2道高数的题1.作变量代换X=lnt简化方程d^2y/dx^2-dy/dx+ye^2x=02.利用函数的凹凸性,证明不等
来源:学生作业帮 编辑:搜搜考试网作业帮 分类:数学作业 时间:2024/08/07 01:01:17
2道高数的题
1.作变量代换X=lnt简化方程d^2y/dx^2-dy/dx+ye^2x=0
2.利用函数的凹凸性,证明不等式:sin(x/2)>x/π(0
1.作变量代换X=lnt简化方程d^2y/dx^2-dy/dx+ye^2x=0
2.利用函数的凹凸性,证明不等式:sin(x/2)>x/π(0
![2道高数的题1.作变量代换X=lnt简化方程d^2y/dx^2-dy/dx+ye^2x=02.利用函数的凹凸性,证明不等](/uploads/image/z/8424553-49-3.jpg?t=2%E9%81%93%E9%AB%98%E6%95%B0%E7%9A%84%E9%A2%981.%E4%BD%9C%E5%8F%98%E9%87%8F%E4%BB%A3%E6%8D%A2X%3Dlnt%E7%AE%80%E5%8C%96%E6%96%B9%E7%A8%8Bd%5E2y%2Fdx%5E2-dy%2Fdx%2Bye%5E2x%3D02.%E5%88%A9%E7%94%A8%E5%87%BD%E6%95%B0%E7%9A%84%E5%87%B9%E5%87%B8%E6%80%A7%2C%E8%AF%81%E6%98%8E%E4%B8%8D%E7%AD%89)
1,X=lnt,那么dx=(1/t)dt,dt/dx=t
dy/dx=(dy/dt)/(dt/dx)=t(dy/dt)
d^2y/dx^2-dy/dx+ye^2x=0
d(dy/dx)/dx-dy/dx+ye^2x=0
d(t(dy/dt))/dx-t(dy/dt)+yt^2=0
t^2(d^2y/dt^2)+t(dy/dt)-t(dy/dt)+yt^2=0
t^2(d^2y/dt^2)+yt^2=0
(d^2y/dt^2)+y=0
2,设f(x)=sin(x/2)-x/π,(0
dy/dx=(dy/dt)/(dt/dx)=t(dy/dt)
d^2y/dx^2-dy/dx+ye^2x=0
d(dy/dx)/dx-dy/dx+ye^2x=0
d(t(dy/dt))/dx-t(dy/dt)+yt^2=0
t^2(d^2y/dt^2)+t(dy/dt)-t(dy/dt)+yt^2=0
t^2(d^2y/dt^2)+yt^2=0
(d^2y/dt^2)+y=0
2,设f(x)=sin(x/2)-x/π,(0
2道高数的题1.作变量代换X=lnt简化方程d^2y/dx^2-dy/dx+ye^2x=02.利用函数的凹凸性,证明不等
作变量代换x=lnt简化方程d^2y/dx^2-dy/dx+e^2x*y=0
做变量代换x=lnt化简方程d的平方y/dx的平方-dy/dx+y*e的2x次幂=0
求由参数方程所确定的函数{x=tlnt y=t^2lnt的导数dy/dx
求由方程xe^y+ye^x=5所确定的函数的导数dy/dx,d^2y/dx^2
用适当的变量代换将微分方程dy/dx=(x+y)^2化为可分离变量的方程,且求通解.
设参数方程x=t方分之(1+lnt),y=t分之(3+2lnt)确定y=y(x),求dx分之dy,dx方分之d方y
d^2y/dx^2=(dy/dx)'×(dy/dx),另外请解释下dx,dy的含义,dx和dy是指x=...和y=...
设参数方程x=t方分之1+lnt,y=t分之3+2int确定y=y(x),求dx分之dy,dx方分之d方y
求微分方程的通解:x^2(d^2y/dx^2)=(dy/dx)^2+2x(dy/dx)
求方程dy/dx+2y=x的通解
dy/dx=(x+y)^2的原函数