Question

高等数学 求下列微分方程满足已给初始条件的特解

Answer 1

特征方程为r²-3r+2=0
得r=1, 2
设特解y*=a, 代入方程得：2a=5, 得a=2.5
故y=C1e^x+C2e^(2x)+2.5
y(0)=C1+C2+2.5=1
y'(0)=C1+2C2=2
解得：C1=-5, C2=3.5
故特解y=-5e^x+3.5e^(2x)+2.5

Answer 2

特征方程为r²-3r+2=0
得r=1, 2
设特解y*=a, 代入方程得：2a=5, 得a=2.5
故y=C1e^x+C2e^(2x)+2.5
y(0)=C1+C2+2.5=1
y'(0)=C1+2C2=2
解得：C1=-5, C2=3.5
故特解y=-5e^x+3.5e^(2x)+2.5

Answer 3

特征方程为r²-3r+2=0
得r=1, 2
设特解y*=a, 代入方程得：2a=5, 得a=2.5
故y=C1e^x+C2e^(2x)+2.5
y(0)=C1+C2+2.5=1
y'(0)=C1+2C2=2
解得：C1=-5, C2=3.5
故特解y=-5e^x+3.5e^(2x)+2.5

Answer 4

为r²-3r+2=0
得r=1, 2
设特解y*=a, 代入方程得：2a=5, 得a=2.5
故y=C1e^x+C2e^(2x)+2.5
y(0)=C1+C2+2.5=1
y'(0)=C1+2C2=2
解得：C1=-5, C2=3.5
故特解y=-5e^x+3.5e^(2x)+2.5

Answer 5

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