微分方程(1+y^2)dx+(xy-genhao1+y^2 cosy)dy=0
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微分方程(1+y^2)dx+(xy-genhao1+y^2 cosy)dy=0
∵(1+y²)dx+(xy-√(1+y²)cosy)dy=0
==>√(1+y²)dx+(xy/√(1+y²)-cosy)dy=0 (等式两端同除√(1+y²))
==>√(1+y²)dx+xydy/√(1+y²)-cosydy=0
==>√(1+y²)dx+xd(√(1+y²))-cosy)dy=0
==>d(x√(1+y²))=d(siny)
==>x√(1+y²)=siny+C (C是积分常数)
∴原方程的通解是x√(1+y²)=siny+C (C是积分常数)
∵当x=2时,y=0
∴代入通解,得C=2
故满足初始条件的解是x√(1+y²)=siny+2.
==>√(1+y²)dx+(xy/√(1+y²)-cosy)dy=0 (等式两端同除√(1+y²))
==>√(1+y²)dx+xydy/√(1+y²)-cosydy=0
==>√(1+y²)dx+xd(√(1+y²))-cosy)dy=0
==>d(x√(1+y²))=d(siny)
==>x√(1+y²)=siny+C (C是积分常数)
∴原方程的通解是x√(1+y²)=siny+C (C是积分常数)
∵当x=2时,y=0
∴代入通解,得C=2
故满足初始条件的解是x√(1+y²)=siny+2.
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