In Problems 1–6 solve the wave equation (1) subject to the given conditions.
3.
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- 3. Find the orthogonal trajectory of y =c cos xarrow_forwardShow that the function a(z, t) = bi sin )cos() + b sin cos( 2) Tct 2nct bị sin COS L + b2 sin L COS L where c, L, b1, b2 are nonzero constants with L > 0 and c > 0, is a solution to the one-dimensional wave equation c2.arrow_forward5C. Under suitable assumptions derive one dimensional wave equation.arrow_forward
- Q.5 Solve the wave equation: U, =U xx-e* –6x, U (0,t) =8, Ux(l,t)= 3, 57 U(x,0) = 4 sin x+e* + x3 – ex+7 %3D .... .... ............arrow_forward4. Solve the wave equation Utt = Uxx, 00 u(0, t) = 0, - u(1, t) = 0 - u(x,0) = x(1 − x), ut(x,0) = 0arrow_forwardPROBLEM (1): If =2 z-x'y and Ā = 2 xî-3 yzj+x z°k, find A.V and A xV4 at the point ( 1,-1,1). PROBLEM (2):arrow_forward
- 6. Simplified equations for an electric motor can be given O"(t) + 20'(t) = u(t) where 0(t) is the motor shaft angle, and u(t) is the voltage applied to the armature windings. a. Write down a state equation for the motor assuming a state vector x(t) = [0(t) O'(t)] and input u(t). b. Transform the state equation to that for a new state variable z(t) so that the new "A-matrix" is diagonal. c. Assuming that (0) = 0'(0) = 0, solve for x(t), t 2 0, when u(t) = e*, t 2 0.arrow_forward(a) If V = x³ + axy², where a is a constant, show that +y = 3V əx ду Find the value of a if V is to satisfy the equation a²v__a²v = 0 ду? əx² (b) Show that wave equation is satisfied when a² = b²c² 1 a²u a²u Wave equestion: c² at² əx² u = cos at sin bx (c) Determine the first three non-zero terms of the Taylor series expansion for the given function. f (x) = e2× cos(x) about x =0 (d) The partial differential equation a²u a²u 3D 16 — х2 — 2у for 0 < x < 4, 0 < y< 2 (1.1) ax² ду? is subject to the boundary conditions u(x, 0) = 0 and u(x, 2) = 2(16 – x²) for 0arrow_forwardChapter 13, Section 13.7, Question 031 Find parametric equations for the tangent line to the curve of intersection of the cylinders x +z = 25 and y + z = 10 at the point (4, –1,3). %3D O x(t) = 12 (t – 4), y (t) = -48 (t + 1), and z(t) = -16 (t – 3) O x(t) = 4, y(t) = -1+ 2t , and z(t) = 3 + 6t O x(t) = 4 + 12t , y(t) = -1 – 48t , and z(t) = -3+ 16t O x(t) = 4 + 3t , y(t) = -1 – 12t , and z(t) = 3 – 4t O x(t) = 4 + 8t , y(t) = -1, and z(t) = 3 + 6tarrow_forwardThe wave equation 1=10 may be studied by separation of variables: u(x, t) = X(x)T(t). If(x) = -k²X(x), what is the ODE obeyed by T(t)? [] Which of the following solutions obey the boundary conditions X(0) = 0 and X (L) = 0? [tick all that are correct □sin() for & integer sin() sin( (2k+1)mz 2L ) for k integer □ sin(2) sin() □ sin() Which of the following is a possible solution of the above wave equation? ○ cos(kx)e-ket O cos(kex) sin(kt) ○ Az + B ○ cos(kx) sin(kt) O None of the choices applyarrow_forward- 11) Calculate the Jacobian, J, for the change of variables x = u cos(0) – v sin(0) and yu sin(0) + v cos(0).arrow_forward2²u 2²u 7. Solve the wave equation = 4 Subject to u(0,t)=0, u(4,t)=0,u(x,0)=x(4-x), taking h=1, k=0.5 up to four steps. at² əx²arrow_forwardarrow_back_iosSEE MORE QUESTIONSarrow_forward_ios
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