Задача 46277 Показать, что функция удовлетворяет...
Условие
Фото)
Решение
[m]=\frac{(sin(x-y))`_{x}\cdot x-x`\cdot sin(x-y)}{x^2}=\frac{x\cdot cos(x-y)\cdot (x-y)`_{x}-1\cdot sin(x-y)}{x^2}[/m]=
=[m]\frac{x\cdot cos(x-y)-sin(x-y)}{x^2}[/m]
∂z/ ∂y=z`_(y)=
[m]=\frac{1}{x}\cdot(sin(x-y))`_{y}=\frac{1}{x}\cdot cos(x-y)\cdot (x-y)`_{y}=\frac{1}{x}\cdot cos(x-y)\cdot (-1)= [/m]
=[m]-\frac{1}{x}\cdot cos(x-y)[/m]
∂^2z/ ∂y^2=(z`_(y))`_(y)=([m]-\frac{1}{x}\cdot cos(x-y)[/m])`_(y)=
=-[m]\frac{1}{x}\cdot (-sin(x-y))\cdot (x-y)`_{y}=-\frac{1}{x}\cdot sin(x-y)[/m]
∂ / ∂ x(x^2* ∂ z/ ∂ x)=[m](x\cdot cos(x-y)-sin(x-y))`_{x}=[/m]
[m]=cos(x-y)+x(-sin(x-y))-cos(x-y)=-xsin(x-y)[/m]
Подставляем в уравнение:
-xsin(x-y) - х* [m](-\frac{1}{x}\cdot sin(x-y))=0[/m] - верно