Differential Equations · Question 3

For all x > 0, let y1(x), y2(x), and y3(x) be the functions satisfying dy1dx - ( x)2y1 = 0, y1(1) = 5, dy2dx - ( x)2y2 = 0, y2(1) = 13, dy3dx - ( 2 - x3x3 )y3 = 0, y3(1) = 35e, respectively. Then x 0+ y1(x)y2(x)y3(x) + 2xe3x x is equal to .

Answer

2

Solution

Separable: y1 = 5e 1x 2t dt, y2 = 13e 1x 2t dt, so y1y2 = 53e 1x 1 dt = 53ex - 1.

dy3y3 = ( 2x3 - 1 )dx y3 = - 1x2 - x + C.

y3(1) = 35e: -2 + C = 35 - 1 y3 = 35e1 - x - 1/x2.

y1y2y3 = e-1/x2.

Limit = x 0+ 1e3x ( e-1/x2 x + 2x x ) = 1 (0 + 2) = 2 (since e-1/x2/x 0).

Answer: 2