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

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The discussion revolves around solving the equation y d²y/dx² + (dy/dx)² - 10y³ = 0 for y = 1/x². Participants calculate y' and y'' as -2x^-3 and 6x^-4, respectively, but struggle to simplify the expression to equal zero. They clarify the calculations for y dy/dx and (dy/dx)², leading to expressions involving x. A participant points out a potential typo, suggesting it should be y cubed. The conversation emphasizes the challenge of achieving the equation's balance with the given values.
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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
 
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markosheehan said:
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?
 
markosheehan said:
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}

 
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