f(0)=0,f'(o)=1,[xy(x+y)-f(x)y]dx+[f'(x)+x^2y]dy]=0为全微分方程,求f(x)微分方程通解.

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f(0)=0,f'(o)=1,[xy(x+y)-f(x)y]dx+[f'(x)+x^2y]dy]=0为全微分方程,求f(x)微分方程通解.
f(0)=0,f'(o)=1,[xy(x+y)-f(x)y]dx+[f'(x)+x^2y]dy]=0为全微分方程,求f(x)微分方程通解.

f(0)=0,f'(o)=1,[xy(x+y)-f(x)y]dx+[f'(x)+x^2y]dy]=0为全微分方程,求f(x)微分方程通解.
所以D[xy(x+y)-f(x)y]/Dy=D[f'(x)+x^2y]/Dx,D表偏导
x^2+2xy-f(x)=f''(x)+2xy
f''(x)+f(x)=x^2 (*)
先找齐次解,f''(x)+f(x)=0,
特征根方程 r^2+1=0,r=±i
f(x)=Acos x+Bsinx
特解,看着就像二次函数,x^2-2满足方程(*),没凑出就直接f=ax^2+bx+c乖乖列方程
f(x)=Acosx+Bsinx+x^2-2
f(0)=0,A-2=0,A=2
f'(0)=1,B=1
f(x)=2cosx+sinx+x^2-2
假设这是一个Q的全微分
dQ/dx=xy(x+y)-yf(x)=x^2y+xy^2-y(2cosx+sinx+x^2-2)=xy^2+2y-2ycosx-ysinx
dQ/dy=f'(x)+x^2y=-2sinx+cosx+2x+x^2y
dQ=(xy^2+2y-2ycosx-ysinx)dx
dQ=(-2sinx+cosx+2x+x^2y)dy
随便用哪个积分(xy^2 dx积分得x^2y^2/2以此类推)
Q=x^2y^2/2+2xy-2ysinx+ycosx+C1
所以全微分方程dQ=0
Q=C2
即x^2y^2/2+2xy-2ysinx+ycosx=C(C=C2-C1)

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