BMTC-132 Solved Assignment 2024

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BMTC-132 Solved Assignment 2024 Sample Solution
bmtc-132-solved-assignment-2024-ss-10cd7422-4b11-4531-9171-a6445826d855

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 = 0x=0.
For a function of three variables, we can construct a similar example. Consider the function
f ( x , y , z ) = x 2 + y 2 + z 2 f ( x , y , z ) = 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)=x2+y2+z2
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 xxx at the origin. The partial derivative is defined as the limit of the difference quotient:
lim h → 0 f ( h , 0 , 0 ) − f ( 0 , 0 , 0 ) h lim h → 0   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}limh→0f(h,0,0)−f(0,0,0)h
Substituting the function, we get:
lim h → 0 h 2 h lim h → 0   h 2 h lim_(h rarr0)(sqrt(h^(2)))/(h)\lim_{h \to 0} \frac{\sqrt{h^2}}{h}limh→0h2h
This simplifies to:
lim h → 0 | h | h lim h → 0   | h | h lim_(h rarr0)(|h|)/(h)\lim_{h \to 0} \frac{|h|}{h}limh→0|h|h
This limit does not exist because it approaches 1 1 111 as h h hhh approaches 0 0 000 from the right and − 1 − 1 -1-1−1 as h h hhh approaches 0 0 000 from the left. Therefore, the function f ( x , y , z ) = x 2 + y 2 + z 2 f ( x , y , z ) = 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)=x2+y2+z2 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 ⁡ ( x + y x ) f ( x , y ) = ln ⁡ x + y x f(x,y)=ln((x+y)/(x))f(x, y)=\ln \left(\frac{x+y}{x}\right)f(x,y)=ln⁡(x+yx) is not homogeneous function.
Answer:
To determine the truth of the statement "The function f ( x , y ) = ln ⁡ ( x + y x ) f ( x , y ) = ln ⁡ x + y x f(x,y)=ln((x+y)/(x))f(x, y) = \ln\left(\frac{x+y}{x}\right)f(x,y)=ln⁡(x+yx) 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 nnn if it satisfies:
f ( λ x , λ y ) = λ n f ( x , y ) f ( λ x , λ y ) = λ 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(λx,λy)=λnf(x,y)
for all scalars λ λ lambda\lambdaλ. If a function satisfies this condition for some degree n n nnn, it is homogeneous; otherwise, it is not.
For the given function f ( x , y ) = ln ⁡ ( x + y x ) f ( x , y ) = ln ⁡ x + y x f(x,y)=ln((x+y)/(x))f(x, y) = \ln\left(\frac{x+y}{x}\right)f(x,y)=ln⁡(x+yx), let’s substitute λ x λ x lambda x\lambda xλx and λ y λ y lambda y\lambda yλy and simplify:
f ( λ x , λ y ) = ln ⁡ ( λ x + λ y λ x ) f ( λ x , λ y ) = ln ⁡ λ x + λ y λ x f(lambda x,lambda y)=ln((lambda x+lambda y)/(lambda x))f(\lambda x, \lambda y) = \ln\left(\frac{\lambda x + \lambda y}{\lambda x}\right)f(λx,λy)=ln⁡(λx+λyλx)
Simplifying this expression will help us compare it with λ n f ( x , y ) λ n f ( x , y ) lambda ^(n)f(x,y)\lambda^n f(x, y)λnf(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 ( λ x , λ y ) = ln ⁡ ( λ x + λ y λ x ) f ( λ x , λ y ) = ln ⁡ λ x + λ y λ x f(lambda x,lambda y)=ln((lambda x+lambda y)/(lambda x))f(\lambda x, \lambda y) = \ln\left(\frac{\lambda x + \lambda y}{\lambda x}\right)f(λx,λy)=ln⁡(λx+λyλx), we find that it simplifies to:
f ( λ x , λ y ) = ln ⁡ ( x + y x ) f ( λ x , λ y ) = ln ⁡ x + y x f(lambda x,lambda y)=ln((x+y)/(x))f(\lambda x, \lambda y) = \ln\left(\frac{x + y}{x}\right)f(λx,λy)=ln⁡(x+yx)
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 ( λ x , λ y ) = f ( x , y ) f ( λ x , λ y ) = f ( x , y ) f(lambda x,lambda y)=f(x,y)f(\lambda x, \lambda y) = f(x, y)f(λx,λy)=f(x,y)
which is the same as:
f ( λ x , λ y ) = λ 0 f ( x , y ) f ( λ x , λ y ) = λ 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(λx,λy)=λ0f(x,y)
since λ 0 = 1 λ 0 = 1 lambda^(0)=1\lambda^0 = 1λ0=1 for any λ λ lambda\lambdaλ. This means that the function f ( x , y ) = ln ⁡ ( x + y x ) f ( x , y ) = ln ⁡ x + y x f(x,y)=ln((x+y)/(x))f(x, y) = \ln\left(\frac{x+y}{x}\right)f(x,y)=ln⁡(x+yx) is indeed homogeneous, but of degree n = 0 n = 0 n=0n = 0n=0.
Thus, the statement "The function f ( x , y ) = ln ⁡ ( x + y x ) f ( x , y ) = ln ⁡ x + y x f(x,y)=ln((x+y)/(x))f(x, y)=\ln \left(\frac{x+y}{x}\right)f(x,y)=ln⁡(x+yx) 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 , π 6 , π 2 ) 8 , π 6 , π 2 (8,(pi)/(6),(pi)/(2))\left(8, \frac{\pi}{6}, \frac{\pi}{2}\right)(8,π6,π2) is ( 8 , π 6 , 0 ) 8 , π 6 , 0 (8,(pi)/(6),0)\left(8, \frac{\pi}{6}, 0\right)(8,π6,0)
Answer:
The statement "The cylindrical coordinates of the point whose spherical coordinates are ( 8 , π 6 , π 2 ) 8 , π 6 , π 2 (8,(pi)/(6),(pi)/(2))\left(8, \frac{\pi}{6}, \frac{\pi}{2}\right)(8,π6,π2) is ( 8 , π 6 , 0 ) 8 , π 6 , 0 (8,(pi)/(6),0)\left(8, \frac{\pi}{6}, 0\right)(8,π6,0)" 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 , θ , ϕ ) ( r , θ , ϕ ) (r,theta,phi)(r, \theta, \phi)(r,θ,ϕ) are related to cylindrical coordinates ( ρ , ϕ , z ) ( ρ , ϕ , z ) (rho,phi,z)(\rho, \phi, z)(ρ,ϕ,z) by the following relationships:
  • ρ = r sin ⁡ ( ϕ ) ρ = r sin ⁡ ( ϕ ) rho=r sin(phi)\rho = r \sin(\phi)ρ=rsin⁡(ϕ)
  • z = r cos ⁡ ( ϕ ) z = r cos ⁡ ( ϕ ) z=r cos(phi)z = r \cos(\phi)z=rcos⁡(ϕ)
  • ϕ ϕ phi\phiϕ (the angle in the xy-plane from the x-axis) remains the same in both coordinate systems.
Given the spherical coordinates ( 8 , π 6 , π 2 ) 8 , π 6 , π 2 (8,(pi)/(6),(pi)/(2))\left(8, \frac{\pi}{6}, \frac{\pi}{2}\right)(8,π6,π2), where r = 8 r = 8 r=8r = 8r=8, θ = π 6 θ = π 6 theta=(pi)/(6)\theta = \frac{\pi}{6}θ=π6, and ϕ = π 2 ϕ = π 2 phi=(pi)/(2)\phi = \frac{\pi}{2}ϕ=π2, we can convert them to cylindrical coordinates:
  • ρ = 8 sin ⁡ ( π 2 ) = 8 ρ = 8 sin ⁡ π 2 = 8 rho=8sin((pi)/(2))=8\rho = 8 \sin\left(\frac{\pi}{2}\right) = 8ρ=8sin⁡(π2)=8 (since sin ⁡ ( π 2 ) = 1 sin ⁡ π 2 = 1 sin((pi)/(2))=1\sin\left(\frac{\pi}{2}\right) = 1sin⁡(π2)=1)
  • z = 8 cos ⁡ ( π 2 ) = 0 z = 8 cos ⁡ π 2 = 0 z=8cos((pi)/(2))=0z = 8 \cos\left(\frac{\pi}{2}\right) = 0z=8cos⁡(π2)=0 (since cos ⁡ ( π 2 ) = 0 cos ⁡ π 2 = 0 cos((pi)/(2))=0\cos\left(\frac{\pi}{2}\right) = 0cos⁡(π2)=0)
  • The angle ϕ ϕ phi\phiϕ in cylindrical coordinates is the same as θ θ theta\thetaθ in spherical coordinates, so ϕ = π 6 ϕ = π 6 phi=(pi)/(6)\phi = \frac{\pi}{6}ϕ=π6.

Conclusion

Therefore, the cylindrical coordinates of the point whose spherical coordinates are ( 8 , π 6 , π 2 ) 8 , π 6 , π 2 (8,(pi)/(6),(pi)/(2))\left(8, \frac{\pi}{6}, \frac{\pi}{2}\right)(8,π6,π2) are ( 8 , π 6 , 0 ) 8 , π 6 , 0 (8,(pi)/(6),0)\left(8, \frac{\pi}{6}, 0\right)(8,π6,0) in terms of ( ρ , ϕ , z ) ( ρ , ϕ , z ) (rho,phi,z)(\rho, \phi, z)(ρ,ϕ,z), not ( 8 , π 6 , 0 ) 8 , π 6 , 0 (8,(pi)/(6),0)\left(8, \frac{\pi}{6}, 0\right)(8,π6,0) in terms of ( r , θ , z ) ( r , θ , z ) (r,theta,z)(r, \theta, z)(r,θ,z). The statement is false because it incorrectly identifies the third component of the cylindrical coordinates as 0 0 000 instead of correctly identifying the second component as π 6 π 6 (pi)/(6)\frac{\pi}{6}π6 and the third component as 0 0 000.
d) The unique solution y ( x ) y ( x ) y(x)y(x)y(x) of an ordinary differential equation:
d y d x = { 0 , for x < 0 1 , for x ≥ 0 d y d x = 0 ,      for x < 0 1 ,      for x ≥ 0 (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 0\end{cases}dydx={0, for x<01, for x≥0
exists ∀ x ∈ R ∀ x ∈ R AA x inR\forall x \in \mathbb{R}∀x∈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)
d y d x = { 0 , for x < 0 1 , for x ≥ 0 d y d x = 0 ,      for x < 0 1 ,      for x ≥ 0 (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}dydx={0,for x<01,for x≥0
for all x ∈ R x ∈ R x inRx \in \mathbb{R}x∈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 yyy on a domain if there exists a constant L L LLL such that for all y 1 , y 2 y 1 , y 2 y_(1),y_(2)y_1, y_2y1,y2 in the domain,
| f ( x , y 1 ) − f ( x , y 2 ) | ≤ 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))| <= L|y_(1)-y_(2)||f(x, y_1) – f(x, y_2)| \leq L |y_1 – y_2||f(x,y1)−f(x,y2)|≤L|y1−y2|

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 < 0x<0 and 1 for x ≥ 0 x ≥ 0 x >= 0x \geq 0x≥0. This function does not depend on y y yyy, 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,y1)−f(x,y2)| is always 0 regardless of the values of y 1 y 1 y_(1)y_1y1 and y 2 y 2 y_(2)y_2y2. Therefore, the function trivially satisfies the Lipschitz condition in y y yyy for any x x xxx, with any Lipschitz constant L ≥ 0 L ≥ 0 L >= 0L \geq 0L≥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 ∈ R x ∈ R x inRx \in \mathbb{R}x∈R.
Hence given statement is True.
e) The differential equation:
[ 1 + ( y ′ ) 2 ] 5 3 = y ′ ′ 1 + y ′ 2 5 3 = y ′ ′ [1+(y^(‘))^(2)]^((5)/(3))=y^(”)\left[1+\left(y^{\prime}\right)^2\right]^{\frac{5}{3}}=y^{\prime \prime}[1+(y′)2]53=y′′
is a second order differential equation of degree 3 .
Answer:
The original equation is:
[ 1 + ( y ′ ) 2 ] 5 3 = y ′ ′ 1 + y ′ 2 5 3 = y ′ ′ [1+(y^(‘))^(2)]^((5)/(3))=y^(”)\left[1+\left(y^{\prime}\right)^2\right]^{\frac{5}{3}}=y^{\prime \prime}[1+(y′)2]53=y′′
Cubing both sides gives:
( [ 1 + ( y ′ ) 2 ] 5 3 ) 3 = ( y ′ ′ ) 3 1 + y ′ 2 5 3 3 = y ′ ′ 3 ([1+(y^(‘))^(2)]^((5)/(3)))^(3)=(y^(”))^(3)\left(\left[1+\left(y^{\prime}\right)^2\right]^{\frac{5}{3}}\right)^3=\left(y^{\prime \prime}\right)^3([1+(y′)2]53)3=(y′′)3
Simplifying the left side:
[ 1 + ( y ′ ) 2 ] 5 = ( y ′ ′ ) 3 1 + y ′ 2 5 = y ′ ′ 3 [1+(y^(‘))^(2)]^(5)=(y^(”))^(3)\left[1+\left(y^{\prime}\right)^2\right]^5=\left(y^{\prime \prime}\right)^3[1+(y′)2]5=(y′′)3
Now, we have a second-order differential equation of degree 3 :
( y ′ ′ ) 3 = [ 1 + ( y ′ ) 2 ] 5 y ′ ′ 3 = 1 + y ′ 2 5 (y^(”))^(3)=[1+(y^(‘))^(2)]^(5)\left(y^{\prime \prime}\right)^3=\left[1+\left(y^{\prime}\right)^2\right]^5(y′′)3=[1+(y′)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 d 2 y d x 2 + d y d x + x y 2 = 0 cos ⁡ x d 2 y d x 2 + d y d x + x y 2 = 0 cos x(d^(2)y)/(dx^(2))+(dy)/(dx)+xy^(2)=0\cos x \frac{d^2 y}{d x^2}+\frac{d y}{d x}+x y^2=0cos⁡xd2ydx2+dydx+xy2=0
in ] 0 , π [ ] 0 , π [ ]0,pi[] 0, \pi[]0,π[ is a linear, homogeneous equation.
Answer:
The statement "The differential equation
cos ⁡ x d 2 y d x 2 + d y d x + x y 2 = 0 cos ⁡ x d 2 y d x 2 + d y d x + x y 2 = 0 cos x(d^(2)y)/(dx^(2))+(dy)/(dx)+xy^(2)=0\cos x \frac{d^2 y}{d x^2} + \frac{d y}{d x} + x y^2 = 0cos⁡xd2ydx2+dydx+xy2=0
in ] 0 , π [ ] 0 , π [ ]0,pi[]0, \pi[]0,π[ 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 ) + ⋯ + a 1 ( x ) y ′ + a 0 ( x ) y = g ( x ) a n ( x ) y ( n ) + a n − 1 ( x ) y ( n − 1 ) + ⋯ + 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’ + a_0(x)y = g(x)an(x)y(n)+an−1(x)y(n−1)+⋯+a1(x)y′+a0(x)y=g(x)
    where y ( n ) y ( n ) y^((n))y^{(n)}y(n) is the n n nnnth derivative of y y yyy with respect to x x xxx, and a i ( x ) a i ( x ) a_(i)(x)a_i(x)ai(x) are functions of x x xxx 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) = 0g(x)=0 for all x x xxx.

Analysis of the Given Differential Equation

The given differential equation is:
cos ⁡ x d 2 y d x 2 + d y d x + x y 2 = 0 cos ⁡ x d 2 y d x 2 + d y d x + x y 2 = 0 cos x(d^(2)y)/(dx^(2))+(dy)/(dx)+xy^(2)=0\cos x \frac{d^2 y}{d x^2} + \frac{d y}{d x} + x y^2 = 0cos⁡xd2ydx2+dydx+xy2=0
  • The term cos ⁡ x d 2 y d x 2 cos ⁡ x d 2 y d x 2 cos x(d^(2)y)/(dx^(2))\cos x \frac{d^2 y}{d x^2}cos⁡xd2ydx2 is linear because it involves the second derivative of y y yyy multiplied by a function of x x xxx only.
  • The term d y d x d y d x (dy)/(dx)\frac{d y}{d x}dydx is linear because it is the first derivative of y y yyy without any multiplication by y y yyy or its derivatives.
  • However, the term x y 2 x y 2 xy^(2)x y^2xy2 is not linear because it involves y y yyy 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^2xy2, 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^2y2 term.
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