Question

4(10pt). Find the general solution for the following equations. (a) y) – 4y(3) + 5y" = 0.

Answer 1

Solution :- y(4) _44 +5y"=0 &2+54"=0 y!4) Aurilliary Equation is n4-4 m3+5 m2=0 m m2-umts=0 m² (m²-4 m+5)=0 m=-(-4) Z √64=4x5x1 m=9,0, 2zł = 42/16-20 2 Complementary function (Cf) = y=(ci teh ponte "(C3 Loshtly sinn 27 cuts-u 2 - utora The General solution of y424 425ylo is = 2 (2 & FT) = ati Y = (Etangente " (<3 Cosattusina) 4 = (t cante22(ez cosa resina) (b) only" - nyl-yzo *?y" - nyl - y =0 - x dy andy-y=0 Pur xzdy D CD-1)y ndy = Dy and x = -) 0 = loga daz 1 DCD-1) Y-DY-Y=0 (D2-D)4 - DY-y=0 (0²-D-D-10 Y=0 (0²-2D-1)4=0

Auxilliary equation is m²-2 m-l-o m=-(-2) I 18272-4 X1 X-1 2x at 4th 2 = 2118 = 2 + 252 F (1782) roots are distinct IJz if 52, 1-62 so gools are The General solution is (1-12ja (+12rate Y =qe