Differiental Equation Question

[cmhex]

Hello, short of it is I just started Calc II at a different school than I took Calc I at, and this Calc II starts off about a chapter ahead of where the last one left off so I'm having a knowledge gap that I'm trying to bridge.

So to the question

Dif Equation: y' = 2xy/x^2-y^2
Solution: x^2 + y^2 = Cy

Check: 2x + 2yy' = Cy'

The 2's I assume came down as a part of derivative, no problem there. Why are we putting that y' in there?

The equation gets rewritten as

y' = -2x/2y - C, no problem here I understand where this came from

Then we go to

y' = -2xy/2y^2-Cy

Multiplying all terms by y obviously, but where did this come from?

Plug in the Cy from the given, subtract it out and such to get -2xy/y^2-x^2, flip the bottom term and cancel the negatives to get 2xy/y^2-x^2, no problem here.

 

[stapel]

Dif Equation: y' = 2xy/x^2-y^2
Solution: x^2 + y^2 = Cy

Check: 2x + 2yy' = Cy'

The 2's I assume came down as a part of derivative, no problem there. Why are we putting that y' in there?

Have you done anything with "implicit differentiation"? Because that was probably what you're needing.

If you remember all the way back to the beginning of differentiation, you learned about differentiating various things according to various rules, always "with respect to x", and then, when you got down to "dx/dx", you ignored this, "because dx/dx is just 1". Well, now you're differentiating with respect to x, but you don't have y by itself. So you can't do "y = (something in terms only of x)" leading to "y' = (something differentiated with respect to x, and in terms only of x)".

Instead, you have to differentiate everything, down to the "variable", even when the "variable" is y(x). And, while you can still ignore dx/dx, you need to pay VERY close attention to dy/dx, because this can not be ignored, any more now than when y was politely by itself on the other side of the "equals" sign.

So the "y'" is there for the exact same reason that it's always been there. It just isn't pretty now; it's messy.

 

[HallsofIvy]

Hello, short of it is I just started Calc II at a different school than I took Calc I at, and this Calc II starts off about a chapter ahead of where the last one left off so I'm having a knowledge gap that I'm trying to bridge.

So to the question

Dif Equation: y' = 2xy/x^2-y^2

I assume you mean y'= 2xy/(x^2- y^2)

Solution: x^2 + y^2 = Cy

Check: 2x + 2yy' = Cy'

The 2's I assume came down as a part of derivative, no problem there. Why are we putting that y' in there?

That's the "chain rule" If y is a function of x and f a function of y then $\displaystyle \frac{df}{dx}=\frac{df}{dy}\frac{dy}{dx}$. In particular, if $\displaystyle f(y)= y^2$ then $\displaystyle \frac{df}{dy}= 2y$ so that $\displaystyle \frac{df}{dx}= \frac{dy}{dx}\frac{dy}{dx}= 2yy'$

The equation gets rewritten as

y' = -2x/2y - C, no problem here I understand where this came from

PLEASE use parentheses! You mean "y'= -2x/(2y- C)

Then we go to

y' = -2xy/(2y^2-Cy)

Multiplying all terms by y obviously, but where did this come from?

What do you mean by "where did this come from"? Obviously they have multiplied numerator and denominator on the right (not "all terms") by y. The reason they did that was to show that it is the same as the y' in the original problem.

plug in the Cy from the given, subtract it out and such to get -2xy/y^2-x^2, flip the bottom term and cancel the negatives to get 2xy/y^2-x^2, no problem here.

 

[cmhex]

Thanks for the help, I'll remember my () from now on. Sorry, bit new around here.