Implicit Dif.

[fgmango]

if x^2+4xy-y^2=8, find y".

i got y'=x+2y/y-2x

then I had problems
y"=(y-2x)^2(1+2y')-(x+2y)(y'-2) all over (y-2x)^2

multiply numorator and denomiator by y-2x

y+2yy"-2x-4xy"-xy"+2x-2yy"+4y all over (y-2x)^2 simplify

y"=(y-2x)^2/5y-5x

 

[stapel]

got y'=x+2y/y-2x

If you mean:

. . . . .y' = (x + 2y)/(y - 2x)

...then I agree.

y"=(y-2x)^2(1+2y')-(x+2y)(y'-2) all over (y-2x)^2

I'm not getting this. Please show your steps.

multiply numorator and denomiator by y-2x

Why?

. . . . .d²y/dx² = [(1 + 2(dy/dx))(y - 2x) - (x + 2y)((dy/dx) - 2)] / [y - 2x]²

. . . . .= [y - 2x + 2y(dy/dx) - 4x(dy/dx) + 2x + 4y - x(dy/dx) - 2y(dy/dx)] / [y - 2x]²

. . . . .= [5y - 5x(dy/dx)] / [y - 2x]²

Now plug in the known expression for "dy/dx", and simplify.

Eliz.

 

[fgmango]

okay, so

(5y-5xy')/((y-2x)^2)

(y-2x)^2=5y-5xy'/5y-5x

(y-2x)^2/(5y-5x)=y'

(y-2x)^2/5(y-x)=y'


I'm not good at fractions, is this even close

 

[stapel]

Your first line is what I gave you. But where is your second line coming from?

Please reply showing all your steps and reasoning. Thank you.

Eliz.

 

[fgmango]

I need to get y' to one side by itself so i multiplied both sides of the equal sign by the denominator (y-2x)^2 then divide by (5y-5x)

I know its wrong. the answer the teacher gave was -40/(y-2x)^3. I don't see how she got that.

 

[stapel]

Why do you need to isolate y', when you are asked to find y"?

I gave you "d²y/dx² = (an expression)". How did you "multiply through" and get rid of the d²y/dx²?

You say the teacher gave you something. What does this something stand for? How does it relate to the posted exercise?

Please show all of your steps and clearly state your definitions and reasoning. Thank you.

Eliz.