Getting Rid of Negative Exponent

[bmed90]

Getting Rid of Negative Exponent!

Homework Statement

So I am solving a Diff Eq ( separable equation) and I am down the the last step. After integrating both sives and solving for y I am left with y^(-2) on the left hand side. How do I finalize it and get rid of negative 2 expononent?

y^(-2)=-2(x+x^(2)/2)

Homework Equations

The Attempt at a Solution

 

[rock.freak667]

y^(-2) is the same as 1/y², so just write it as that and invert both sides.

 

[bmed90]

So you invert both sides, then what? do you take the square root of both sides and invert back to y^(-1)??

 

[bmed90]

Also, is y' understood to be d/dy or dy/dx?

 

[rock.freak667]

So you invert both sides, then what? do you take the square root of both sides and invert back to y^(-1)??

if you invert 1/y² you will get y² then you can take the square root if you wish.

Also, is y' understood to be d/dy or dy/dx?

Usually it is dy/dx.

 

[bmed90]

Ok I don't think you understand

Say you have y^(-2)=(something), how do I get to... y=(something)

If you invert y^(-2) you get 1/y^(+2) ...yes ...but then what ...take square root? that would leave you with 1/y on left... Then invert both sides again? so it would end up being y= sqrt(something)also how do you know when y' is indeed dy/dx and not d/dy

 

[rock.freak667]

Ok I don't think you understand

Say you have y^(-2)=(something), how do I get to... y=(something)

If you invert y^(-2) you get 1/y^(+2) ...yes ...but then what ...take square root? that would leave you with 1/y on left... Then invert both sides again? so it would end up being y= sqrt(something)

I meant that y^-2 is the same as 1/y², so you have

1/y² = -2(x+x^(2)/2) (I am not exactly sure how you have a negative sign, so you may need to check that part over)

and if you have 1/y² = A, then y² = A and so y = √ A

also how do you know when y' is indeed dy/dx and not d/dy

y' is usually used to denote the first derivative. d/dy is just write as that, d/dy.

 

[bmed90]

How is this possible1/y2 = A, then y2 = A and so y = √ A

 

[rock.freak667]

How is this possible


1/y2 = A, then y2 = A and so y = √ A

oops, sorry about that, typo, it should read:


1/y² = A, then y² = 1/A and so y = √(1/A)