Let y=y(x) be the solution of the differential equation : dydx+ ( 6x2+ (3x2+2x3+4 )e-2x (x3+2 ) (2+e-2x ) )y=2+e-2x, x (-1, 2), satisfying y(0)= 32. If y(1)= (2+e-2 ), then is equal to :
- 138
- 613
- 1213
- 1312
Answer
(4) 1312
Solution
It is linear: dydx+P(x) y=Q(x) with Q=2+e-2x.
Numerator of P: 6x2+3x2e-2x+2 (x3+2 )e-2x=3x2 (2+e-2x )+2e-2x (x3+2 )
P(x)= 3x2x3+2+ 2e-2x2+e-2x
P dx= e (x3+2 )- e (2+e-2x ) (both x3+2>0, 2+e-2x>0 on (-1, 2))
I.F. = x3+22+e-2x
y x3+22+e-2x= (2+e-2x ) x3+22+e-2x dx= x44+2x+C
y(0)= 32: 32 23=C C=1
At x=1: y(1) 32+e-2= 14+2+1= 134
y(1)= 1312 (2+e-2 )
= 1312
Answer: (4)