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

  • Thread starter Thread starter markosheehan
  • Start date Start date
  • Tags Tags
    Zero
Click For Summary
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.
markosheehan
Messages
133
Reaction score
0
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
 
Physics news on Phys.org
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}

 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

Similar threads

  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 18 ·
Replies
18
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 20 ·
Replies
20
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K