Solving y' and y'' for y=1/x²: How to Equal Zero?

[markosheehan]

if y=1/x² show that y d²y/dx² +(dy/dx)² -10y³=0

i can work out y′=-2x^-3 and y″=6x^-4 but when i sub these in i can't make the whole expression equal zero

 

[Deveno]

if y=1/x² show that y d²y/dx² +(dy/dx)² -10y³=0

i can work out y′=-2x^-3 and y″=6x^-4 but when i sub these in i can't make the whole expression equal zero

Note that $y\dfrac{dy}{dx} = \left(\dfrac{1}{x^2}\right)\left(\dfrac{6}{x^4}\right)$, and that

$\left(\dfrac{dy}{dx}\right)^2 = \left(\dfrac{-2}{x^3}\right)^2 = \dfrac{4}{x^6}$,

while $10y^3 = 10\left(\dfrac{1}{x^2}\right)^3 = \dfrac{10}{(x^2)^3}$. Does this help?

 

[soroban]

if $$y=\frac{1}{x^2}$$
show that: $$y \frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 -10y^{\color{red}3} \;=\;0$$

i can work out $$y′=-2x^{-3}$$ and $$y″=6x^{-4}$$
but when i sub these in i can't make the whole expression equal zero

There is a typo . . . it should be $$y-\text{cubed}$$