laplace变换 求解微分方程 y"-2y'+5y=5x+8 y(0)=3 y'(0)=-1 请写出步骤.

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第一步：两边同时做Laplace变换得：
L[y"-2y'+5y]=L[5x+8]
L[y"]-2*L[y]'+5*L[y]=5*L[x]+8*L[1]
第二步：求解出L[y]:
p^2L[y]-p*y(0)-y'(0)-2*[p*L[y]-y(0)]+5*L[y]=5/(p^2)+8/p
p^2L[y]-p*3-(-1)-2*[p*L[y]-(-1)]+5*L[y]=5/(p^2)+8/p
(p^2-2p+5)*L[y]-3p+1-2=5/(p^2)+8/p
(p^2-2p+5)*L[y]-3p+1-2=5/(p^2)+8/p
(p^2-2p+5)*L[y]=5/(p^2)+8/p+3p+1
L[y]=5/[(p^2)*(p^2-2p+5)]+8/[p*(p^2-2p+5)]+(3p+1)/(p^2-2p+5)
第三步：作Laplace逆变换,求出y
5/[(p^2)*(p^2-2p+5)]=5*L[x]*L[(e^t)*sin(2t)]的Laplace逆变换为
y1(x)=5*(x与(e^t)*sin(2t)的卷积),
8/(p*(p^2-2p+5)]=5*L[1]*L[(e^t)*sin(2t)]的Laplace逆变换为
y2(x)=8*(1与(e^t)*sin(2t)的卷积),
(3p+1)/(p^2-2p+5)的Laplace逆变换为
y3(x)=2(e^t)*[3cos(2t)+2sin(2t)],
则y(t)=y1(x)+y2(x)+y3(x).