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

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

There is a typo . . . it should be [tex]y-\text{cubed}[/tex]