二阶常系数齐次线性微分方程通解
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发布时间：2022-04-28 18:18
共1个回答
热心网友
时间：2022-06-22 19:32
设y = e^[f(x)],则
y' = e^[f(x)]*f'(x),
y''= e^[f(x)]*[f'(x)]^2 + e^[f(x)]*f''(x).
0 = y'' - 2y' + 5y = e^[f(x)]*[f'(x)]^2 + e^[f(x)]*f''(x) - 2e^[f(x)]*f'(x) + 5e^[f(x)],
0 = [f'(x)]^2 + f''(x) - 2f'(x) + 5,
当f(x) = ax + b,a,b是常数时.
f''(x) = 0,
f'(x) = a.
0 = a^2 - 2a + 5.
2^2 - 4*5 = -16 < 0.(2^2-4*5)^(1/2)=4i.
a = [2 + 4i]/2 = 1 + 2i或a = [2-4i]/2 = 1 - 2i.
y = e^[f(x)] = e^[ax+b] = e^[(1+2i)x + b] = e^[x+b]*e^(2ix)
或
y = e^[f(x)] = e^[ax+b] = e^[(1-2i)x + b] = e^[x+b]*e^(-2ix)
因2个解都满足微分方程.所以,微分方程的实函数解为,
y = e^[x+b]*e^(2ix) + e^[x+b]*e^(-2ix) = e^[x+b][e^(2ix)+e^(-2ix)] = 2e^[x+b][cos(2x)]
或
y = e^[x+b]*e^(2ix) - e^[x+b]*e^(-2ix) = e^[x+b][e^(2ix)-e^(-2ix)] = 2e^[x+b][sin(2x)]
微分方程的实函数的通解为,
y = 2c1e^[x+b][cos(2x)] + 2c2e^[x+b][sin(2x)]
= e^x[2c1e^bcos(2x) + 2c2e^bsin(2x)]
其中,c1,c2 是任意常数.
记
C1 = 2c1e^b,C2 = 2c2e^b,
有
y = e^x[C1cos(2x) + C2sin(2x)]
C1,C2为任意常数.
