anonymous
  • anonymous
find D.E x^2/a^2+y^2/b^2=1 ans. xyy''+xy'^2=yy'
Differential Equations
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
diff. twice and eliminate a^2 and b^2 from three equations.
anonymous
  • anonymous
how @surjithayer
anonymous
  • anonymous
help please

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More answers

ganeshie8
  • ganeshie8
Have you used the surjithayer's hint and differentiated twice ?
anonymous
  • anonymous
yeah
ganeshie8
  • ganeshie8
show you work please
anonymous
  • anonymous
wait
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
number 14
ganeshie8
  • ganeshie8
whats the derivative of \(\dfrac{x^2}{a^2}\) with respect to \(x\) ?
anonymous
  • anonymous
but i think , may solution is not right
anonymous
  • anonymous
yes
ganeshie8
  • ganeshie8
Keep in mind, \(a\) and \(b\) are constants here
ganeshie8
  • 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
  • anonymous
okay :)
anonymous
  • anonymous
but , i did it first , but my answer was not right
ganeshie8
  • 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
  • anonymous
can you help please
ganeshie8
  • 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
  • anonymous
@ganeshie8 gets :)) thank you very much :))
ganeshie8
  • ganeshie8
yw, are you able to differentiate it again and arrive at the answer ?
anonymous
  • anonymous
@ganeshie8 im very confused
ganeshie8
  • ganeshie8
remember product rule ?
anonymous
  • anonymous
yeah i remember
anonymous
  • anonymous
how about the b^2 ?
ganeshie8
  • ganeshie8
b^2 is just a constant, like, 3^2 or 5^2
ganeshie8
  • ganeshie8
it doesn't change, so the derivative is 0
anonymous
  • anonymous
\[-x y y'' . -1 + x y^2.y'=b^2\]
ganeshie8
  • 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
  • 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
  • anonymous
@ganeshie8 i got the answer :))))
anonymous
  • anonymous
how about x^2+axy+bx+c=0
ganeshie8
  • ganeshie8
Give it a try first
anonymous
  • anonymous
i have a answer
anonymous
  • anonymous
@ganeshie8 number 11
1 Attachment
ganeshie8
  • ganeshie8
Looks you forgot to use product rule here ? |dw:1436688898948:dw|
UsukiDoll
  • 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
  • anonymous
@ganeshie8
UsukiDoll
  • UsukiDoll
x and y's are variables treat a b c as constants which when the derivative is taken it goes to 0
anonymous
  • anonymous
ai derivative of x is 1 = a (1) ?
ganeshie8
  • ganeshie8
@UsukiDoll this is like constructing the differential equation from the given general solution by eliminating the arbitrary constants
UsukiDoll
  • UsukiDoll
oh I see so we're not verifying... we're finding a DE
ganeshie8
  • ganeshie8
yeah we can think of the equations of given ellipses/parabolas as general solutions
anonymous
  • anonymous
yeah , were finding D.E
ganeshie8
  • ganeshie8
@jacalneaila did you find ur mistake in #11 ?
anonymous
  • anonymous
yeah, i try to solve ut again , but may answer is different now :(
anonymous
  • anonymous
*it
ganeshie8
  • ganeshie8
show ur work
anonymous
  • anonymous
wait , i dont have a cellphone
ganeshie8
  • ganeshie8
#11 should be fairly simple compared to the earlier one
anonymous
  • anonymous
\[ax+yy'+a(1)? \]
anonymous
  • anonymous
@ganeshie8
ganeshie8
  • ganeshie8
|dw:1436690038639:dw|
anonymous
  • anonymous
ugh
anonymous
  • anonymous
im still doing it , and still different
anonymous
  • anonymous
@ganeshie8 Thank you very much , its very helpful :)) .
anonymous
  • anonymous
@ganeshie8 Thank you again ^_^
UsukiDoll
  • 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
  • ganeshie8
the missing part is \(=0\) in every line :)
UsukiDoll
  • UsukiDoll
oh boy xD! I was doing this while dealing with a migraine.
ganeshie8
  • ganeshie8
From personal experience, regular diet+good sleep are the best medicine for migraine
UsukiDoll
  • UsukiDoll
yeah...that's what I lot of people say...but I'm such a nighthawk and it runs in the family.
ganeshie8
  • ganeshie8
Also there are some ellergetic foods which you need to avoid eating
UsukiDoll
  • UsukiDoll
if it's chocolate, then I can't =(
ganeshie8
  • ganeshie8
For me they are potato and some other vegetable whose name idk in english
ganeshie8
  • ganeshie8
they depend on the person i think..
ganeshie8
  • ganeshie8
this one http://www.mommyscuisine.com/wp-content/uploads/chikkudukayatomato1-300x225.jpg

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