Question

What does it mean when asked 'solve the equation d^2y/dx^2 = 0 for x' like in the equation y = x^2-6x^2+12x-1. Should I just solve for the derivatives up to the second order?

Answer 1

You find its second derivative and set it to zero

Answer 2

y'' = 2. What do you mean by setting it to zero?

Answer 3

Did you mean y = x^3-6x^2+12x-1 ?

Answer 4

Yeah, in here y = x^2-6x^2+12x-1

Answer 5

You sure it is x^2 instead of x^3?

Answer 6

Yes

Answer 7

Then y = -5x^2+12x-1 ?

Answer 8

No, I made a mistake. The equation is x^2 - 5x - 1. Sorry. So y'' = 2

Answer 9

y = x^2 - 5x - 1
y' = 2x - 5
y'' = 2

Answer 10

yes

Answer 11

Your steps are all correct

Answer 12

Either you made a mistake while copying the question or the question maker is an a-hole

Answer 13

how about when setting it to 0?

Answer 14

you mean the final answer is y'' = 2 ?

Answer 15

I made a mistake while copying the question.

Answer 16

No, the equation should be cubic, for example y = 4x^3+3x^2+2x+1, then y'' = 24x+6, and then you set it to zero 24x+6=0 and x=-0.25

Answer 17

bye

Answer 18

Ah sure, I get it. Thanks.

Answer 19

But how do you solve for x if y'' = 2?

Answer 20

i think you already solve the y'' equation for x, hence to solve for its value (x), you will need to substitute 2 to y'' and therefore you can easily solve for x :)

Answer 21

you can see that y'' = 2

so y''=0 is not a valid equation.

y'' is independent of x in this case (and all 2nd order equations or less)

So - as said before - the equation is wrong, or has no solution.

@Yttrium has misread the question I think.