Calculate d2y/dx2 from dy/dx ?!

y + 2x (dy/dx) = x - 2y

diff wrt x

How do I do this - to get d2y/dx2 do I manipulate equation and do product rule.

This is to calculate the nature of the stationary points.

Stationary points at (4,2) and (-4,-2).

I know what to do in the later stages, but how do I differentiate a differential?

After doing diff wrt x I get: d2y/dx2 + 2 dy/dx = 1 - (2 * dy/dx).

Can someone please explain where I have gone wrong, I think it is to do with the LHS.

Thanks

Reply 1

Notnek

Original post by Moordland

What happened to the first y when you differentiated? You seemed to have ignored it and started by differentiating

2x\frac{dy}{dx}

Reply 2

username1599987

Original post by Moordland

y + 2x (dy/dx) = x - 2y

diff wrt x.

Simplify first:

2xy'=x-3y

Then differentiate

2y'+2xy''=1-3y'

[use the product rule on the RHS (2x)(y')]

2xy''=1-5y'

(edited 9 years ago)

Reply 3

Moordland

Original post by notnek

What happened to the first y when you differentiated? You seemed to have ignored it and started by differentiating

2x\frac{dy}{dx}

Original post by gagafacea1

Simplify first:

2xy'=x-3y

Then differentiate

2y'+2xy''=1-3y'

[use the product rule on the RHS (2x)(y')]

2xy''=1-5y'

Sorry I probably should have added brackets:

I meant to write (dy/dx) [ y+2x] = x-2y

Find d2y/dx2 should I use product rule?

Reply 4

KitikatStrider

Original post by Moordland

Sorry I probably should have added brackets:

I meant to write (dy/dx) [ y+2x] = x-2y

Find d2y/dx2 should I use product rule?

You should divide by (y+2x) so you have (dy/dx) = (x-2y)/(y+2x) and differentiate with quotient rule

Reply 5

username1599987

Original post by Moordland

Sorry I probably should have added brackets:

I meant to write (dy/dx) [ y+2x] = x-2y

Find d2y/dx2 should I use product rule?

yup product rule. Or if you don't want to have dy/dx in the answer, then you could divide by [y+2x] and us the quotient rule.

Reply 6

Moordland

Original post by KitikatStrider

You should divide by (y+2x) so you have (dy/dx) = (x-2y)/(y+2x) and differentiate with quotient rule

Original post by gagafacea1

yup product rule. Or if you don't want to have dy/dx in the answer, then you could divide by [y+2x] and us the quotient rule.

Thanks guys, I will give it a go!

Reply 7

Moordland

Original post by KitikatStrider

You should divide by (y+2x) so you have (dy/dx) = (x-2y)/(y+2x) and differentiate with quotient rule

Original post by gagafacea1

yup product rule. Or if you don't want to have dy/dx in the answer, then you could divide by [y+2x] and us the quotient rule.

Not to be a pain in the ass but can you guys tell me where I went wrong.

http://imgur.com/5gZffmu

I get d2y/dx2 = 1 and -1 and it should be 1/10 and -1/10 respectively :cry:

Reply 8

TenOfThem

Original post by Moordland

Not to be a pain in the ass but can you guys tell me where I went wrong.

http://imgur.com/5gZffmu

I get d2y/dx2 = 1 and -1 and it should be 1/10 and -1/10 respectively :cry:

To start ... When you differentiated v you got +3 instead of +2

I could not be bothered to check your multiplication of brackets as it is totally unnecessary ... You just need to put your values into the differentiated bit ... Since dy/dx is zero it should be easy

Then you have some odd bit where you have set the numerator to equal zero ... That was nonsense ..