. (1)
Put
. (2)
Then
=>.
. (3)
Using (3) in (1)
=>.
=>.
=>.
=>.
by (2)
=>.
As
So
=>.
Option g is correct.
The solution of the IVP dy dx = (ax+by+112 - 8 yO=0, where a ER and...
The solution of the IVP dy dx (ax +by+12-6: YO)=0, where a ER and b ERVO) is Select one: a. ax+by=1 b. (ax+by+1)(1-bx)= 3 c. (ax+by-1)(1-bx) = 1 d. (ax+by+1)(1-bx)2 = 1 e. (ax+by+1)(1 - bx)= 1 2 f. (ax+by+1) = 1- by o g. (ax +by+1) (1 - bx)=1 n. (ax +by+1)(1+bx)= 1
The solution of the IVP dy dx = (ax+by+ 1)2 - ; y(0)=0, where a ER and b ERVO) is Select one: a. (ax+by+1)(1 + bx)= 1 2 b. (ax+by+1)= 1-bx c. ax + by=1 d. (ax +by+1)(1 - bx)2 = 1 e. (ax+by-1)(1 - bx)= 1 f. (ax+by+1) (1 - bx)= 1 g. (ax +by+1)(1-bx)= 1 h. (ax+by+1)(1 - bx)=3
The solution of the IVP dy dx = (ax+by+1)2 - Ø; y(D)=0, where a ER and b ERVO) is Select one: a. ax+by=1 b. (ax+by+1)(1-bx) - 3 c. (ax+by-1)(1 - bx)= 1 d. (ax +by+1)(1 - bx)2 = 1 e. (ax+by+1)(1-bx)= 1 2 f. (ax+by+1)= 1-bx g. (ax+by+1) (1 - bx)= 1 h. (ax+by+ 1)(1+bx) = 1 O O
The solution of the IVP dy dx = (ax+by+1)2 – 6; y(0)=0, where a € R and b ERVO} Select one: a. (ax +by+1)(1+x)= 1 O b. (ax+by+1)(1-x)=3 O c. (ax+by+1)2(1 - bx)=1 2 O d. (ax +by+1)= 1- bx e. (ax+by+1)(1-bx)= 1 of. (ax +by+1) (1 -bx)2 = 1 о g. (ax +by-1)(1-bx) = 1 O h. ax + by=1
Flag question Marked out of 6.00 Question 1 dy The solution of the IVP = (ax+by+1) where a ER and b ERVO) is bi y0)=0, Select one a. (ax + by - 1)(1-x-1 b. (ax+by+1)(1+x)=1 clax+by + 1) (1 -bx=1 dax+by = 1 lox+by+1 (1 -bx) = 1 2 Olax+by+1)- 1-bx (x+by+ 1X1 - bx) = 3 (ax+by+ 1X1 -bx) = 1 Ne
Please fast back, i don't have enough time The solution of the IVP dy - (ax + by+1)2-1(0)=0, where a ER and b ERYO) 15 Select one: e a. (ax+by + 1)(1-x) = 1 2 b. (ax+by+1) (ax+by+1)1 -bx) = di ax+by 1 e Cax+by+1211 -bx)= 1 1. (ax+by- 1)(1-x)=1 Cox+by+1)(1+x)=1 hax+by+1)(1-bx)= 3
The solution of the equation [ax? +(6+1)y?]dx– xydy=0, where a and b are constant is Select one: O a. ax” +by? =C b. nlatok 3) = 2 = 2bln(x) o cby? = x26+2 O d. ax? +by? = cx20 +2 e (a+b)x?+by? = x22 o fax +(6+1)/2 = c x2 +2 og Inax? +by?)=2bin(x)+c = -2bln(x)+c
Find a solution of the IVP dy/dx=xy^3(1+x^2)^-1/2, y(0)=1, and give the interval where the solution is defined.
this question is only this The solution of the equation [ax? +(6+1)y2]dx– xydy=0, where a and b are constant is Select one: a. In(a+b ) = 2bln(x) b. (a+b)x2 + by2 = x26+2 c. ax? +by= c x26+2 O d. by2 = c x26+2 e. ax2 + y2 = 0 - 2bln(x)+c g. ax? +(6+1)y2 = C x26 +2 h. In(ax2 +by?)=2bln(x)+c
The solution of the equation [ax? +(+1)y2]dx - xydy=0, where a and b are constant is Select one: O O a. ax+(6+1)y2 = c X20 +2 b. (a+b)x2 +by2 = c x26+2 c. by2 = c x2b +2 d. ax? +by2 = c x26+2 O O O en( - ) - - 2bumba) + 6 o f. ax? +by2=C O g. In(ax2 + by2)=2bln(x)+C a info o *) – 2016)