxiliary equation:
$m^2+3m+2=0$ $(m+1)(m+2)=0$ $m=-1,\ -2.$

So,
$y_c=C_1e^{-x}+C_2e^{-2x}.$

For P.I.,
$f(D)=D^2+3D+2.$ Since
$f(2)=4+6+2=12\ne 0,$ $\text{P.I.}=\frac{e^{2x}}{12}.$

Therefore,
$y=C_1e^{-x}+C_2e^{-2x}+\frac{e^{2x}}{12}.$

Using $y(0)=1$:
$C_1+C_2+\frac{1}{12}=1$ $C_1+C_2=\frac{11}{12}. \quad (1)$

Now,
$y'=-C_1e^{-x}-2C_2e^{-2x}+\frac{1}{6}e^{2x}.$

Using $y'(0)=0$:
$-C_1-2C_2+\frac{1}{6}=0$ $C_1+2C_2=\frac{1}{6}. \quad (2)$

Subtracting (1) from (2):

$$C_2=\frac{1}{6}-\frac{11}{12} =\frac{2}{12}-\frac{11}{12} =-\frac{3}{4}.$$

Then from (1):

$$C_1=\frac{11}{12}+\frac{3}{4} =\frac{11}{12}+\frac{9}{12} =\frac{5}{3}.$$

Hence the required solution is
$\boxed{y=\frac{5}{3}e^{-x}-\frac{3}{4}e^{-2x}+\frac{1}{12}e^{2x}}$

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

First evaluate the inner integral 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) = x^2\left(\frac{3}{2}\right) = \frac{3}{2}x^2 ]

Now evaluate the outer integral 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}} ]

So,

[ \int_0^1 \int_1^2 x^2 y , dy , dx = \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 (\frac{d^2y}{dx^2}), so
  [ \boxed{\text{Order}=2} ]

• The equation is polynomial in all its derivatives. The highest derivative (\frac{d^2y}{dx^2}) has power (2), so
  [ \boxed{\text{Degree}=2} ]

Hence, order (=2), degree (=2).

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

• Highest derivative present is (\frac{dy}{dx}), so
  [ \boxed{\text{Order}=1} ]

• The equation contains (\cos\left(\frac{dy}{dx}\right)), which is not a polynomial expression in (\frac{dy}{dx}). Therefore, the differential equation is not polynomial in its derivative, so its degree is not defined.

[ \boxed{\text{Degree does not exist}} ]

Hence, order (=1), degree is not defined.