Question

Home > My courses > Y-MATT1-258 » 27 july-Z August > term Question 2 Not yet answered Marked out of 6.00 p Flag question The solution of the IVP bERKO} is e x = (ax+by+12- V(0)=0, where a ER and N Select one: O a. (ax+by-1) (1 -bx)=1 O b. (ax+by+1) (1 - bx)=1 o c.(ax +by+1)(1 - bx)2 = 1 2 O d. (ax +by+1) = 1-bx O e. (ax +by+1)-(1-bx) = 1 O f. (ax +by+1)(1 - bx) = 3 8. ax+by = 1 h. (ax +by+1)(1+bx) = 1

Answer 1

Solution: Given differential equation is dy a = (ax +by+ 1)2 = dx b (1) y(0)=0 Put ax +by+1=u=a+b dy du dx dx b dy_du dx dx a > dy 1 du a dx bdx b Eqn(1)= a 1 du a b dx b =u2 b du dx = bu du dx bu2 1 du = bdx =du = bſdx u = bx +C u = bx +C ax+by+1 (1) Using y(0) = 0 On eqn(1),we get = 0+C 0+0+1 C=-1 Substituting C value in eqn(1),we get 1 bx -1 ax+by+1 1 -bx +1 ax +by+1 1 (1-bx) (ax + by +1) 31=(ax +by+1)(1-bx) :. Solution for the given D.E is (ax + by + 1)(1-bx) = 1 Hence option (a) is correct. . Select one: a. (ax +by+1X1 -bx)= 1 o b. (ax +by+1X1 + bx)= 1 c. (ax +by+1X1 -bx)= 3 d. (ax +by-1X1 - bx)= 1 e. (ax +by+1%(1 - bx)= 1 of. (ax +by+1X1 - bx = 1 a g. ax + y = 1 oh.(ax +by+1)= 2 1-ox