2. 考研
  3. 设f(u,v)具有连续偏导数，且满足f’u(u,v)+f’v(u,v)=uv,则函数y(x)=e-2xf(x,x)满足条件y(0)=1的表达式为________.

设f(u,v)具有连续偏导数，且满足f’u(u,v)+f’v(u,v)=uv,则函数y(x)=e-2xf(x,x)满足条件y(0)=1的表达式为________.

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问题 设f(u,v)具有连续偏导数，且满足f’u(u,v)+f’v(u,v)=uv,则函数y(x)=e-2xf(x,x)满足条件y(0)=1的表达式为________.
选项
答案y=([*]+1)e-2x
解析 y’=－2e-2xf(x,x)+e-2xf’u(x,x)+e-2xf’v(x,x)=－2y+x2e-2x
因此y(x)满足一阶线性微分方程y’+2y=x2e-2x,解得
y=e-∫2dx(∫x2e-2xe∫2dxdx+C)=(+C)e-2x
由y(0)=1,得C=1,所以y=(+1)e-2x.
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考研数学三

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