find D.E
x^2/a^2+y^2/b^2=1 ans. xyy''+xy'^2=yy'

- anonymous

find D.E
x^2/a^2+y^2/b^2=1 ans. xyy''+xy'^2=yy'

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- anonymous

diff. twice and eliminate a^2 and b^2 from three equations.

- anonymous

how @surjithayer

- anonymous

help please

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## More answers

- ganeshie8

Have you used the surjithayer's hint and differentiated twice ?

- anonymous

yeah

- ganeshie8

show you work please

- anonymous

wait

- anonymous

##### 1 Attachment

- anonymous

##### 1 Attachment

- anonymous

number 14

- ganeshie8

whats the derivative of \(\dfrac{x^2}{a^2}\) with respect to \(x\) ?

- anonymous

but i think , may solution is not right

- anonymous

yes

- ganeshie8

Keep in mind, \(a\) and \(b\) are constants here

- ganeshie8

\[\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\tag{1}\]
differentiating implicitly with respect to \(x\) gives
\[\dfrac{2x}{a^2} + \dfrac{2yy'}{b^2} = 0\tag{2}\]
yes ?

- anonymous

okay :)

- anonymous

but , i did it first , but my answer was not right

- ganeshie8

your work in that picture is wrong
you're not treating a^2 as constant, instead trying to differentiate it and writing 2a.. which is wrong

- anonymous

can you help please

- ganeshie8

\[\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\tag{1}\]
differentiating implicitly with respect to \(x\) gives
\[\dfrac{2x}{a^2} + \dfrac{2yy'}{b^2} = 0 \implies a^2 = -\dfrac{b^2x}{yy'} \]
Plug this in first equation and get
\[\dfrac{x^2}{-b^2x/(yy')}+\dfrac{y^2}{b^2}=1 \]
\[\implies -xyy' +y^2 = b^2 \tag{2}\]
differentiate this again and you're done!

- anonymous

@ganeshie8 gets :)) thank you very much :))

- ganeshie8

yw, are you able to differentiate it again and arrive at the answer ?

- anonymous

@ganeshie8 im very confused

- ganeshie8

remember product rule ?

- anonymous

yeah i remember

- anonymous

how about the b^2 ?

- ganeshie8

b^2 is just a constant, like, 3^2 or 5^2

- ganeshie8

it doesn't change, so the derivative is 0

- anonymous

\[-x y y'' . -1 + x y^2.y'=b^2\]

- ganeshie8

\[\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\tag{1}\]
differentiating implicitly with respect to \(x\) gives
\[\dfrac{2x}{a^2} + \dfrac{2yy'}{b^2} = 0 \implies a^2 = -\dfrac{b^2x}{yy'} \]
Plug this in first equation and get
\[\dfrac{x^2}{-b^2x/(yy')}+\dfrac{y^2}{b^2}=1 \]
\[\implies -xyy' +y^2 = b^2 \tag{2}\]
differentiating w.r.t \(x\) again
\[-(yy' +y'xy' + y''xy) + 2yy'= 0\]
\[yy' = xyy'' + x{y'}^2\]

- ganeshie8

Recall below product rules
\[\left(f(x)*g(x)\right)' = f'(x)g(x) + f(x)g'(x)\]
\[\left(f(x)*g(x)*h(x)\right)' = f'(x)g(x)h(x) + f(x)g'(x)h(x) + f(x)g(x)h'(x)\]

- anonymous

@ganeshie8 i got the answer :))))

- anonymous

how about
x^2+axy+bx+c=0

- ganeshie8

Give it a try first

- anonymous

i have a answer

- anonymous

@ganeshie8 number 11

##### 1 Attachment

- ganeshie8

Looks you forgot to use product rule here ?
|dw:1436688898948:dw|

- UsukiDoll

isn't that we have to find the second and third derivatives and then plug them back into the equation to check if the equation we are given (that ans. thing) holds true?

- anonymous

@ganeshie8

- UsukiDoll

x and y's are variables
treat a b c as constants which when the derivative is taken it goes to 0

- anonymous

ai derivative of x is 1 = a (1) ?

- ganeshie8

@UsukiDoll this is like constructing the differential equation from the given general solution by eliminating the arbitrary constants

- UsukiDoll

oh I see so we're not verifying... we're finding a DE

- ganeshie8

yeah we can think of the equations of given ellipses/parabolas as general solutions

- anonymous

yeah , were finding D.E

- ganeshie8

@jacalneaila did you find ur mistake in #11 ?

- anonymous

yeah, i try to solve ut again , but may answer is different now :(

- anonymous

*it

- ganeshie8

show ur work

- anonymous

wait , i dont have a cellphone

- ganeshie8

#11 should be fairly simple compared to the earlier one

- anonymous

\[ax+yy'+a(1)? \]

- anonymous

@ganeshie8

- ganeshie8

|dw:1436690038639:dw|

- anonymous

ugh

- anonymous

im still doing it , and still different

- anonymous

@ganeshie8 Thank you very much , its very helpful :)) .

- anonymous

@ganeshie8 Thank you again ^_^

- UsukiDoll

so for this part axy
to make it less confusing I would put a () like this
a(xy) and then take the product rule of xy
a(xy'+y)
then I have to take the product rule for xy'
a(xy''+y'+y')
and then another product rule for xy''
a(xy'''+y''+y''+y'')
so that leaves
(xy'''+3y'')
I think I'm missing a step.. that a is supposed to go away x.x

- ganeshie8

the missing part is \(=0\) in every line :)

- UsukiDoll

oh boy xD! I was doing this while dealing with a migraine.

- ganeshie8

From personal experience,
regular diet+good sleep are the best medicine for migraine

- UsukiDoll

yeah...that's what I lot of people say...but I'm such a nighthawk and it runs in the family.

- ganeshie8

Also there are some ellergetic foods which you need to avoid eating

- UsukiDoll

if it's chocolate, then I can't =(

- ganeshie8

For me they are potato and some other vegetable whose name idk in english

- ganeshie8

they depend on the person i think..

- ganeshie8

this one
http://www.mommyscuisine.com/wp-content/uploads/chikkudukayatomato1-300x225.jpg

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