Задача 71675 Доказать, что функция удовлетворяет...
Условие
y^3y^n + 1=0, y= sqrt(2-x^2)
Решение
[m]y`=\frac{1}{2\sqrt{2x-x^2}}\cdot (2-2x)[/m]
[m]y`=\frac{ (1-x)}{\sqrt{2x-x^2}}[/m]
[m]y``=\frac{ (1-x)`\cdot(\sqrt{2x-x^2})-(1-x)\cdot (\sqrt{2x-x^2})` }{(\sqrt{2x-x^2})^2}[/m]
[m]y``=\frac{ -\sqrt{2x-x^2}-(1-x)\cdot (\frac{1}{2\sqrt{2x-x^2}}\cdot (2x-x^2)` }{2x-x^2}[/m]
[m]y``=\frac{ -\sqrt{2x-x^2}-\frac{(1-x)^2}{\sqrt{2x-x^2}}}{2x-x^2}[/m]
[m]y``=\frac{ -(2x-x^2)-(1-x)^2}{\sqrt{2x-x^2}(2x-x^2)}[/m]
[m]y``=\frac{ -2x+x^2-1+2x-x^2}{\sqrt{2x-x^2}(2x-x^2)}[/m]
[m]y``=\frac{ -1}{\sqrt{2x-x^2}(2x-x^2)}[/m]
Подставляем в уравнение
[m](\sqrt{2x-x^2})^3\cdot( \frac{ -1}{\sqrt{2x-x^2}(2x-x^2)})+1=0[/m]
[m]-1+1=0[/m]- верно