设f(x)二阶连续可导,f(0)=0,f’(0)=1,且[xy(x+y)一f(x)y]dx+[f’(x)+x2y]dy=0为全微分方程,求f(x)及该全微分方程的通解.

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问题 设f(x)二阶连续可导,f(0)=0,f(0)=1,且[xy(x+y)一f(x)y]dx+[f(x)+x2y]dy=0为全微分方程,求f(x)及该全微分方程的通解.

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答案令P(x,y)=xy(x+y)-f(x)y,Q(x,y)=f(x)+x2y,因为[xy(x+y)一f(x)y]dx+[f(x)+x2y]dy=0为全微分方程,所以[*],即f’’(x)+f(x)=x2, 解得f(x)=C1cosx+C2sinx+x2一2,由f(0)=0,f(0)=1得C1=2,C2=1, 所以f(x)=2cosx+sinx+x2一2. 原方程为[xy2-(2cosx+sinx)y+2y]dx+(一2sinx+cosx+2x+x2y)dy=0,整理得 (xy2dx+x2ydy)+2(ydx+xdy)一2(ycosxdx+sinxdy)+(-ysinxdx+cosxdy)=0, 即d([*]x2y2+2xy一2ysinx+ycosx)=0, 原方程的通解为[*]x2y2+2xy-2ysinx+ycosx=C.

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