TPDE MCQ.pdf - Department of Mechanical Engineering MCQ for Regulations 2017 T MA8353 Transforms and Partial Differential Equations which implies T =

# TPDE MCQ.pdf - Department of Mechanical Engineering MCQ for...

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This preview shows page 1 out of 44 pages. Unformatted text preview: Department of Mechanical Engineering MCQ for Regulations 2017 T MA8353 Transforms and Partial Differential Equations which implies T = c’ ekt ′ = k T C O M Therefore, u(x,t) = cc’ e(1+6k)x ekt Now, u(x,0) = 10 e-x = cc’e(1+6k)x Therefore, cc’ = 10 and k = -1⁄3 Therefore, u(x,t) = 10 e-x e-t/3. 2. Find the solution of = 36 + 10u if (t = 0) = 3e using the method of separation of variables. a) e e b) 3e e c) e e d) 3e e ∂u ∂x ∂u −2x ∂x −3 −2x −t/3 2 x 3 −t/3 2x −t/3 −x −t/3 ∂t T. Mechanical Engineering - Third Semester ∂u O 2 SP Answer: a Explanation: u(x, t) = X(x)T (t) Substituting in the given equation, ′ ′ X T = 36T X + 10XT which implies X ′ = k LO G UNIT I PARTIAL DIFFERENTIAL EQUATIONS X 17 .B TOPIC 1.1 FORMATION OF PARTIAL DIFFERENTIAL EQUATIONS - SINGULAR INTEGRALS 1. Solve = 6 + u using the method of separation of variables if u(x,0) = 10 e-x. a) 10 e-x e-t/3 b) 10 ex e-t/3 c) 10 ex/3 e-t d) 10 e-x/3 e-t ∂u ∂x ∂t C SE -R ∂u Answer: a Explanation: u(x,t) = X(x) T(t) Substituting in the given equation, X’T = 6 T’X + XT ′ X −X 6X X = T T T ′ = T ∂u ∂x (k−10) −2x (t = 0) = 3e = 1 + 6k ′ ′ = c e k−10 36 t kx = cc ke Therefore k = -2 and cc’ = e e Hence, u(x, t) = −3 −2x 2 −3 2 −t/3 . 3. Solve the partial differential equation x + y = 0 using method of 3 ∂u 2 ∂u ∂x ∂y separation of variables if u(0, y) = 10 e a) 10e b) 10e c) 10e d) 10e 5 5 y . 5 2x2 e y e x −5 5 2y2 −5 2 2y −5 e x −5 2x2 5 e y Answer: d Explanation: u(x, t) = X(x)T (t) 3 ′ 2 ′ x X Y + y Y X = 0 X ′ X Y Y = ′ = k 3 x −k 2 y k which implies X = ce which implies Y = c e k ′ u(x, t) = cc e u(0, y) = 10e y ′ 2 2x k y k 2x2 5 = k kx = ce which implies T 36 ′ ′ X X e y ′ = cc e k y which implies X = ce(1+6k)x Downloaded From: 1 Department of Mechanical Engineering MCQ for Regulations 2017 Therefore k = 5 and cc’ = 10 Hence, u(x, t) = 10e e . −5 5 2 2x y the product of two functions which depend on one variable only. a) True b) False ∂u ∂u ∂x ∂y separation of variables if u(0, y) = 9e a) 9e e b) 9e e c) 9e e d) 9e e 17 5 13 5 x −5y x −5y −17 5 x −5y x −5y −13 5 Answer: a Explanation: If we have a function u(x,t), then the function u depends on both x and t. For using the variable separable method we assume that it can be written as u(x,t) = X(x).T(t) where X depends only on x and T depends only on t. . T. −5y C O M 4. Solve the differential equation 5 + 3 = 2u using the method of Answer: a Explanation: u(x, t) = X(x)T (t) ′ ′ which implies X = ce = 2 − k/3 which implies Y Therefore u(x, t) = cc e e ′ X Y k = k 5 x 5 ′ k ′ ′ 2−k = e 3 y 2−k x 3 3 y y −5y 9e Hence cc’ = 9 and k = 17 Therefore, u(x, t) = 9 e 17 5 x LO G u(0, y) = cc 5 2−k ′ = c e Y SP X O 5X Y + 3Y X = 2XY −5y e . 5. Solve the differential equation x + y = u using the method of 2 ∂u 2 ∂u ∂x ∂y 2 separation of variables if u(0, y) = e . a) e e b) e e c) e e d) e e y −3 2 y 2 y x 2 −3 y x 2 17 3 x .B x 3 y Answer: c Explanation: u(x, t) = X(x)T (t) 2 ′ 2 ′ -R x X Y + y Y X = XY −k X = ce ′ C SE Y = c e x k−1 y ′ −k u(x, t) = cc e u(0, y) = e 2 y x e 7. While solving a partial differential equation using a variable separable method, we equate the ratio to a constant which? a) can be positive or negative integer or zero b) can be positive or negative rational number or zero c) must be a positive integer d) must be a negative integer Answer: b Explanation: The constant can be any rational number. For example, we use a positive rational number to solve a 1Dimensional wave equation, we use a negative rational number to solve 1Dimensional heat equation, 0 when we have steady state. The choice of constant depends on the nature of the given problem. 8. When solving a 1-Dimensional wave equation using variable separable method, we get the solution if _____________ a) k is positive b) k is negative c) k is 0 d) k can be anything k−1 y ′ = cc e k−1 y k = 3 and cc’ = 1 Therefore u(x, t) = e 2 −3 x e y . 6. While solving a partial differential equation using a variable separable method, we assume that the function can be written as Downloaded From: Answer: b Explanation: Since the given problem is 1Dimensional wave equation, the solution should be periodic in nature. If k is a positive number, then the solution comes out to be (c7 epx⁄c+e-px⁄cc8)(c7 ept+e-ptc8) and if k is positive the solution comes out to be 2 Department of Mechanical Engineering MCQ for Regulations 2017 Answer: a Explanation: The calculus of variations is a type of analysis in the field of mathematics (branch of calculus) which is used to find maxima and minima of definite integrals. C O M (ccos(px/c) + c’sin(px/c))(c’’cospt + c’’’sinpt). Now, since it should be periodic, the solution is (ccos(px/c) + c’sin(px/c))(c’’cospt + c’’’sinpt). 9. When solving a 1-Dimensional heat equation using a variable separable method, we get the solution if ______________ a) k is positive b) k is negative c) k is 0 d) k can be anything c’’’sinpx) (c e -c2 p2 t T. O Answer: a Explanation: Partial derivatives are indicated by the symbol ∂. This was first used in mathematics by Marquis de Condorcet who used it for partial differences. SP Answer: b Explanation: Since this is a heat equation, the solution must be a transient solution, that is it should decay as time increases. This happens only when k is negative and the solution comes out to be (c’’cospx + 2. The symbol used for partial derivatives, ∂, was first used in mathematics by Marquis de Condorcet. a) True b) False 3. What is the order of the equation, xy ( ) + yx + = 0? a) Third Order b) Second Order c) First Order d) Zero Order 3 ). ∂y 2 LO G 10. While solving any partial differentiation equation using a variable separable method which is of order 1 or 2, we use the formula of fourier series to find the coefficients at last. a) True b) False 17 .B Answer: a Explanation: After using the boundary conditions, when we are left with only one constant and one boundary condition, then we use Fourier series coefficient formula to find the constant. -R TOPIC 1.2 SOLUTIONS OF STANDARD TYPES OF FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS SE C 2 ∂x 1. First order partial differential equations arise in the calculus of variations. a) True b) False Downloaded From: ∂y ∂x Answer: c Explanation: The equation having only first derivative, i.e., are said to be first order differential equation. Since the given equation satisfies this condition, it is of first order. ∂y ∂x 4. In the equation, y= x2+c,c is known as the parameter and x and y are known as the main variables. a) True b) False Answer: a Explanation: Given: y= x2+c, where c is known as an arbitrary constant. It is also referred to as the parameter to differentiate it from the main variables x and y. 5. Which of the following is one of the criterions for linearity of an equation? a) The dependent variable and its derivatives should be of second order 3 Department of Mechanical Engineering MCQ for Regulations 2017 Answer: d Explanation: Lagrange’s linear equation contains only the first-order partial derivatives which appear only with first power; hence the equation is of first-order and first-degree. Answer: d Explanation: The two criterions for linearity of an equation are: 9. Find the general solution of the linear partial differential equation, yzp+zxq=xy. a) φ(x2-y2 – z2 )=0 b) φ(x2-y2, y2-z2 )=0 c) φ(x2-y2, y2-x2 )=0 d) φ(x2-z2, z2-x2 )=0 C O M b) The dependent variable and its derivatives should not be of same order c) Each coefficient does not depend on the independent variable d) Each coefficient depends only on the independent variable SP O Answer: b Explanation: Given: yzp+zxq=xy Here, the subsidiary equations are, dx yz = dy = zx dz xy From the first two and last two terms, we get, respectively, = or xdx-ydy=0, and dx dy yz zx LO G 6. Which of the following is a type of Iterative method of solving non-linear equations? a) Graphical method b) Interpolation method c) Trial and Error methods d) Direct Analytical methods T. The dependent variable y and its derivatives are of first degree. Each coefficient depends only on the independent variable or ydx-zdy=0, Integrating these we get two solutions x2-y2=a , y2-z2=b Hence, the general solution of the given equation is, φ(x2-y2, y2-z2 )=0. dx .B Answer: b Explanation: There are 2 types of Iterative methods, (i) Interpolation methods (or Bracketing methods) and (ii) Extrapolation methods (or Open-end methods). -R 17 7. Which of the following is an example for first order linear partial differential equation? a) Lagrange’s Partial Differential Equation b) Clairaut’s Partial Differential Equation c) One-dimensional Wave Equation d) One-dimensional Heat Equation C SE Answer: a Explanation: Equations of the form Pp + Qq = R , where P, Q and R are functions of x, y, z, are known as Lagrange’s linear equation. 8. What is the nature of Lagrange’s linear partial differential equation? a) First-order, Third-degree b) Second-order, First-degree c) First-order, Second-degree d) First-order, First-degree Downloaded From: ′ zx = dy ′ xy 10. The equation 2 – xy = y , is an example for Bernoulli’s equation. a) False b) True dy −2 dx Answer: b Explanation: A first order, first degree differential equation of the form, + P (x). y = Q(x). y , is known as Bernoulli’s equation. dy a dx 11. A particular solution for an equation is derived by eliminating arbitrary constants. a) True b) False 4 Department of Mechanical Engineering MCQ for Regulations 2017 Answer: b Explanation: A particular solution for an equation is derived by substituting particular values to the arbitrary constants in the complete solution. 2. Obtain the general solution for the equation xp2+px-py+1-y=0 where p= . a) y=cx+ b) x=cy-(c+1) c) x=cyd) y=cx+(c+1) dy dx 1 C O M c+1 1 c+1 12. A partial differential equation is one in which a dependent variable (say ‘y’) depends on one or more independent variables (say ’x’, ’t’ etc.) a) False b) True ∂y ∂y ∂x ∂t ……(1) thus (1) is in the Clairaut’s equation form y=px+f(p), thus general solution is y=cx+ . dy c LO G ′ x a 2 .B a) 2 + y a 2 = 1 17 b) y2=-4ax c) y2=4ax d) x2=-2ay C SE -R ′ 0 = x − a 2 c → c 2 = hence (1) becomes y = √ y2=4ax a x dy dx c a x + a√ x a a x → c = √ c+1 1 2 2c+1 c 2 2c+2 Answer: b Explanation: X 2 Y = y → now p = 1 p = Answer: c Explanation: Let p = y = Given equation is of the form y = px + f(p), whose general solution is y = cx + f(c) thus the general solution is y = cx + …………..(1), to find the value of c we differentiate (1) partially w.r.t ‘c’ i.e 2c 2 2 1. Singular solution for the Clairaut’s equation y = y x + is given by _______ 2 c+1 2 2 a 1 c+1 dx TOPIC 1.3 LAGRANGE'S LINEAR EQUATION y 1 p+1 3. Find the general solution for the equation (px-py)(py+x)=2p by reducing into Clairaut’s form by using the substitution X=x2, Y=y2 where p= . a) y = x + b) y = cx – c) x = cy – d) x = y + 2 ′ or y = px + O p+1 T. xp(p+1)+1 y = SP Answer: b Explanation: A differential equation is divided into two types, ordinary and partial differential equations. A partial differential equation is one in which a dependent variable depends on one or more independent variables. Example: F (x, t, y, , , … …) = 0 Answer: a Explanation: xp2+px-py+1-y=0 xp2+px+1=y(p+1) dY dy dy dx = = x 2 dy dY dX dY dX dx → y = 2√ax is the singular solution. = 2x dY and let P = x P i. e p = √ y dx X Y P now consider (px-py)(py+x)=2p substituting the value of p we get (√ X Y P √ X– √ Y )(√ (P X−Y ) √Y X Y (P + 1)√ X = 2√ is in the Clairaut’s form hence general solution is y 2 a x dx = 2y ∗ P ∗ 2x or p = 2y dX → P √ Y + √ X) = 2 X Y P → (P X − Y ) 2 = cx – 2c c+1 . 4. Find the general solution of the D.E 2y4xy’-log y’=0. a) y(p) = – 1 + b) y(p) = – 2 + logp 2c logp p 2 c 2p Downloaded From: 5 Department of Mechanical Engineering −1 d) x(p) = 1 2p c + p p TOPIC 1.4 LINEAR PARTIAL DIFFERENTIAL EQUATIONS OF SECOND AND HIGHER ORDER WITH CONSTANT COEFFICIENTS OF BOTH HOMOGENEOUS AND NONHOMOGENEOUS TYPES 2 c + p 1/2 Answer: a Explanation: Let y’=p and hence given equation is in Lagrange equation form i.e 2y=4xp+log p …..(1) differentiating both sides of the equation 2dy=4xdp+4pdx+ and dy=pdx C O M c) x(p) = MCQ for Regulations 2017 dp p dp p p dx 1 dx 2 −1 dp p dp p 2p dp 2 logp 2p −1 p get 2y = 4p( y(p) = p 2 −1 c + 2p p logp –1 + 2 2 . c 2p 2 2c p c 2 .B p d) x(p) = 2p + p 2 2 p x(p) = ∫ 6p 2 ∫ p dp 2 = p dp + c → x(p) = 2p + c 2 p ….substituting in (1) we get y(p) = 2(2p + c 2 p c) lnx d) lnx xy + x 2 )p − 3p = c lny = c y lny – x = c y 2 = p + 2c p dy y(x−ylny) = dx x(xlnx−y) –> x2 lnx dy-xy dy=xy dx – y2 lny dx …….dividing by x2 y2 then lny dy – 2 y d( lnx y 1 xy 1 (lnx( 1 dy = ) + d( lny x dx – xy 1 dy) + 2 −y lnx ∫ d( 17 -R dp (2) is a linear D.E whose I.F=e hence its solution is SE C 2 xlnx−ylny 2 Answer: b Explanation: Let y’=p –> y = 2xp – 3p2 …. (1) is in the Lagrange equation form now differentiating we get dy=2xdp+2pdx6pdp and dy=pdx thus pdx=2xdp+2pdx-6pdp→-pdx=2xdp6pdp –> + x– 6 = 0…(2) dx b) = c xy lny 2 x dx dx) + (lny( xy 1 2 −x dx) + 1 xy ) = 0 on integrating we get 2c p xlnx+ylny Answer: a Explanation: ) + logp 5. Find the general solution of the D.E y = 2xy’ – 3(y’)2. a) y(p) = p + b) y(p) = p + c) x(p) = −cp + 1/2 a) 2 c 2p y(x−ylny) x(xlnx−y) 2 −1 2 2c 2 dy dx SP 2 p 2 LO G ∫ 1. Solution of the differential equation = is _____________ T. dp O –> 2pdx=4xdp+4pdx+ –> -2pdx=4pdx+ -2p = 4x + → + x = (p≠0)…….this is a linear D.E for the function x(p) I.F is e = e = p and solution is x(p) p = ∫ p ∗ dp + c x(p) = + substituting back in (1) we . Sanfoundry Global Education & Learning Series – Ordinary Differential Equations. Downloaded From: lny ) + ∫ d( y x ) = c…. where c is a constant of integration. lnx lny + y x 2. Solution of the differential equation = e + x e is ______ = + + c a) dy 3x−2y 2 −2y dx e 2y e 3 b) c) d) e 3x 2 x 3 3y (e 2x 2 3 +x ) + c 6 e 2y (e 3x 3 +x ) + c 6 e 2y 2 = e 3x 3 + 3 Answer: d Explanation: dy dx −2y = e 3 dy dx 3x (e x + c 3x−2y = e 2 −2y + x e 2 + x ) separating the variable 6 d Department of Mechanical Engineering MCQ for Regulations 2017 e 2 3x 1 3 3 1 3 Answer: c Explanation: sec2 x tany dx + sec2 y tanx dy=0 dividing throughout by tan y.tan x we get dx + dy = 0……separating the variable now integrating we get sec y tany sec x tanx dx + ∫ sec y tany dy t dt + ∫ p y b) .B √2 1 ( 4x+2y+1 d) √2 dy 1 dt 2 dx dt dx 2 dy = dt dx x → dy dx = 1 2 2 2 dx + ∫ 1 dx dy– log x– x = c dy dx 2 2 x +y = 3xy 2 dy = dx 1+ 2 x +y 3xy = 3 2 y we can x2 y x dx dv = v + x dx 2 = 1+v 3v separating the variables and integrating –2 = t = 2t dy y = vx → dx dx 1+y x clearly see that it is an homogeneous equation hence substituting Answer: a Explanation: = (4x + 2y + 1) here we use substitution for 4x + 2y + 1 = t → 4 + 2 1 Explanation: ) = c cot-1(4x+2y+1)=x+c SE C √2 1 dy = ∫ 1+y x Answer: b (4x + 2y + 1) = x + c −1 tan x 6. Solve the differential equation is _______ a) xp=(x2+2y2)-3 b) x2 p=(x2-2y2)3 c) x4 p=(x2-2y2)-3 d) x6 p=(x2+2y2)3 √2 −1 cot -R c) 1 1+x y – log(1+y) – log x – x = c (y-x) – log(x(1+y)) = c is the solution. 4x+2y+1 −1 (1+y)−1 ∫ 1 dy– ∫ 17 1 = 1 + x + y + xy dx x dy = ∫ 1+y ∫ 2 2√ 2 1+x dy = y ∫ 4. Solution of the differential equation = (4x + 2y + 1) is ______ a) tan ( ) = x + c dy dy dx = (1 + x) + y(1 + x) = (1 + x)(1 + y) dx 1+y dy = c dp = c is the solution. separating the variables & hence integrating log t + log p = c –>log(tan x)+log(tan y) = c = log k….since it is an unknown constant log(tan x .tan y) = log k (tan x tan y) = k is the solution. dx dy LO G → ∫ 1 ) = x + c √2 dx xy substituting tan x = t & tan y=p→sec2 x dx=dt & sec2 y dy=dp 1 ( Answer: a Explanation: xy 2 2 ∫ tan 4x+2y+1 SP 2 tanx = x + c √2 5. Solution of the differential equation xy = 1 + x + y + xy is ______ a) (y-x)-log(x(1+y))=c b) log(x(1+y))=c c) (y+x)-log(x)=c d) (y-x)-log(y(1+x))=c 2 sec x tan −1 2√ 2 3. Solution of the differential equation sec2 x tany dx + sec2 y tanx dy=0 is _______ a) (sec x. sec y)=k b) (sec x .tany)=k c) (tan x. tany)=k d) (sec x .tan x)+(sec y .tan y)=k t −1 2√ 2 x dt = ∫ dx T. 2y 2 2t +4 O e 1 ∫ C O M e2y dy = (e3x+x2)dx…..integrating ∫ e2y dy =∫ (e3x+x2)dx = + + c. + 4 separating the variable and integrating Downloaded From: dv dx dt ∫ dx 2 = –3v 2 2 1−2v 2 1+v –v = 3v dv = ∫ 1 x 1−2v dx 3v …….substituting 1- 2v2=t→-4v dv=dt we get −3 4 log t = log x + log c → 2 −3log( −3 4 x −2y 2 x 2 log(1 − 2v ) = 2 2 ) = 4log cx → log( 2 x −2y 2 x −3 ) = 7 Department of Mechanical Engineering 6 x 4 2 2 = kx 3 (x −2y ) → x 2 p = (x 2 2 3 − 2y ) MCQ for Regulations 2017 2 is the (v +1)x solution where p is constant. 2 → y dx x y x x +y x+y = c → k(x 2 x y x y 2 x y UNIT II FOURIER SERIES x dx y y x x = v + x dv = v + tan v dx 1. How many dirichlet’s conditions are there? a) One b) Two c) Three d) Four 1 1 dv = ∫ tanv x SP separating the variables and integrating we get ∫ dx log(sin v) = log x + log c where c is constant. y x ) = xc is the solution LO G sin v = xc → sin( TOPIC 2.1 DIRICHLET'S CONDITIONS O dy y = vx → dy T. Answer: d Explanation: = + tan we can clearly see that it is an homogeneous equation substituting dx 8. Particular solution of the differential = equation given y=-1 at x=1. a) y=x b) y+x=2 c) y=-x d) y-x=2 2 2 2 2 y −2xy−x y +2xy−x .B dy dx 17 Answer: c Explanation: dy dx = 2 2 2 2 y −2xy−x y +2xy−x 2 y = -R ……. is a homogeneous equation thus put y = vx → = v + x – 2 x 2 y x2 + 2y x 2y x –1 −1 2 dv v −2v−1 = dx 2 v +2v−1 separating the variables and integrating we get SE C dy dx x dv = dx 3 2 v −2v−1 2 v +2v−1 –v = − 2 v +2v−1 ∫ ∫ dv = ∫ 2 (v +1)(v+1) 2 (v +1)(v+1) 2v 2 v +1 − x 2 (v +v +v+1) 2 v +2v−1 2 = − 1 v+1 Answer: c Explanation: There are three dirichlet’s conditions. These conditions are certain conditions that a signal must possess for its fourier series to converge at all points where the signal is continuous. 2. What are the Dirichlet’s conditions? a) Conditions required for fourier series to diverge b) Conditions required for fourier series to converge if continuous c) Conditions required for fourier series to converge d) Conditions required for fourier series to diverge if continuous Answer: b Explanation: Dirichlet’s conditions are Conditions required for fourier series to converge. That is there are certain conditions that a signal must posses for its fourier series to converge at all points where the signal is continuous. (v +1)(v+1) 2 v +2v−1 dx 2 2v(v+1)−(v +1) ∫( −1 2 + y ) = (x + y) C O M y 2 where k=1/c at x=1, y=-1 substituting we get 2k=0→k=0 thus the particular solution is y=-x. 7. The solution of differential equation = + tan is ______ a) cot( ) = xc b) cos( ) = xc c) sec ( ) = xc d) sin( ) = xc dy = c (v+1) dv = log c– log x ) dv = log c– log x log(v2+1) – log(v+1) + log x = log c –> Downloaded From: 8 Department of Mechanical Engineering MCQ for Regulations 2017 3. What is the first Dirichlet’s condition? a) Over any period, signal x(t) must be integrable b) Multiplication of the signals must be continuous c) x(t) should be continuous only d) A signal can be integrable except break points d) In majority finite interval, x(t) is of unbounded variation Answer: a Explanation: In the case of Dirichlet’s conditions, the first property leads to the integration of signal. It states that over any period, signal x(t) must be integrable. That is ∫|x(t)|dt<∞. 7. There are maxima and minima not possible in dirichlet’s conditions. a) True b) False C O M T. O LO G Answer: b Explanation: Dirichlet’s conditions is not possible in case of discrete signals. That is these are certain conditions that a signal must posses for its fourier series to converge at all points where the signal is continuous only. Answer: b Explanation: Maxima and minima are possible if they are infinite number as stated by the second dirichlet’s condition. In any finite interval, x(t) is of bounded variation. That is there are no more than a finite number of maxima and minima during a single period of the signal. SP 4. Is dirichlet’s condition possible in case of discrete signals...
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