Trial solution of the first part of RHS:
This is a first degree polynomial,
Trial solution is:
Trial solution of the second part of RHS:
This is a second degree polynomial with exponential and cosine term.
Trial solution is:
Trial solution of the third part of RHS:
This is exponential and cosine term.
Trial solution is:
The actual trial solution is sum of all three parts.
1. Find a Trial Solution for the following DE: D (D2 -4D +13)2 2r +3r e cos (3r) 4e sin (3r) 1. Find a Trial Solut...
QUESTION 1 To find a particular solution of the differential equation (D - 1)?(D – 2)(D2 + 1)y = e* + cos – 2 sin z one can use the following trial solution OA Ao + A, cos x + A2 sin OB Anx?et + x² (A, COS I+ A2 sin I) Ос. Aoxe" + x(A, COS X + A2 sin x) OD. Age* + x(A, COS I+ A2 sin x) OC Apx?et + ](A, COS I + A2 sin...
de 14) 55+ Acos e sin de 5+ 4 cos e 13) 2 - cos 15) cos 2 de 1-2k cos A + k? < 1)
d1= 3 and d2= 2 Question 1 ch- 3, d2 - 2 (a) Find the most general solution u(x, y) of the two PDEs Lt +1) y cos y +(di +1)x cos ((d, +1)xy)+2x d2 (b) Find the solution that satisfies initial condition u(0,0) Question 1 ch- 3, d2 - 2 (a) Find the most general solution u(x, y) of the two PDEs Lt +1) y cos y +(di +1)x cos ((d, +1)xy)+2x d2 (b) Find the solution that satisfies...
40 Show the following results. 1-e2 (e) lim(2+3-12)tan(/4) (24.3)-4/ 2 (a) lim -+0 isin(3r) エ→2 3 (f) lim(cos x)In | = 1 エ→0 1+ tanz1/sin z 1+ tanh r (b) lim = 1 -+0 nT nT (g) lim cos no0 +sin 6n+1 (c) lim (sin r)1/(2r-) - 1 エ→/2 = e 3n+1 2 + sin r 1 (h) lim 0. (d) lim エー→0 1 In (1 - V-1) - . 2 In(cos x) r+1+ 40 Show the following results. 1-e2...
5. Find all degree solutions to cos (SA-10°) = 2 Solve sin x =-cos 2r if n
l. Find the general solution: (a) r" -4x +4x-0 (b) "-2r (c) 0. (d)4r' +3r-0
Find the Jacobian of the transformation. r = 3er sin(20), y=e-3r cos(20) a (x, y) a(r, o) - Need Help? Read it Master It Talk to a Tutor
d2y d2y dy +6 da2 (h) +13y 2sin x +9y = 18x -+3 +6 dx da d2y (i) d2y (j d2 18x3 4y = 2 sin x dæ2 d2y ,dy .dy 9y 9x2 +21x - 10 dc (k) (1)2 7 + - 4y = e-4x +6 'da2 da2 d2y dy dy (m) 2 dæ2 (n) 4 7y= e 6 cos x 9y = 4e-3r dr2 dr dx d2y d2y (p*) dy + da2 dy (o* 2a COS I 2y 2...
Replace the polar equation with an equivalent Cartesian equation. r2 + 2r -2 sin e cos 0 = 144 O X + y = $12. O x + y = 12 x² + y² = 144 O x2 + 3y2 = 144
Use variation of parameters to find a particular solution to the given DE -3pt 11.)y" - 2y'+y- tet 13.) y'', + 4y'--8 [cos (2t) + sin(2t)] 15.) хту-xy' + y = x3 6e Use variation of parameters to find a particular solution to the given DE -3pt 11.)y" - 2y'+y- tet 13.) y'', + 4y'--8 [cos (2t) + sin(2t)] 15.) хту-xy' + y = x3 6e