Question

请教一个微分方程的问题设函数y=y(x)满足微分方程y"-3y'+2y=2*e的x次方,而且图形在点(0,1)处的切线方

Answer 1

y''[x] - 3 y'[x] + 2 y[x] = 2*e^x,

y[x]=-2 e^x (1 + x) + e^ x C[1] + e^(2 x) C[2]

当x=0时,y=1,

于是-2 + C[1] + C[2]=1,

当x=0,y=1,时,斜率等于D[x^2-x+1]/D[x]=2 x - 1 = -1.

所以当x=0时,

y'[x]=-2 e^x - 2 e^x (1 + x) + e^x C[1] + 2 e^(2 x) C[2]

=-4 + C[1] + 2 C[2] = -1.

由

-2 + C[1] + C[2]=1,

-4 + C[1] + 2 C[2]=-1

得C[1] = 3,C[2] = 0

于是

y[x] = 3 e^x - 2 e^x (1 + x)