Need Help moving X from one side of equation to other -

[CrazyFrog]

Hi everyone, I would like to know if the following algebraic operation I performed is correct? The operation I performed was move the X to the right side of the equation ( Differential Equation , hence why there is a Y')

 

[stapel]

Hi everyone, I would like to know if the following algebraic operation I performed is correct? The operation I performed was move the X to the right side of the equation ( Differential Equation , hence why there is a Y')

You appear to be saying that the following two statements are equivalent:

[imath]\qquad \textsf{(a) }\; \cfrac{\frac{dy}{dx}}{y^3} - y^{-2}x = 1 + \ln(x)[/imath]

[imath]\qquad \textsf{(b) }\; \cfrac{\frac{dy}{dx}}{y^3} - y^{-2} = \dfrac{1 + \ln(x)}{x}[/imath]

On the right-hand side, it looks as though you've divided through by [imath]x[/imath]. But that's not possible, because then you would have divided the [imath]x[/imath] into only part of the expression on the left-hand side, which you learned in pre-algebra or beginning algebra is not okay (and now you're in calculus, so you've been doing this for years).

So I must be misunderstanding what you're doing. Kindly please reply with clarification. Thank you!

Eliz.

 

[CrazyFrog]

You appear to be saying that the following two statements are equivalent:

[imath]\qquad \textsf{(a) }\; \cfrac{\frac{dy}{dx}}{y^3} - y^{-2}x = 1 + \ln(x)[/imath]

[imath]\qquad \textsf{(b) }\; \cfrac{\frac{dy}{dx}}{y^3} - y^{-2} = \dfrac{1 + \ln(x)}{x}[/imath]

On the right-hand side, it looks as though you've divided through by [imath]x[/imath]. But that's not possible, because then you would have divided the [imath]x[/imath] into only part of the expression on the left-hand side, which you learned in pre-algebra or beginning algebra is not okay (and now you're in calculus, so you've been doing this for years).

So I must be misunderstanding what you're doing. Kindly please reply with clarification. Thank you!

Eliz.

Hi eliz,thank you for your prompt reply, suppose you wanted to move the x that is on the left hand side of the equation i showed, to the right , how would you go about doing that?
You appear to be saying that the following two statements are equivalent:

[imath]\qquad \textsf{(a) }\; \cfrac{\frac{dy}{dx}}{y^3} - y^{-2}x = 1 + \ln(x)[/imath]

[imath]\qquad \textsf{(b) }\; \cfrac{\frac{dy}{dx}}{y^3} - y^{-2} = \dfrac{1 + \ln(x)}{x}[/imath]

On the right-hand side, it looks as though you've divided through by [imath]x[/imath]. But that's not possible, because then you would have divided the [imath]x[/imath] into only part of the expression on the left-hand side, which you learned in pre-algebra or beginning algebra is not okay (and now you're in calculus, so you've been doing this for years).

So I must be misunderstanding what you're doing. Kindly please reply with clarification. Thank you!

 

[pka]

Hi everyone, I would like to know if the following algebraic operation I performed is correct? The operation I performed was move the X to the right side of the equation ( Differential Equation , hence why there is a Y')
View attachment 35839

From the given [imath]\dfrac{y^{\prime}}{y^3}-y^{-2}x=1+\log(x)[/imath] it is the case that [imath]\dfrac{y^{\prime}}{y^3}-y^{-2}\ne\dfrac{1+\log(x)}{x}[/imath].
But this is the case that: [imath]\dfrac{y^{\prime}}{xy^3}-y^{-2}=\dfrac{1+\log(x)}{x}[/imath] multiply all by [imath]\frac{1}{x},\;x\ne 0[/imath]
It seems that you need a good reviewing of middle school algebra.
[imath][/imath]

 

[CrazyFrog]

From the given [imath]\dfrac{y^{\prime}}{y^3}-y^{-2}x=1+\log(x)[/imath] it is the case that [imath]\dfrac{y^{\prime}}{y^3}-y^{-2}\ne\dfrac{1+\log(x)}{x}[/imath].
But this is the case that: [imath]\dfrac{y^{\prime}}{xy^3}-y^{-2}=\dfrac{1+\log(x)}{x}[/imath] multiply all by [imath]\frac{1}{x},\;x\ne 0[/imath]
It seems that you need a good reviewing of middle school algebra.
[imath][/imath]

Thank you for the reply, I have one additional question from the last line you wrote, , if I wanted to have No Xs in the left side of the equation, what would I then have to do, such that all Y variables are on the left side of the equation and all X variables on the right side of the equation.

 

[Steven G]

Thank you for the reply, I have one additional question from the last line you wrote, View attachment 35841, if I wanted to have No Xs in the left side of the equation, what would I then have to do, such that all Y variables are on the left side of the equation and all X variables on the right side of the equation.

If your goal is to really just have No X's in the left hand side then you can simply move the term with an x in it to the rhs--that is move \(\displaystyle \dfrac{y'}{xy^3}\) to the rhs OR move everything from the lhs to the rhs--this will leave 0 on the lhs and 0 has no x's in it!


Regarding your question: Does \(\displaystyle \dfrac{30+10}{5}=50+ \dfrac{10}{5}\)? NO! You can't divide the entire rhs by 5 and only part of the lhs by 5 and expect equality.

 

[topsquark]

Thank you for the reply, I have one additional question from the last line you wrote, View attachment 35841, if I wanted to have No Xs in the left side of the equation, what would I then have to do, such that all Y variables are on the left side of the equation and all X variables on the right side of the equation.

It can't be done, unless you can think of a clever substitution that I am missing.

-Dan

 

[nasi112]

I understand why you want to move the \(\displaystyle x\) to the right side. You want to have a familiar differential equation with which you can work easily. You don't need to do that or may be you cannot (as topsquark said). Instead look at the equation closely, you will notice that it is a Bernoulli differential equation, and it can be solved when you let \(\displaystyle z = y^{1-n}\).

 

[topsquark]

I understand why you want to move the \(\displaystyle x\) to the right side. You want to have a familiar differential equation with which you can work easily. You don't need to do that or may be you cannot (as topsquark said). Instead look at the equation closely, you will notice that it is a Bernoulli differential equation, and it can be solved when you let \(\displaystyle z = y^{1-n}\).

So it is! Good catch! (It's an ugly looking one, too.)

-Dan