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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0diff. twice and eliminate a^2 and b^2 from three equations.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5Have you used the surjithayer's hint and differentiated twice ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5show you work please

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5whats the derivative of \(\dfrac{x^2}{a^2}\) with respect to \(x\) ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but i think , may solution is not right

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5Keep in mind, \(a\) and \(b\) are constants here

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5\[\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\tag{1}\] differentiating implicitly with respect to \(x\) gives \[\dfrac{2x}{a^2} + \dfrac{2yy'}{b^2} = 0\tag{2}\] yes ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but , i did it first , but my answer was not right

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5your 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

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5\[\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\tag{1}\] 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
 one year ago
Best ResponseYou've already chosen the best response.0@ganeshie8 gets :)) thank you very much :))

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5yw, are you able to differentiate it again and arrive at the answer ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@ganeshie8 im very confused

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5remember product rule ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5b^2 is just a constant, like, 3^2 or 5^2

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5it doesn't change, so the derivative is 0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[x y y'' . 1 + x y^2.y'=b^2\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5\[\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\tag{1}\] 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
 one year ago
Best ResponseYou've already chosen the best response.5Recall 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
 one year ago
Best ResponseYou've already chosen the best response.0@ganeshie8 i got the answer :))))

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how about x^2+axy+bx+c=0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@ganeshie8 number 11

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5Looks you forgot to use product rule here ? dw:1436688898948:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1isn'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?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1x and y's are variables treat a b c as constants which when the derivative is taken it goes to 0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ai derivative of x is 1 = a (1) ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5@UsukiDoll this is like constructing the differential equation from the given general solution by eliminating the arbitrary constants

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1oh I see so we're not verifying... we're finding a DE

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5yeah we can think of the equations of given ellipses/parabolas as general solutions

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah , were finding D.E

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5@jacalneaila did you find ur mistake in #11 ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah, i try to solve ut again , but may answer is different now :(

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wait , i dont have a cellphone

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5#11 should be fairly simple compared to the earlier one

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5dw:1436690038639:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0im still doing it , and still different

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@ganeshie8 Thank you very much , its very helpful :)) .

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@ganeshie8 Thank you again ^_^

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1so 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
 one year ago
Best ResponseYou've already chosen the best response.5the missing part is \(=0\) in every line :)

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1oh boy xD! I was doing this while dealing with a migraine.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5From personal experience, regular diet+good sleep are the best medicine for migraine

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1yeah...that's what I lot of people say...but I'm such a nighthawk and it runs in the family.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5Also there are some ellergetic foods which you need to avoid eating

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1if it's chocolate, then I can't =(

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5For me they are potato and some other vegetable whose name idk in english

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5they depend on the person i think..

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5this one http://www.mommyscuisine.com/wpcontent/uploads/chikkudukayatomato1300x225.jpg
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