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Let y = y(x) be the solution of the differential equation (2x e x) dydx + 2y = 3x e x, x > 0 and y (e-1 ) = 0. Then, y(e) is equal to
(1) - 3e
Divide by 2x e x: dydx + yx e x = 32x2 (linear)
I.F. = e dxx e x = e e( e x) = e x
y e x = 32 e xx2dx
= 32 (- e xx - 1x ) + C (by parts)
y(e-1) = 0: 0 = 32(e - e) + C C = 0
At x = e: y(e) 1 = 32 (- 1e - 1e ) = - 3e
Answer: (1)
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