Question

求微分方程y″+y′-2y=e^-x和y″-2y′+2y=e^xcosx的通解

求y″+y′-2y=e^-x和y″-2y′+2y=e^xcosx的通解，要详细解答过程啊，谢谢了。。。

Answer 1

y=C2*exp(x) - 1/(3*e^x*(log(e) + 1)) + 1/(3*e^x*(log(e) - 2)) + C3*exp(-2*x)
y=C5*exp(x)*cos(x) + C6*exp(x)*sin(x) + (cosx*e^x*cos(x)*(cos(x) + sin(x) - log(e)*sin(x)))/(log(e)^2 - 2*log(e) + 2) + (cosx*e^x*sin(x)*(sin(x) - cos(x) + log(e)*cos(x)))/(log(e)^2 - 2*log(e) + 2)通过拉普拉斯变换，然后再反变换