Question

f(x)具有二阶连续导数,f(0)=1,f'(0)=-1,且[xy(x+y)-f(x)y]dx+[f'(x)+x^2y]

Answer 1

设该二元函数为g(x,y),则

g'x(x,y)=xy(x+y)-f(x)y

两边对x求积分

g(x,y)=x³y/3+x²y²/2-y∫f(x)dx

g'y(x,y)=f'(x)+x²y

两边对y求积分

g(x,y)=f'(x)y+x²y²/2+C

∴x³y/3-y∫f(x)dx=f'(x)y+C

两边对x求导

x²y-yf(x)=f''(x)y

x²-f(x)=f''(x)

f''(x)+f(x)=x²

解微分方程,得通解为

f(x)=C1*cosx+C2*sinx+x²-2

又f(0)=1,f'(0)=-1

∴C1-2=1 => C1=3

C2=-1

∴f(x)=3cosx-sinx+x²-2