∵3yy'=(1+y)cosx
==>3ydy/(1+y)=cosxdx
==>3[1-1/(1+y)]dy=cosxdx
=3(y-ln│1+y│)=sinx+C (C是常数)
∴原方程的通解是3(y-ln│1+y│)=sinx+C.
求方程3yy'=(1+y)cosx的通解
3ydy/(1+y)=cosxdx==>3[1-1/(1+y)]dy=cosxdx=3(y-ln│1+y│)=si"}}}'>
∵3yy'=(1+y)cosx
==>3ydy/(1+y)=cosxdx
==>3[1-1/(1+y)]dy=cosxdx
=3(y-ln│1+y│)=sinx+C (C是常数)
∴原方程的通解是3(y-ln│1+y│)=sinx+C.