If u(x,t) = Σ [An cos(nct) + Bn sin(nct)] sin(nx) is the general solution of then=1a²u a²u1дх2д+2'conditions u(0,t)=0, u(π,t)=0, t> 0 and initial conditionsinitial boundary value problem c².u(x,0)=0.5sin(3x),A1 + A₂ + A3 + B₁ + B₂ is equal to:012.53.50.51.500 with boundaryu₁(x,0) = 0. Then the sum of constants

(.) Given , u(x,t) = ∑n=1∞An cos(nct) + Bn sin(nct) sinnπxa is the general solution of initial…

Answered: If u(x,t) = Σ [An cos(nct) + Bn…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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B\ Suppose that = 4 cos² t , x = d²y =- V3 sin(2t) and 0
The solution to the IVP U=4 Uxi -00 0 u(x, 0)= cos x u,(x, 0) = 1, is 1 u(x, t) =t+ [sin(x+20)+sin(x-21)] a. 1 u(x, t) = x+-[sin(x+2t)-sin(x-2t)] O b. 4 OC ux, 1) = x– [cos (x+21)- cos(x-2t)] lcos(x+2t)- cos(x-2t)] d. u(x, t) = x+ [cos(x+20)-cos(x-2t)] x+[cos(x+20-cos(x-20)] 4 O e. u(x, t) =t+[cos(x+2t)+ cos(x-20] 2 Of. u(x, t)=2x+[-cos(x+21)+cos(x-20)] 4 O g. u(x, t) =2x+-[sin(x+2t)- sin(x-20]
1,2 and 3 please
13) The integer n = a) If n = 3 (mod 9), then what is x ? b) If n = 3 (mod 11), then what is x ? Hint for #13: 12700140x3243243 is missing the digit x. Consider the number 1234. By writing each digit in scientific notation, we have 1234 1 x 10³ +2× 10² +3 × 10¹ + 4 × 10⁰ Then notice that 10 = 1 (mod 9). When we reduce mod 9, we obtain 1234 1 x 10³ +2× 10² +3 x 10¹ + 4 x 10⁰ = = 1x 1³ + 2 x 1² + 3 × 1¹ + 4 × 1⁰ = 1+ 2+ 3+ 4 = 10 = 1 (mod 9). Since 10 = -1 (mod 11), when we reduce mod 11, we obtain 1234 1 x 10³ +2× 10² +3 x 10¹ + 4 x 10⁰ = = 1x (-1)³ +2× (-1)² +3 × (-1)¹ +4× (-1)⁰ = -1 + 2-3 + 4 = 2 (mod 11).
If u(x, t) = D=1 fn(t) sin(nx) is the solution Utt – 9upa = 2e2t sin(3x) + e-2t sin(4x) u(0, t) = 0 = u(, t) u(x,0) = sin(3x) 4(x,0) = 2 sin(4x). Then the equation for y = f3(t) is y" + 81y = 2e2t O y" – 81y = 2e2t O %3D y" – 81y = 2e-2t O
Show that an implicit solution of 2x sin²(y) dx - (x² + 17) cos(y) dy = 0 is given by In(x² +17) + csc(y) = C. Differentiating In(x² + 17) + csc(y) = C we get y = csc 2x x² + 17 X cos (y) sin ² (y) dy dx = 0 or 2x sin²(y) dx + - (x² +17) Find the constant solutions, if any, that were lost in the solution of the differential equation. (Let k represent an arbitrary integer.) ;−¹(k − ln(x² +17)) -17) cos(y) Jay - U. 0.
5. (2xy-3) dx+ (2ry+ 4) dy =0 6. 2y + cos 3x 4r + 3y sin 3x 7. - y) dx + (- 2ry) dy = 0 8. 1 + In x + In x) dy 9. (x – y + y² sin x) dx = (3ry + 2y cos x) dy 10. (x + y*) dx + 3ry² dy = 0 11. (y In y *) dx + x In y dy = 0 In Problems 21-26 solve the given initial-value nrmblem
Q5) Consider y(5) +y"" = 3 sint + cos² t. Then a suitable form of the particular solution can be written as yp(t) = (A) At² + Bt cost + Ct sint + D cos(2t) + E sin(2t) (B) At+ Bt cost + Ct sint + D cos(2t) + E sin(2t) (C) At³+ Bt cost + Ct sint + D cos(2t) + E sin(2t) (D) At³ + B cost + C sint + D cos(2t) + E sin(2t) (E) At³ + B cost + C sint + Dt cos(2t) + Et sin(2t)
3. Solve of the Boundary-Value Problem (BVP) by the method of seperation of variables. ди at 9mx u,(0,t) = 0, u(2, t) = 0, u(x,0) = 8cos 6cos SHAN (d) u = 8e-9m²t/16 cos 3Tx sin 5x 6e-81m³t/16 cos 3mx 4 - The answer 97x 4
3. Solve of the Boundary-Value Problem (BVP) by the method of seperation of variab u, (0, 1) = 0, u(2. t) = 0, u(x,0) = 8cos-6cos Source answer ↓ u= 8e-9³t/16 cos 372 - The answer 6e-81n³t/16 cos 9x Ge

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