Details For BMTC-132 Solved Assignment

IGNOU BMTC-132 Assignment Question Paper 2024

bmtc-132-solved-assignment-2024-10cd7422-4b11-4531-9171-a6445826d855

BMTC-132 Solved Assignment 2024

1. State whether the following statements are true or false. Give a short proof or a counterexample in support of your Answer:

a) A real-valued function of three variables which is continuous everywhere is differentiable.
b) The function $f(x,y)=\ln \left(\frac{x+y}{x}\right)$$f(x,y)=\ln \left(\frac{x+y}{x}\right)$f(x,y)=ln((x+y)/(x))f(x, y)=ln left(frac{x+y}{x}right)$f(x,y)=\ln \left(\frac{x+y}{x}\right)$ is not homogeneous function.
c) The cylindrical coordinates of the point whose spherical coordinates is $(8,\frac{\pi }{6},\frac{\pi }{2})$$\left(8,\frac{\pi }{6},\frac{\pi }{2}\right)$(8,(pi)/(6),(pi)/(2))left(8, frac{pi}{6}, frac{pi}{2}right)$(8,\frac{\pi }{6},\frac{\pi }{2})$ is $(8,\frac{\pi }{6},0)$$\left(8,\frac{\pi }{6},0\right)$(8,(pi)/(6),0)left(8, frac{pi}{6}, 0right)$(8,\frac{\pi }{6},0)$
d) The unique solution $y(x)$$y(x)$y(x)y(x)$y(x)$ of an ordinary differential equation:
$$\frac{dy}{dx}=\{\begin{array}{ll}0,& \text{for}x<0\\ 1,& \text{for}x\ge 0\end{array}$$$$\frac{dy}{dx}=\left\{\begin{array}{l}0,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{for}x<0\\ 1,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{for}x\ge 0\end{array}\right.$$(dy)/(dx)={[0″,”,” for “x < 0],[1″,”,” for “x >= 0]:}frac{d y}{d x}= begin{cases}0, & text { for } x<0 \ 1, & text { for } x geq 0end{cases}$$\frac{dy}{dx}=\{\begin{array}{ll}0,& \text{for}x<0\\ 1,& \text{for}x\ge 0\end{array}$$
exists $\mathrm{\forall }x\in \mathbb{R}$$\mathrm{\forall }x\in \mathbb{R}$AA x inRforall x in mathbb{R}$\mathrm{\forall }x\in \mathbb{R}$.
e) The differential equation:
$${[1+{\left(y^{\mathrm{\prime }}\right)}^2]}^{\frac{5}{3}}=y^{\mathrm{\prime }\mathrm{\prime }}$$$${\left[1+{\left(y^{\mathrm{\prime }}\right)}^2\right]}^{\frac{5}{3}}=y^{\mathrm{\prime }\mathrm{\prime }}$$[1+(y^(‘))^(2)]^((5)/(3))=y^(”)left[1+left(y^{prime}right)^2right]^{frac{5}{3}}=y^{prime prime}$${[1+{\left(y^{\mathrm{\prime }}\right)}^2]}^{\frac{5}{3}}=y^{\mathrm{\prime }\mathrm{\prime }}$$
is a second order differential equation of degree 3 .
f) Differential equation:
$$\cos x\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}+x{y}^2=0$$$$\cos x\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}+x{y}^2=0$$cos x(d^(2)y)/(dx^(2))+(dy)/(dx)+xy^(2)=0cos x frac{d^2 y}{d x^2}+frac{d y}{d x}+x y^2=0$$\cos x\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}+x{y}^2=0$$
in $]0,\pi [$$]0,\pi [$]0,pi[] 0, pi[$]0,\pi [$ is a linear, homogeneous equation.
g) The total differential equation corresponding to the family of surfaces $x^{3}z+x^{2}y=c$$x^{3}z+x^{2}y=c$x^(3)z+x^(2)y=cx^3 z+x^2 y=c$x^{3}z+x^{2}y=c$, where $c$$c$cc$c$ is a parameter is $3x^{2}dz+x(ydx+xdy)=0$$3x^{2}dz+x(ydx+xdy)=0$3x^(2)dz+x(ydx+xdy)=03 x^2 d z+x(y d x+x d y)=0$3x^{2}dz+x(ydx+xdy)=0$.
h) Differential equation:
$$5x^2{y}^2{z}^2=2p{x}^2{y}^2+5q{x}^2{y}^3+2p{z}^2+9x^2{y}^2$$$$5x^2{y}^2{z}^2=2p{x}^2{y}^2+5q{x}^2{y}^3+2p{z}^2+9x^2{y}^2$$5x^(2)y^(2)z^(2)=2px^(2)y^(2)+5qx^(2)y^(3)+2pz^(2)+9x^(2)y^(2)5 x^2 y^2 z^2=2 p x^2 y^2+5 q x^2 y^3+2 p z^2+9 x^2 y^2$$5x^2{y}^2{z}^2=2p{x}^2{y}^2+5q{x}^2{y}^3+2p{z}^2+9x^2{y}^2$$
is a semi-linear partial differential equation of first order.
i) The differential equation:
$$v\frac{du}{dv}=e^{2v}+uv-u$$$$v\frac{du}{dv}=e^{2v}+uv-u$$v(du)/(dv)=e^(2v)+uv-uv frac{d u}{d v}=e^{2 v}+u v-u$$v\frac{du}{dv}=e^{2v}+uv-u$$
has an integrating factor $v\exp (-v)$$v\exp (-v)$v exp(-v)v exp (-v)$v\exp (-v)$.
j) The simultaneous differential equation of simple harmonic motion of a particle in phase -plane is:
$$\frac{dx}{y}=\frac{dy}{-w^{2}x}=dt\text{with}y\left(x_0\right)=y_0$$$$\frac{dx}{y}=\frac{dy}{-w^{2}x}=dt\text{with}y\left(x_0\right)=y_0$$(dx)/(y)=(dy)/(-w^(2)x)=dt” with “y(x_(0))=y_(0)frac{d x}{y}=frac{d y}{-w^2 x}=d t text { with } yleft(x_0right)=y_0$$\frac{dx}{y}=\frac{dy}{-w^{2}x}=dt\text{with}y\left(x_0\right)=y_0$$

2. a) Solve the differential equation:

$$\frac{dy}{dx}\frac{-\tan y}{(1+x)}=(1+x)e^x\sec y$$$$\frac{dy}{dx}\frac{-\tan y}{(1+x)}=(1+x)e^x\sec y$$(dy)/(dx)(-tan y)/((1+x))=(1+x)e^(x)sec yfrac{d y}{d x} frac{-tan y}{(1+x)}=(1+x) e^x sec y$$\frac{dy}{dx}\frac{-\tan y}{(1+x)}=(1+x)e^x\sec y$$
b) Show that the differential equation:
$$(y^2+yx)dx+(z^2+zx)dy+(y^2-xy)dz=0$$$$\left(y^2+yx\right)dx+\left(z^2+zx\right)dy+\left(y^2-xy\right)dz=0$$(y^(2)+yx)dx+(z^(2)+zx)dy+(y^(2)-xy)dz=0left(y^2+y xright) d x+left(z^2+z xright) d y+left(y^2-x yright) d z=0$$(y^2+yx)dx+(z^2+zx)dy+(y^2-xy)dz=0$$
is integrable and find its integral.

3. a) Solve the differential equation:

$$x^2\frac{d^{2}y}{d{x}^2}-2x\frac{dy}{dx}-4y=x^2+2\ln x$$$$x^2\frac{d^{2}y}{d{x}^2}-2x\frac{dy}{dx}-4y=x^2+2\ln x$$x^(2)(d^(2)y)/(dx^(2))-2x(dy)/(dx)-4y=x^(2)+2ln xx^2 frac{d^2 y}{d x^2}-2 x frac{d y}{d x}-4 y=x^2+2 ln x$$x^2\frac{d^{2}y}{d{x}^2}-2x\frac{dy}{dx}-4y=x^2+2\ln x$$
b) Using Charpit’s method, find the complete integral of the differential equation:
$$p(1+q^2)+(b-z)q=0,$$$$p\left(1+q^2\right)+(b-z)q=0,$$p(1+q^(2))+(b-z)q=0,pleft(1+q^2right)+(b-z) q=0,$$p(1+q^2)+(b-z)q=0,$$
$b$$b$bb$b$ being a constant.

4. a) Find all the first order partial derivatives of the following function:

$$h(x,y,t)=e^{x-t}\cos (y+t)$$$$h(x,y,t)=e^{x-t}\cos (y+t)$$h(x,y,t)=e^(x-t)cos(y+t)h(x, y, t)=e^{x-t} cos (y+t)$$h(x,y,t)=e^{x-t}\cos (y+t)$$
What is the value of $\frac{\mathrm{\partial }h}{\mathrm{\partial }t}$$\frac{\mathrm{\partial }h}{\mathrm{\partial }t}$(del h)/(del t)frac{partial h}{partial t}$\frac{\mathrm{\partial }h}{\mathrm{\partial }t}$ at $(0,\frac{\pi }{2},0)$$\left(0,\frac{\pi }{2},0\right)$(0,(pi)/(2),0)left(0, frac{pi}{2}, 0right)$(0,\frac{\pi }{2},0)$ ?
b) Find the limit of:
$$f(x,y)=\frac{x^2-y^2}{x^2+y^2}$$$$f(x,y)=\frac{x^2-y^2}{x^2+y^2}$$f(x,y)=(x^(2)-y^(2))/(x^(2)+y^(2))f(x, y)=frac{x^2-y^2}{x^2+y^2}$$f(x,y)=\frac{x^2-y^2}{x^2+y^2}$$
as $(x,y)\to (0,0)$$(x,y)\to (0,0)$(x,y)rarr(0,0)(x, y) rightarrow(0,0)$(x,y)\to (0,0)$ along:
i) $y=3x$$y=3x$y=3xy=3 x$y=3x$
ii) $y=5x$$y=5x$y=5xy=5 x$y=5x$
What can you conclude about $\underset{(x,y)\to (0,0)}{lim}f(x,y)$$\underset{(x,y)\to (0,0)}{lim} f(x,y)$lim_((x,y)rarr(0,0))f(x,y)lim _{(x, y) rightarrow(0,0)} f(x, y)$\underset{(x,y)\to (0,0)}{lim}f(x,y)$ ? Justify your answer.
c) If $\cos \alpha ,\cos \beta$$\cos \alpha ,\cos \beta$cos alpha,cos betacos alpha, cos beta$\cos \alpha ,\cos \beta$ and $\cos \gamma$$\cos \gamma$cos gammacos gamma$\cos \gamma$ are the direction cosines of a line, then show that:
$${\sin }^2\alpha +{\sin }^2\beta +{\sin }^2\gamma =2$$$${\sin }^2\alpha +{\sin }^2\beta +{\sin }^2\gamma =2$$sin^(2)alpha+sin^(2)beta+sin^(2)gamma=2sin ^2 alpha+sin ^2 beta+sin ^2 gamma=2$${\sin }^2\alpha +{\sin }^2\beta +{\sin }^2\gamma =2$$

5. a) Solve the differential equation:

$$2x\frac{dy}{dx}+y(6y^2-x-1)=0$$$$2x\frac{dy}{dx}+y\left(6y^2-x-1\right)=0$$2x(dy)/(dx)+y(6y^(2)-x-1)=02 x frac{d y}{d x}+yleft(6 y^2-x-1right)=0$$2x\frac{dy}{dx}+y(6y^2-x-1)=0$$
b) The rate of change of the price of a commodity is proportional to the difference between the demand $D$$D$DD$D$ and the supply $\mathrm{S}$$\mathrm{S}$Smathrm{S}$\mathrm{S}$. If $D=\alpha -bP$$D=\alpha -bP$D=alpha-bPD=alpha-b P$D=\alpha -bP$ and $S=c\sin \beta t$$S=c\sin \beta t$S=c sin beta tS=c sin beta t$S=c\sin \beta t$, where $a,b,c$$a,b,c$a,b,ca, b, c$a,b,c$ and $\beta$$\beta$betabeta$\beta$ are constants, determine $P(t)$$P(t)$P(t)P(t)$P(t)$. It is given that at $t=0,P=P_0$$t=0,P=P_0$t=0,P=P_(0)t=0, P=P_0$t=0,P=P_0$.

6. a) Find the differential equations of the space curve in which the two families of surfaces:

$$u=x^2+y^2+3z=c_1$$$$u=x^2+y^2+3z=c_1$$u=x^(2)+y^(2)+3z=c_(1)u=x^2+y^2+3 z=c_1$$u=x^2+y^2+3z=c_1$$
and $v=zx+x^2+y^2=c_2$$v=zx+x^2+y^2=c_2$v=zx+x^(2)+y^(2)=c_(2)v=z x+x^2+y^2=c_2$v=zx+x^2+y^2=c_2$
intersect.
b) Transform the given equation to Clairaut’s form and hence find its general solution:
$$xy(y-px)=x-py$$$$xy(y-px)=x-py$$xy(y-px)=x-pyx y(y-p x)=x-p y$$xy(y-px)=x-py$$
Also find its singular solution, if it exists.

7. a) Find the envelope and the characteristic curves of the family of curves:

$$x^2(y-c{)}^2+z^2=c^2{\cos }^2\alpha ,$$$$x^2(y-c{)}^2+z^2=c^2{\cos }^2\alpha ,$$x^(2)(y-c)^(2)+z^(2)=c^(2)cos^(2)alpha,x^2(y-c)^2+z^2=c^2 cos ^2 alpha,$$x^2(y-c{)}^2+z^2=c^2{\cos }^2\alpha ,$$
$c$$c$cc$c$ and $\alpha$$\alpha$alphaalpha$\alpha$ are constants.
b) Using Lagrange’s method, solve the differential equation:
$$(x^2-y^2-z^2)p+2xq=2xz$$$$\left(x^2-y^2-z^2\right)p+2xq=2xz$$(x^(2)-y^(2)-z^(2))p+2xq=2xzleft(x^2-y^2-z^2right) p+2 x q=2 x z$$(x^2-y^2-z^2)p+2xq=2xz$$

8. a) Using the method of variation of parameters, solve the differential equation:

$$\frac{d^{2}y}{d{x}^2}+y={\sec }^{3}x$$$$\frac{d^{2}y}{d{x}^2}+y={\sec }^{3}x$$(d^(2)y)/(dx^(2))+y=sec^(3)xfrac{d^2 y}{d x^2}+y=sec ^3 x$$\frac{d^{2}y}{d{x}^2}+y={\sec }^{3}x$$
b) Solve the differential equation:
$$(2x{e}^y{y}^4+2x{y}^{3}y)dx+(x^2{y}^4{e}^y-x^2{y}^2-3x)dy=0$$$$\left(2x{e}^y{y}^4+2x{y}^{3}y\right)dx+\left(x^2{y}^4{e}^y-x^2{y}^2-3x\right)dy=0$$(2xe^(y)y^(4)+2xy^(3)y)dx+(x^(2)y^(4)e^(y)-x^(2)y^(2)-3x)dy=0left(2 x e^y y^4+2 x y^3 yright) d x+left(x^2 y^4 e^y-x^2 y^2-3 xright) d y=0$$(2x{e}^y{y}^4+2x{y}^{3}y)dx+(x^2{y}^4{e}^y-x^2{y}^2-3x)dy=0$$

9. a) Show that the limit of the function $f(x,y)$$f(x,y)$f(x,y)f(x, y)$f(x,y)$ exists at the origin, where:

$$f(x,y)=\{\begin{array}{cc}x\cos \frac{1}{y}+y\cos \frac{1}{x},& (x,y)\ne (0,0)\\ 0,& (x,y)=(0,0)\end{array}$$$$f(x,y)=\left\{\begin{array}{cc}x\cos \frac{1}{y}+y\cos \frac{1}{x},& (x,y)\ne (0,0)\\ 0,& (x,y)=(0,0)\end{array}\right.$$f(x,y)={[x cos (1)/(y)+y cos (1)/(x)”,”,(x”,”y)!=(0″,”0)],[0″,”,(x”,”y)=(0″,”0)]:}f(x, y)=left{begin{array}{cc}
x cos frac{1}{y}+y cos frac{1}{x}, & (x, y) neq(0,0) \
0, & (x, y)=(0,0)
end{array}right.$$f(x,y)=\{\begin{array}{cc}x\cos \frac{1}{y}+y\cos \frac{1}{x},& (x,y)\ne (0,0)\\ 0,& (x,y)=(0,0)\end{array}$$
Do the repeated limits of $f(x,y)$$f(x,y)$f(x,y)f(x, y)$f(x,y)$ exist? Justify your answer.
b) For the function:
$$f(x,y)=x^3+xy-2y^2,$$$$f(x,y)=x^3+xy-2y^2,$$f(x,y)=x^(3)+xy-2y^(2),f(x, y)=x^3+x y-2 y^2,$$f(x,y)=x^3+xy-2y^2,$$
Find the polynomial given by:
$$f_{xx}(1,2)(x-2{)}^2+f_{xy}(1,2)(x-2)(y-1)+f_{yy}(1,2)(y-1{)}^2$$$$f_{xx}(1,2)(x-2{)}^2+f_{xy}(1,2)(x-2)(y-1)+f_{yy}(1,2)(y-1{)}^2$$f_(xx)(1,2)(x-2)^(2)+f_(xy)(1,2)(x-2)(y-1)+f_(yy)(1,2)(y-1)^(2)f_{x x}(1,2)(x-2)^2+f_{x y}(1,2)(x-2)(y-1)+f_{y y}(1,2)(y-1)^2$$f_{xx}(1,2)(x-2{)}^2+f_{xy}(1,2)(x-2)(y-1)+f_{yy}(1,2)(y-1{)}^2$$


[stray-random categories=trigonometry]

BMTC-132 Sample Solution 2024

bmtc-132-solved-assignment-2024-ss-10cd7422-4b11-4531-9171-a6445826d855

BMTC-132 Solved Assignment 2024

1. State whether the following statements are true or false. Give a short proof or a counterexample in support of your answer:

a) A real-valued function of three variables which is continuous everywhere is differentiable.
Answer:
The statement "A real-valued function of three variables which is continuous everywhere is differentiable" is false.

Justification

Continuity and differentiability are related but distinct properties of functions. While differentiability implies continuity, the converse is not necessarily true. A function can be continuous at a point but not differentiable at that point. This concept holds for functions of any number of variables, including functions of three variables.

Counterexample

A classic counterexample in the context of functions of a single variable is the absolute value function, $f(x)=|x|$$f(x)=|x|$f(x)=|x|f(x) = |x|$f(x)=|x|$, which is continuous everywhere but not differentiable at $x=0$$x=0$x=0x = 0$x=0$.
For a function of three variables, we can construct a similar example. Consider the function
$$f(x,y,z)=\sqrt{x^2+y^2+z^2}$$$$f(x,y,z)=\sqrt{x^2+y^2+z^2}$$f(x,y,z)=sqrt(x^(2)+y^(2)+z^(2))f(x, y, z) = sqrt{x^2 + y^2 + z^2}$$f(x,y,z)=\sqrt{x^2+y^2+z^2}$$
This function represents the distance from the origin in three-dimensional space and is continuous everywhere. However, it is not differentiable at the origin $(0,0,0)$$(0,0,0)$(0,0,0)(0, 0, 0)$(0,0,0)$.
To see why, consider the partial derivative with respect to $x$$x$xx$x$ at the origin. The partial derivative is defined as the limit of the difference quotient:
$$\underset{h\to 0}{lim}\frac{f(h,0,0)-f(0,0,0)}{h}$$$$\underset{h\to 0}{lim} \frac{f(h,0,0)-f(0,0,0)}{h}$$lim_(h rarr0)(f(h,0,0)-f(0,0,0))/(h)lim_{h to 0} frac{f(h, 0, 0) – f(0, 0, 0)}{h}$$\underset{h\to 0}{lim}\frac{f(h,0,0)-f(0,0,0)}{h}$$
Substituting the function, we get:
$$\underset{h\to 0}{lim}\frac{\sqrt{h^2}}{h}$$$$\underset{h\to 0}{lim} \frac{\sqrt{h^2}}{h}$$lim_(h rarr0)(sqrt(h^(2)))/(h)lim_{h to 0} frac{sqrt{h^2}}{h}$$\underset{h\to 0}{lim}\frac{\sqrt{h^2}}{h}$$
This simplifies to:
$$\underset{h\to 0}{lim}\frac{|h|}{h}$$$$\underset{h\to 0}{lim} \frac{|h|}{h}$$lim_(h rarr0)(|h|)/(h)lim_{h to 0} frac{|h|}{h}$$\underset{h\to 0}{lim}\frac{|h|}{h}$$
This limit does not exist because it approaches $1$$1$11$1$ as $h$$h$hh$h$ approaches $0$$0$00$0$ from the right and $-1$$-1$-1-1$-1$ as $h$$h$hh$h$ approaches $0$$0$00$0$ from the left. Therefore, the function $f(x,y,z)=\sqrt{x^2+y^2+z^2}$$f(x,y,z)=\sqrt{x^2+y^2+z^2}$f(x,y,z)=sqrt(x^(2)+y^(2)+z^(2))f(x, y, z) = sqrt{x^2 + y^2 + z^2}$f(x,y,z)=\sqrt{x^2+y^2+z^2}$ is not differentiable at the origin, even though it is continuous everywhere.

Conclusion

Thus, the statement is false. A real-valued function of three variables that is continuous everywhere is not necessarily differentiable everywhere.
b) The function $f(x,y)=\ln \left(\frac{x+y}{x}\right)$$f(x,y)=\ln \left(\frac{x+y}{x}\right)$f(x,y)=ln((x+y)/(x))f(x, y)=ln left(frac{x+y}{x}right)$f(x,y)=\ln \left(\frac{x+y}{x}\right)$ is not homogeneous function.
Answer:
To determine the truth of the statement "The function $f(x,y)=\ln \left(\frac{x+y}{x}\right)$$f(x,y)=\ln \left(\frac{x+y}{x}\right)$f(x,y)=ln((x+y)/(x))f(x, y) = lnleft(frac{x+y}{x}right)$f(x,y)=\ln \left(\frac{x+y}{x}\right)$ is not a homogeneous function", we need to verify if the function is homogeneous of any degree. A function $f(x,y)$$f(x,y)$f(x,y)f(x, y)$f(x,y)$ is homogeneous of degree $n$$n$nn$n$ if it satisfies:
$$f(\lambda x,\lambda y)={\lambda }^{n}f(x,y)$$$$f(\lambda x,\lambda y)={\lambda }^{n}f(x,y)$$f(lambda x,lambda y)=lambda ^(n)f(x,y)f(lambda x, lambda y) = lambda^n f(x, y)$$f(\lambda x,\lambda y)={\lambda }^{n}f(x,y)$$
for all scalars $\lambda$$\lambda$lambdalambda$\lambda$. If a function satisfies this condition for some degree $n$$n$nn$n$, it is homogeneous; otherwise, it is not.
For the given function $f(x,y)=\ln \left(\frac{x+y}{x}\right)$$f(x,y)=\ln \left(\frac{x+y}{x}\right)$f(x,y)=ln((x+y)/(x))f(x, y) = lnleft(frac{x+y}{x}right)$f(x,y)=\ln \left(\frac{x+y}{x}\right)$, let’s substitute $\lambda x$$\lambda x$lambda xlambda x$\lambda x$ and $\lambda y$$\lambda y$lambda ylambda y$\lambda y$ and simplify:
$$f(\lambda x,\lambda y)=\ln \left(\frac{\lambda x+\lambda y}{\lambda x}\right)$$$$f(\lambda x,\lambda y)=\ln \left(\frac{\lambda x+\lambda y}{\lambda x}\right)$$f(lambda x,lambda y)=ln((lambda x+lambda y)/(lambda x))f(lambda x, lambda y) = lnleft(frac{lambda x + lambda y}{lambda x}right)$$f(\lambda x,\lambda y)=\ln \left(\frac{\lambda x+\lambda y}{\lambda x}\right)$$
Simplifying this expression will help us compare it with ${\lambda }^{n}f(x,y)$${\lambda }^{n}f(x,y)$lambda ^(n)f(x,y)lambda^n f(x, y)${\lambda }^{n}f(x,y)$ to determine if the function is homogeneous and, if so, of which degree. Let’s perform this calculation.
Upon simplifying the function $f(\lambda x,\lambda y)=\ln \left(\frac{\lambda x+\lambda y}{\lambda x}\right)$$f(\lambda x,\lambda y)=\ln \left(\frac{\lambda x+\lambda y}{\lambda x}\right)$f(lambda x,lambda y)=ln((lambda x+lambda y)/(lambda x))f(lambda x, lambda y) = lnleft(frac{lambda x + lambda y}{lambda x}right)$f(\lambda x,\lambda y)=\ln \left(\frac{\lambda x+\lambda y}{\lambda x}\right)$, we find that it simplifies to:
$$f(\lambda x,\lambda y)=\ln \left(\frac{x+y}{x}\right)$$$$f(\lambda x,\lambda y)=\ln \left(\frac{x+y}{x}\right)$$f(lambda x,lambda y)=ln((x+y)/(x))f(lambda x, lambda y) = lnleft(frac{x + y}{x}right)$$f(\lambda x,\lambda y)=\ln \left(\frac{x+y}{x}\right)$$
This is identical to the original function $f(x,y)$$f(x,y)$f(x,y)f(x, y)$f(x,y)$. Therefore, the function satisfies the condition:
$$f(\lambda x,\lambda y)=f(x,y)$$$$f(\lambda x,\lambda y)=f(x,y)$$f(lambda x,lambda y)=f(x,y)f(lambda x, lambda y) = f(x, y)$$f(\lambda x,\lambda y)=f(x,y)$$
which is the same as:
$$f(\lambda x,\lambda y)={\lambda }^{0}f(x,y)$$$$f(\lambda x,\lambda y)={\lambda }^{0}f(x,y)$$f(lambda x,lambda y)=lambda^(0)f(x,y)f(lambda x, lambda y) = lambda^0 f(x, y)$$f(\lambda x,\lambda y)={\lambda }^{0}f(x,y)$$
since ${\lambda }^0=1$${\lambda }^0=1$lambda^(0)=1lambda^0 = 1${\lambda }^0=1$ for any $\lambda$$\lambda$lambdalambda$\lambda$. This means that the function $f(x,y)=\ln \left(\frac{x+y}{x}\right)$$f(x,y)=\ln \left(\frac{x+y}{x}\right)$f(x,y)=ln((x+y)/(x))f(x, y) = lnleft(frac{x+y}{x}right)$f(x,y)=\ln \left(\frac{x+y}{x}\right)$ is indeed homogeneous, but of degree $n=0$$n=0$n=0n = 0$n=0$.
Thus, the statement "The function $f(x,y)=\ln \left(\frac{x+y}{x}\right)$$f(x,y)=\ln \left(\frac{x+y}{x}\right)$f(x,y)=ln((x+y)/(x))f(x, y)=ln left(frac{x+y}{x}right)$f(x,y)=\ln \left(\frac{x+y}{x}\right)$ is not a homogeneous function" is false, as the function is homogeneous of degree 0.
c) The cylindrical coordinates of the point whose spherical coordinates is $(8,\frac{\pi }{6},\frac{\pi }{2})$$\left(8,\frac{\pi }{6},\frac{\pi }{2}\right)$(8,(pi)/(6),(pi)/(2))left(8, frac{pi}{6}, frac{pi}{2}right)$(8,\frac{\pi }{6},\frac{\pi }{2})$ is $(8,\frac{\pi }{6},0)$$\left(8,\frac{\pi }{6},0\right)$(8,(pi)/(6),0)left(8, frac{pi}{6}, 0right)$(8,\frac{\pi }{6},0)$
Answer:
The statement "The cylindrical coordinates of the point whose spherical coordinates are $(8,\frac{\pi }{6},\frac{\pi }{2})$$\left(8,\frac{\pi }{6},\frac{\pi }{2}\right)$(8,(pi)/(6),(pi)/(2))left(8, frac{pi}{6}, frac{pi}{2}right)$(8,\frac{\pi }{6},\frac{\pi }{2})$ is $(8,\frac{\pi }{6},0)$$\left(8,\frac{\pi }{6},0\right)$(8,(pi)/(6),0)left(8, frac{pi}{6}, 0right)$(8,\frac{\pi }{6},0)$" is false. To demonstrate this, let’s convert the given spherical coordinates to cylindrical coordinates and compare.

Spherical Coordinates to Cylindrical Coordinates Conversion

The spherical coordinates $(r,\theta ,\varphi )$$(r,\theta ,\varphi )$(r,theta,phi)(r, theta, phi)$(r,\theta ,\varphi )$ are related to cylindrical coordinates $(\rho ,\varphi ,z)$$(\rho ,\varphi ,z)$(rho,phi,z)(rho, phi, z)$(\rho ,\varphi ,z)$ by the following relationships:

• $\rho =r\sin (\varphi )$$\rho =r\sin (\varphi )$rho=r sin(phi)rho = r sin(phi)$\rho =r\sin (\varphi )$
• $z=r\cos (\varphi )$$z=r\cos (\varphi )$z=r cos(phi)z = r cos(phi)$z=r\cos (\varphi )$
• $\varphi$$\varphi$phiphi$\varphi$ (the angle in the xy-plane from the x-axis) remains the same in both coordinate systems.

Given the spherical coordinates $(8,\frac{\pi }{6},\frac{\pi }{2})$$\left(8,\frac{\pi }{6},\frac{\pi }{2}\right)$(8,(pi)/(6),(pi)/(2))left(8, frac{pi}{6}, frac{pi}{2}right)$(8,\frac{\pi }{6},\frac{\pi }{2})$, where $r=8$$r=8$r=8r = 8$r=8$, $\theta =\frac{\pi }{6}$$\theta =\frac{\pi }{6}$theta=(pi)/(6)theta = frac{pi}{6}$\theta =\frac{\pi }{6}$, and $\varphi =\frac{\pi }{2}$$\varphi =\frac{\pi }{2}$phi=(pi)/(2)phi = frac{pi}{2}$\varphi =\frac{\pi }{2}$, we can convert them to cylindrical coordinates:

• $\rho =8\sin \left(\frac{\pi }{2}\right)=8$$\rho =8\sin \left(\frac{\pi }{2}\right)=8$rho=8sin((pi)/(2))=8rho = 8 sinleft(frac{pi}{2}right) = 8$\rho =8\sin \left(\frac{\pi }{2}\right)=8$ (since $\sin \left(\frac{\pi }{2}\right)=1$$\sin \left(\frac{\pi }{2}\right)=1$sin((pi)/(2))=1sinleft(frac{pi}{2}right) = 1$\sin \left(\frac{\pi }{2}\right)=1$)
• $z=8\cos \left(\frac{\pi }{2}\right)=0$$z=8\cos \left(\frac{\pi }{2}\right)=0$z=8cos((pi)/(2))=0z = 8 cosleft(frac{pi}{2}right) = 0$z=8\cos \left(\frac{\pi }{2}\right)=0$ (since $\cos \left(\frac{\pi }{2}\right)=0$$\cos \left(\frac{\pi }{2}\right)=0$cos((pi)/(2))=0cosleft(frac{pi}{2}right) = 0$\cos \left(\frac{\pi }{2}\right)=0$)
• The angle $\varphi$$\varphi$phiphi$\varphi$ in cylindrical coordinates is the same as $\theta$$\theta$thetatheta$\theta$ in spherical coordinates, so $\varphi =\frac{\pi }{6}$$\varphi =\frac{\pi }{6}$phi=(pi)/(6)phi = frac{pi}{6}$\varphi =\frac{\pi }{6}$.

Conclusion

Therefore, the cylindrical coordinates of the point whose spherical coordinates are $(8,\frac{\pi }{6},\frac{\pi }{2})$$\left(8,\frac{\pi }{6},\frac{\pi }{2}\right)$(8,(pi)/(6),(pi)/(2))left(8, frac{pi}{6}, frac{pi}{2}right)$(8,\frac{\pi }{6},\frac{\pi }{2})$ are $(8,\frac{\pi }{6},0)$$\left(8,\frac{\pi }{6},0\right)$(8,(pi)/(6),0)left(8, frac{pi}{6}, 0right)$(8,\frac{\pi }{6},0)$ in terms of $(\rho ,\varphi ,z)$$(\rho ,\varphi ,z)$(rho,phi,z)(rho, phi, z)$(\rho ,\varphi ,z)$, not $(8,\frac{\pi }{6},0)$$\left(8,\frac{\pi }{6},0\right)$(8,(pi)/(6),0)left(8, frac{pi}{6}, 0right)$(8,\frac{\pi }{6},0)$ in terms of $(r,\theta ,z)$$(r,\theta ,z)$(r,theta,z)(r, theta, z)$(r,\theta ,z)$. The statement is false because it incorrectly identifies the third component of the cylindrical coordinates as $0$$0$00$0$ instead of correctly identifying the second component as $\frac{\pi }{6}$$\frac{\pi }{6}$(pi)/(6)frac{pi}{6}$\frac{\pi }{6}$ and the third component as $0$$0$00$0$.
d) The unique solution $y(x)$$y(x)$y(x)y(x)$y(x)$ of an ordinary differential equation:
$$\frac{dy}{dx}=\{\begin{array}{ll}0,& \text{for}x<0\\ 1,& \text{for}x\ge 0\end{array}$$$$\frac{dy}{dx}=\left\{\begin{array}{l}0,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{for}x<0\\ 1,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{for}x\ge 0\end{array}\right.$$(dy)/(dx)={[0″,”,” for “x < 0],[1″,”,” for “x >= 0]:}frac{d y}{d x}= begin{cases}0, & text { for } x<0 \ 1, & text { for } x geq 0end{cases}$$\frac{dy}{dx}=\{\begin{array}{ll}0,& \text{for}x<0\\ 1,& \text{for}x\ge 0\end{array}$$
exists $\mathrm{\forall }x\in \mathbb{R}$$\mathrm{\forall }x\in \mathbb{R}$AA x inRforall x in mathbb{R}$\mathrm{\forall }x\in \mathbb{R}$.
Answer:
To determine whether the unique solution $y(x)$$y(x)$y(x)y(x)$y(x)$ exists for the given ordinary differential equation (ODE)
$$\frac{dy}{dx}=\{\begin{array}{ll}0,& \text{for}x<0\\ 1,& \text{for}x\ge 0\end{array}$$$$\frac{dy}{dx}=\left\{\begin{array}{l}0,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{for}x<0\\ 1,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{for}x\ge 0\end{array}\right.$$(dy)/(dx)={[0″,”,”for “x < 0],[1″,”,”for “x >= 0]:}frac{dy}{dx} = begin{cases} 0, & text{for } x < 0 \ 1, & text{for } x geq 0 end{cases}$$\frac{dy}{dx}=\{\begin{array}{ll}0,& \text{for}x<0\\ 1,& \text{for}x\ge 0\end{array}$$
for all $x\in \mathbb{R}$$x\in \mathbb{R}$x inRx in mathbb{R}$x\in \mathbb{R}$, we can use the Lipschitz condition.

The Lipschitz Condition

The Lipschitz condition is a criterion used to determine the existence and uniqueness of solutions to ODEs. A function $f(x,y)$$f(x,y)$f(x,y)f(x, y)$f(x,y)$ satisfies the Lipschitz condition in $y$$y$yy$y$ on a domain if there exists a constant $L$$L$LL$L$ such that for all $y_1,y_2$$y_1,y_2$y_(1),y_(2)y_1, y_2$y_1,y_2$ in the domain,
$$|f(x,y_1)-f(x,y_2)|\le L|y_1-y_2|$$$$|f(x,y_1)-f(x,y_2)|\le L|y_1-y_2|$$|f(x,y_(1))-f(x,y_(2))| <= L|y_(1)-y_(2)||f(x, y_1) – f(x, y_2)| leq L |y_1 – y_2|$$|f(x,y_1)-f(x,y_2)|\le L|y_1-y_2|$$

Applying the Lipschitz Condition to the Given ODE

In our case, the function $f(x,y)$$f(x,y)$f(x,y)f(x, y)$f(x,y)$ representing the right-hand side of the ODE is piecewise constant: it is 0 for $x<0$$x<0$x < 0x < 0$x<0$ and 1 for $x\ge 0$$x\ge 0$x >= 0x geq 0$x\ge 0$. This function does not depend on $y$$y$yy$y$, so the difference $|f(x,y_1)-f(x,y_2)|$$|f(x,y_1)-f(x,y_2)|$|f(x,y_(1))-f(x,y_(2))||f(x, y_1) – f(x, y_2)|$|f(x,y_1)-f(x,y_2)|$ is always 0 regardless of the values of $y_1$$y_1$y_(1)y_1$y_1$ and $y_2$$y_2$y_(2)y_2$y_2$. Therefore, the function trivially satisfies the Lipschitz condition in $y$$y$yy$y$ for any $x$$x$xx$x$, with any Lipschitz constant $L\ge 0$$L\ge 0$L >= 0L geq 0$L\ge 0$.

Conclusion

Since the function $f(x,y)$$f(x,y)$f(x,y)f(x, y)$f(x,y)$ satisfies the Lipschitz condition, the Picard-Lindelöf theorem (or the existence and uniqueness theorem) guarantees that a unique solution $y(x)$$y(x)$y(x)y(x)$y(x)$ exists for each initial value problem defined by this ODE. This holds for all $x\in \mathbb{R}$$x\in \mathbb{R}$x inRx in mathbb{R}$x\in \mathbb{R}$.
Hence given statement is True.
e) The differential equation:
$${[1+{\left(y^{\mathrm{\prime }}\right)}^2]}^{\frac{5}{3}}=y^{\mathrm{\prime }\mathrm{\prime }}$$$${\left[1+{\left(y^{\mathrm{\prime }}\right)}^2\right]}^{\frac{5}{3}}=y^{\mathrm{\prime }\mathrm{\prime }}$$[1+(y^(‘))^(2)]^((5)/(3))=y^(”)left[1+left(y^{prime}right)^2right]^{frac{5}{3}}=y^{prime prime}$${[1+{\left(y^{\mathrm{\prime }}\right)}^2]}^{\frac{5}{3}}=y^{\mathrm{\prime }\mathrm{\prime }}$$
is a second order differential equation of degree 3 .
Answer:
The original equation is:
$${[1+{\left(y^{\mathrm{\prime }}\right)}^2]}^{\frac{5}{3}}=y^{\mathrm{\prime }\mathrm{\prime }}$$$${\left[1+{\left(y^{\mathrm{\prime }}\right)}^2\right]}^{\frac{5}{3}}=y^{\mathrm{\prime }\mathrm{\prime }}$$[1+(y^(‘))^(2)]^((5)/(3))=y^(”)left[1+left(y^{prime}right)^2right]^{frac{5}{3}}=y^{prime prime}$${[1+{\left(y^{\mathrm{\prime }}\right)}^2]}^{\frac{5}{3}}=y^{\mathrm{\prime }\mathrm{\prime }}$$
Cubing both sides gives:
$${\left({[1+{\left(y^{\mathrm{\prime }}\right)}^2]}^{\frac{5}{3}}\right)}^3={\left(y^{\mathrm{\prime }\mathrm{\prime }}\right)}^3$$$${\left({\left[1+{\left(y^{\mathrm{\prime }}\right)}^2\right]}^{\frac{5}{3}}\right)}^3={\left(y^{\mathrm{\prime }\mathrm{\prime }}\right)}^3$$([1+(y^(‘))^(2)]^((5)/(3)))^(3)=(y^(”))^(3)left(left[1+left(y^{prime}right)^2right]^{frac{5}{3}}right)^3=left(y^{prime prime}right)^3$${\left({[1+{\left(y^{\mathrm{\prime }}\right)}^2]}^{\frac{5}{3}}\right)}^3={\left(y^{\mathrm{\prime }\mathrm{\prime }}\right)}^3$$
Simplifying the left side:
$${[1+{\left(y^{\mathrm{\prime }}\right)}^2]}^5={\left(y^{\mathrm{\prime }\mathrm{\prime }}\right)}^3$$$${\left[1+{\left(y^{\mathrm{\prime }}\right)}^2\right]}^5={\left(y^{\mathrm{\prime }\mathrm{\prime }}\right)}^3$$[1+(y^(‘))^(2)]^(5)=(y^(”))^(3)left[1+left(y^{prime}right)^2right]^5=left(y^{prime prime}right)^3$${[1+{\left(y^{\mathrm{\prime }}\right)}^2]}^5={\left(y^{\mathrm{\prime }\mathrm{\prime }}\right)}^3$$
Now, we have a second-order differential equation of degree 3 :
$${\left(y^{\mathrm{\prime }\mathrm{\prime }}\right)}^3={[1+{\left(y^{\mathrm{\prime }}\right)}^2]}^5$$$${\left(y^{\mathrm{\prime }\mathrm{\prime }}\right)}^3={\left[1+{\left(y^{\mathrm{\prime }}\right)}^2\right]}^5$$(y^(”))^(3)=[1+(y^(‘))^(2)]^(5)left(y^{prime prime}right)^3=left[1+left(y^{prime}right)^2right]^5$${\left(y^{\mathrm{\prime }\mathrm{\prime }}\right)}^3={[1+{\left(y^{\mathrm{\prime }}\right)}^2]}^5$$
Which shows that this differential equation is of second order and degree-3.
Hence, Given statement is True.
f) Differential equation:
$$\cos x\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}+x{y}^2=0$$$$\cos x\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}+x{y}^2=0$$cos x(d^(2)y)/(dx^(2))+(dy)/(dx)+xy^(2)=0cos x frac{d^2 y}{d x^2}+frac{d y}{d x}+x y^2=0$$\cos x\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}+x{y}^2=0$$
in $]0,\pi [$$]0,\pi [$]0,pi[] 0, pi[$]0,\pi [$ is a linear, homogeneous equation.
Answer:
The statement "The differential equation
$$\cos x\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}+x{y}^2=0$$$$\cos x\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}+x{y}^2=0$$cos x(d^(2)y)/(dx^(2))+(dy)/(dx)+xy^(2)=0cos x frac{d^2 y}{d x^2} + frac{d y}{d x} + x y^2 = 0$$\cos x\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}+x{y}^2=0$$
in $]0,\pi [$$]0,\pi [$]0,pi[]0, pi[$]0,\pi [$ is a linear, homogeneous differential equation" is false.

Justification

To determine whether the differential equation is linear and homogeneous, let’s recall the definitions:

1. Linear Differential Equation: A differential equation is linear if it can be written in the form
   $$a_n(x)y^{(n)}+a_{n-1}(x)y^{(n-1)}+\cdots +a_1(x)y^{\prime }+a_0(x)y=g(x)$$$$a_n(x)y^{(n)}+a_{n-1}(x)y^{(n-1)}+\cdots +a_1(x)y^{\prime }+a_0(x)y=g(x)$$a_(n)(x)y^((n))+a_(n-1)(x)y^((n-1))+cdots+a_(1)(x)y^(‘)+a_(0)(x)y=g(x)a_n(x)y^{(n)} + a_{n-1}(x)y^{(n-1)} + cdots + a_1(x)y’ + a_0(x)y = g(x)$$a_n(x)y^{(n)}+a_{n-1}(x)y^{(n-1)}+\cdots +a_1(x)y^{\prime }+a_0(x)y=g(x)$$
   where $y^{(n)}$$y^{(n)}$y^((n))y^{(n)}$y^{(n)}$ is the $n$$n$nn$n$th derivative of $y$$y$yy$y$ with respect to $x$$x$xx$x$, and $a_i(x)$$a_i(x)$a_(i)(x)a_i(x)$a_i(x)$ are functions of $x$$x$xx$x$ only. The function $g(x)$$g(x)$g(x)g(x)$g(x)$ on the right-hand side is the non-homogeneous term.
2. Homogeneous Differential Equation: A linear differential equation is homogeneous if $g(x)=0$$g(x)=0$g(x)=0g(x) = 0$g(x)=0$ for all $x$$x$xx$x$.

Analysis of the Given Differential Equation

The given differential equation is:
$$\cos x\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}+x{y}^2=0$$$$\cos x\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}+x{y}^2=0$$cos x(d^(2)y)/(dx^(2))+(dy)/(dx)+xy^(2)=0cos x frac{d^2 y}{d x^2} + frac{d y}{d x} + x y^2 = 0$$\cos x\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}+x{y}^2=0$$

• The term $\cos x\frac{d^{2}y}{d{x}^2}$$\cos x\frac{d^{2}y}{d{x}^2}$cos x(d^(2)y)/(dx^(2))cos x frac{d^2 y}{d x^2}$\cos x\frac{d^{2}y}{d{x}^2}$ is linear because it involves the second derivative of $y$$y$yy$y$ multiplied by a function of $x$$x$xx$x$ only.
• The term $\frac{dy}{dx}$$\frac{dy}{dx}$(dy)/(dx)frac{d y}{d x}$\frac{dy}{dx}$ is linear because it is the first derivative of $y$$y$yy$y$ without any multiplication by $y$$y$yy$y$ or its derivatives.
• However, the term $x{y}^2$$x{y}^2$xy^(2)x y^2$x{y}^2$ is not linear because it involves $y$$y$yy$y$ raised to the second power, which is a non-linear term.

Conclusion

Since the equation includes the non-linear term $x{y}^2$$x{y}^2$xy^(2)x y^2$x{y}^2$, it is not a linear differential equation. Consequently, it cannot be a homogeneous linear differential equation either. Therefore, the statement is false. The differential equation is non-linear due to the presence of the $y^2$$y^2$y^(2)y^2$y^2$ term.


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