Question

what is the differential equation of all circles along x-axis with radius 2

Answer 1

so first find the equation of all circles along x-axis with radius 2

Answer 2

What will be the equation of the circle along x-axis??
Is it??

\[(x-a)^2 = r^2\]

Answer 3

I think I am wrong...
Not sure...

Answer 4

Or it can be:

\[(x-a)^2 + y^2 = r^2\]

Answer 5

that second one looks right @waterineyes

Answer 6

now r=2

Answer 7

I also think so..

Because y coordinate of centre is 0 along x axis..

Answer 8

i also think this

Answer 9

so solve for y^2

Answer 10

Use this formula also:

\[(a-b)^2 = a^2 + b^2 - 2ab\]

Answer 11

(h,0) r=2
\[(x-h)^{2} + y ^{2}= r ^{2}\]

Answer 12

Now apply the square formula that I have written above..

Answer 13

but can i integrate it now

Answer 14

You are dealing with differential equations and not intergral equations..
You have to eliminate the constants like h...

Answer 15

yeah by integration...

Answer 16

i will remove the arbitrary constants

Answer 17

im wrong i must differentiate it rather.

Answer 18

with resprct to x

Answer 19

Yes you must differentiate it...

Yes by doing so remove the arbitrary constant..

Yeh with respect to x take the derivative...

Answer 20

\[y \prime \left[(x-h)^{2}+y ^{2}=4 \right] \rightarrow 2(x-h)(1-0) + 2y y \prime = 0\]

Answer 21

is this correct @waterineyes

Answer 22

Very true..

But you know h is there now also..

So, you will take the derivative again to eliminate h...

Just go for it..

Answer 23

yeah right\[\left[ 2(x-h)+2y y \prime = 0 \right] \div 2 \rightarrow (x-h)+y y \prime = 0 \rightarrow 1- 0 + y y'' + y' y' = 0 \rightarrow 1 + y y'' + y'^{2} = 0 \]

Answer 24

\[1 + y y'' + (y')^{2} = 0\]

Answer 25

done is this right??

Answer 26

Yes absolutely...

So, this is the answer..

Answer 27

thanks

Answer 28

Welcome dear..