Question

大一数学：求微分方程x*dy/dx+x+sin(x+y)=0的通解

我数学很差，麻烦步骤详细点 谢谢

Answer 1

xdy+xdx+sin(x+y)dx=0令z=x+y，则dy=dz-dxxdz+sinzdx=0即1/x dx=1/sinz dz两边同时做不定积分∫1/x dx=Ln(x)+C1 C1为常数令t=tan(z/2) z=2acrtan(t) 即dz=2/(1+t^2)而sinz=2sin(z/2)cos(z/2)/((cos(z/2))^2+(sin(z/2))^2)=2tan(z/2)/(1+(tan(z/2))^2)∫1/sinz dz=∫(1+t^2)/(2t)*2/(1+t^2) dt=Ln(t)+C2=Ln(tan(z/2))+C2 C2为常数因此，Ln(x)+C1=Ln(tan(z/2))+C2即Ln(x/tan(z/2))=C2-C1 取指数 x/tan(z/2)=e^(C2-C1) 记C=e^(C2-C1)>0 代入z=x+y可得x=C*tan[(x+y)/2] 则y=2acrtan(x/C)-x C>0即为通解

Answer 2

解：∵(x+y)dy+(x-y)dx=0
==>(1+y/x)dy+(1-y/x)dx=0
设y=xt,则dy=tdx+xdt
∴(x+y)dy+(x-y)dx=0
==>(1+t)(tdx+xdt)+(1-t)dx=0
==>(t�0�5+1)dx+x(t+1)dt=0
==>dx/x+(t+1)/(t�0�5+1)dt=0
==>ln|x|+∫t/(t�0�5+1)dt+∫1/(t�0�5+1)dt=ln|C| (C是积分常数)
==>ln|x|+1/2∫d(t�0�5+1)/(t�0�5+1)+arctant=ln|C|
==>ln|x|+1/2ln(t�0�5+1)+arctant=ln|C|
==>x√(t�0�5+1)=Ce^(arctant)
==>√(x�0�5+y�0�5)=Ce^(arctant) (C是积分常数).

Answer 3

这头像、你是月儿吗、我会长

Answer 4

宝