求通解(1+x)y'-y=(1+x)∧2·y∧(-1)
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求通解(1+x)y'-y=(1+x)∧2·y∧(-1)
![求通解(1+x)y'-y=(1+x)∧2·y∧(-1)](/uploads/image/z/2148473-65-3.jpg?t=%E6%B1%82%E9%80%9A%E8%A7%A3%281%2Bx%29y%27-y%3D%281%2Bx%29%E2%88%A72%C2%B7y%E2%88%A7%28-1%29)
∵(1+x)y'-y=(1+x)^2/y
==>(1+x)yy'-y^2=(1+x)^2
==>(1+x)ydy-(y^2+(1+x)^2)dx=0
==>ydy/(1+x)^2-y^2dx/(1+x)^3=dx/(1+x) (等式两端同除(1+x)^3)
==>d(y^2)/(1+x)^2+y^2d(1/(1+x)^2)=2dx/(1+x)
==>d(y^2/(1+x)^2)=2dx/(1+x)
==>y^2/(1+x)^2=2ln│1+x│+C (C是常数)
==>y^2=(2ln│1+x│+C)(1+x)^2
∴原方程的通解是y^2=(2ln│1+x│+C)(1+x)^2.
==>(1+x)yy'-y^2=(1+x)^2
==>(1+x)ydy-(y^2+(1+x)^2)dx=0
==>ydy/(1+x)^2-y^2dx/(1+x)^3=dx/(1+x) (等式两端同除(1+x)^3)
==>d(y^2)/(1+x)^2+y^2d(1/(1+x)^2)=2dx/(1+x)
==>d(y^2/(1+x)^2)=2dx/(1+x)
==>y^2/(1+x)^2=2ln│1+x│+C (C是常数)
==>y^2=(2ln│1+x│+C)(1+x)^2
∴原方程的通解是y^2=(2ln│1+x│+C)(1+x)^2.