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Separable differentiation

ckfeister

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(3i) , so as usual, where did I mess up?

Original post by ckfeister

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(3i) , so as usual, where did I mess up?

You had

\displaystyle \frac{\mathrm dy}{\mathrm dx} - y^2 = 1

, not

\displaystyle -y^2\frac{\mathrm dy}{\mathrm dx} = 1

(as you treated it)

Instead, divide both sides by

1+y^2

, to obtain,

\displaystyle \frac 1 {1+y^2}\frac{\mathrm dy}{\mathrm dx} = 1

. Should be relatively clear where this is going.

Reply 2

atsruser

Original post by ckfeister

(3i) , so as usual, where did I mess up?

To be honest, I'm struggling to make any sense of that at all. What does the (x dx) (-y^2) bit mean?

The first step is to separate the variables i.e. formally write:

\frac{dy}{1+y^2} = dx

then integrate.

Reply 3

ckfeister

Original post by _gcx

You had

\displaystyle \frac{\mathrm dy}{\mathrm dx} - y^2 = 1

, not

\displaystyle -y^2\frac{\mathrm dy}{\mathrm dx} = 1

(as you treated it)

Instead, divide both sides by

1+y^2

, to obtain,

\displaystyle \frac 1 {1+y^2}\frac{\mathrm dy}{\mathrm dx} = 1

. Should be relatively clear where this is going.

I don't get where this is meant to go, whats the link with tanx?

Reply 4

_gcx

Study Forum Helper

Original post by ckfeister

I don't get where this is meant to go, whats the link with tanx?

Are you familiar with how to approach

\displaystyle \int \frac 1 {1+y^2} \mathrm dy

Reply 5

ckfeister

Original post by _gcx

Are you familiar with how to approach

\displaystyle \int \frac 1 {1+y^2} \mathrm dy

Nope... first time. I would try sub but I'm sure it won't get tan(x)

Reply 6

_gcx

Study Forum Helper

Original post by ckfeister

Nope... first time.

Have you covered integration by substitution? Try substituting

y = \tan\theta

Reply 7

Notnek

Original post by ckfeister

I don't get where this is meant to go, whats the link with tanx?

Which module are you doing currently?

Reply 8

ckfeister

Original post by notnek

which module are you doing currently?

aqa - fp2

Reply 9

Notnek

Original post by ckfeister

aqa - fp2

Then you'll need to know how to integrate that using a trig substitution and you should also be aware that is in your fomula book.

Do you have any ideas which substitution you can use?

Reply 10

ckfeister

Original post by Notnek

Then you'll need to know how to integrate that using a trig substitution and you should also be aware that is in your fomula book.

Do you have any ideas which substitution you can use?

x/a tan-1 x/a?

(edited 6 years ago)

Reply 11

Notnek

Original post by ckfeister

u = 1+y^2

du/dy= 2y

du/2y = dy

Only one I know, I'm going to watch a video soon on khan academy but this hasn't been on the lessons yet... strange.

It seems like you haven't covered trig substitutions so you shouldn't be doing this question yet. _gcx has given you the correct substitution above.

Reply 12

_gcx

Study Forum Helper

Original post by ckfeister

u = 1+y^2

du/dy= 2y

du/2y = dy

Only one I know, I'm going to watch a video soon on khan academy but this hasn't been on the lessons yet... strange.

Have you covered integration by substitution? Try substituting

y = \tan\theta

It's not an obvious substitution, but it works quite well as you'll see. Watching a khanacademy video on trig subs, or similar, might be beneficial as you suggested.

Reply 13

ckfeister

Original post by Notnek

It seems like you haven't covered trig substitutions so you shouldn't be doing this question yet. _gcx has given you the correct substitution above.

Original post by _gcx

It's not an obvious substitution, but it works quite well as you'll see. Watching a khanacademy video on trig subs, or similar, might be beneficial as you suggested.

Is it x/a tan-1 x/a?

Reply 14

_gcx

Study Forum Helper

Original post by ckfeister

Is it x/a tan-1 x/a?

Not quite, you might have made a slip. Can you show your working? (Also, where has a come from? I'd dissuade you from copying from a formula booklet, it's more beneficial to do it yourself first)

(edited 6 years ago)

Reply 15

ckfeister

Original post by _gcx

Not quite, you might have made a slip. Can you show your working? (Also, where has a come from? I'd dissuade you from copying from a formula booklet, it's more beneficial to do it yourself first)