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Hey everyone I came across this problem:

$$\dfrac{y^{-2}}{-2} = \dfrac{x^2}{2} + C$$ where $C$ is a constant.

The problem was solved without providing any steps as: $$y^{2} = \dfrac{-1}{x^2 + C}$$

My attempt at the solution came out to: $$y^2 = \dfrac{1}{-x^2-2C}$$

How does the $2$ in front of the $C$ go away for their solution?

Thank you.

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We have $$y^2 = \dfrac{1}{-x^2-2C}=\frac{-1}{x^2+2C}=\frac{-1}{x^2+C_1}$$ where $C_1=2C$ is a constant. Note that the question states that $C$ is just a constant, so multiplying a constant by a constant doesn't change the fact that it is still a constant.

This is a very common problem when you're doing indefinite integration...

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An equation like yours $$\frac {y^{-2}}{-2}=\frac {x^2}2+C$$ often comes from solving a differential equation where $C$ can be any real number because it is the constant of integration. As you say, you can transform that to $$y^2=\frac 1{x^2-2C}$$ as you say. We can define $C'=-2C$ and the last becomes $$y^2=\frac 1{x^2+C'}$$ If we ignore the distinction between $C$ and $C'$ this is the form you saw. As $C$ was arbitrary it is not useful to keep track of the constants we multiply it by.

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  • $\begingroup$ Ahh ok, so if it was an intial value problem, I would have the 2C? Thanks! $\endgroup$ – Vlad Metodiev Jun 29 '18 at 19:12
  • $\begingroup$ If $C$ is a supplied parameter you would keep its value and keep track of the multiplicative constants. There is a $-1$ as well as a $2$ here. $\endgroup$ – Ross Millikan Jun 29 '18 at 19:19

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