Question

求高阶微分方程的通解

Answer 1

令u=y'，则u'=y''
u'=u^3+u
du/u(1+u^2)=dx
∫[1/u-u/(1+u^2)]du=∫dx
ln|u|-(1/2)*ln|1+u^2|=x+C
ln|u/√(1+u^2)|=x+C
u/√(1+u^2)=C*e^x
u^2/(1+u^2)=C^2*e^(2x)
1/u^2=C^(-2)*e^(-2x)-1
u^2=C^2*e^(2x)/[1-C^2*e^(2x)]
u=C*e^x/√[1-C^2*e^(2x)]
y'=C*e^x/√[1-C^2*e^(2x)]
y=∫C*e^x/√[1-C^2*e^(2x)]dx
=∫d(C*e^x)/√[1-(C*e^x)^2]
=arcsin(C*e^x)+C1，其中C和C1都是任意常数