Question

微分方程式问题 dy/dx=tan(x+y)

Answer 1

令x+y=u,则
dx+dy=du,代入换掉y,得
du/dx=tanu+1,分离变量,得
cosudu/(sinu+1)=dx,两边同时积分,得
ln(sinu+1)=x+lnc
所以
通解为
ln[sin(x+y)+1]=x+lnc化得
sin(x+y)+1=c*e^x

Answer 2

解：微分方程为dy/dx=tan(x+y)，设x+y=u，微分方程化为d(u-x)/dx=tanu，du/dx-1=tanu，du/dx=1+tanu，du/(1+tanu)=dx，设tanu=z，u=arctanz，微分方程化为dz/[(1+z)(1+z²)]=dx，dz/(1+z)-(z-1)dz/(1+z²)=2dx，dz/(1+z)-zdz/(1+z²)+dz/(z²+1)=2dx，ln||1+z|-0.5ln(1+z²)+arctanz=2x+ln|c|(c为任意非零常数)，(1+z)/√(1+z²)=ce²ˣ⁻ᵃʳᶜᵗᵃⁿᶻ，微分方程的通解为(1+tan(x+y))=secx×ceˣ⁻ʸ