Question

b) i. Form partial differential equation from z = ax - 4y+b [4 marks] a +1 ii. Solve the partial differential equation 18xy2 + sin(2x - y) = 0 дх2ду c) i. Solve the Lagrange equation [4 Marks] az -zp + xzq = y2 where p az and q = ду [5 Marks] x ax ii. A special form of the second order partial differential equation of the function u of the two independent variables x and t is given by au + Q at² + P axat apu au a.x2 au du +f f(x, t, u, öz öe = 0 (2) At what condition is equation (2) elliptic, parabolic or hyperbolic? [3 Marks] iii. Classify the PDE (1 + x)?Uxx – 4xUxy + Uyy = x. [3 Marks] d) Solve using method of separation of variables the equation 2x – 3y = 0 [8 Marks]

Answer 1

6 aytb Z=an apl 02 =a » P=a az ax az ay a q= afl ati Thus a el q=-(a -Core 2 - PI =0 a qt pel 7 q (pel) + po 1 IPL+P+q=0 az to is the that is da a form of PDE, T 02 82 +02 JI azz = 18x3 + Sim (22-4) and (D²6)z = 18xy + Sim (21-4) Put Dam and d'al then we get the auxiliany equation z m=0,0 to the corresponding Complementary function is Q,C4 +0.2772.02 (9 40.x) ire Q, (Y) + 2O2(4) mao

DD' Sim (22-y) D D I For Particular integral leay't sim (22 o leayer to 1825)+ sin(21-43 Sin (22-4) (18x.y?) sin (22-4) (6ay? 6y?(3%) 국 Sim (22-y) ay By? x + Sin (2n-y) a (2x-4) À - [-eos (22-y)] 그 b² 40' .1 62 1 x3y² + 2 3.3 cos (2a-4 b to the solution is izcay) = Cift P. I 2 (919) = 0, (4) 42.82 (9) 72343– Cos(221-4 4

C) 2 .y a using ZP 4x zq = yr Lagrange equation dz dy alx yr = N Z y z 2 Now 리 a a dz dy 2.da Gr y z xz from 1st and 3rd we get x-da dz ytz x-da zz-dz 222 = 22 te x²-z² = 9 D and from 1st and 2nd we get roda dy y z XZ x² daa y.dy C2 3 23_y3 = C2 Hence F(x22², 22 y)) o is required solution. 43 여 23 t 3

(13) ou. aar +4(2,4,4,m4a up olu anat P. TQ. Ô t? -0 Equation ② is (a) Parcabolic if p_40 =0 (6 elliptie if p²-4Qco (²) Hyperbolic if p²4070 civo here & we Given p.de (reauxx - 4Xuny tayy=a. compute (-42)-4. (1+x)? > 427-4(24+24+1) 2 462 462-8n-4 2 -82-4 @ parabolic if - 82-450 2 82504 2- 6 elliptie if al- 10 Hyperbolie if a>-

az 39. co 23 24 - 27-p-34.q=0 22 p = 39.9 = k (consfont a il za =) p = =Q09) zap zu 39 2Qy) and 34.q=U =) q= row, dzia o(a)oda + Q (4) .dy k dy dat 3y dz ñ 22 da it 39 dz и 21 [ 22 hohen on integration z = " logat "zlogy te IZ of Solution Which is the required The given ode.