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________.

admin2022-03-23  38
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.
转载请注明原文地址:https://kaotiyun.com/show/dBR4777K
0

考研数学三

相关试题推荐
随机试题
最新回复(0)