Find the differential equations of the following family of curves.
Submitted by Bingo on July 8, 2017 - 5:29pm
Differential Equations
Find the differential equations of the following family of curves.
1. Parabolas with axis parallel to the y – axis with distance vertex to focus fixed as a.
2. Parabolas with axis parallel to the x – axis with distance vertex to focus fixed as a.
3. All ellipses with center at the origin and axes on the coordinate axes.
4. Family of cardioids.
5. Family of 3 – leaf roses.
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For number 1.
1
(1) Upwards and downwards
(1) Upwards and downwards parabolas with latus rectum equal to 4a.
$y - k = \pm 4a(x - h)^2$
$y' = \pm 8a(x - h)$
$y'' = \pm 8a$
y−k=±4a(x−h)2 is this the
y−k=±4a(x−h)2 is this the equation for parabolas?
isn't it 4a(y-k)=(x-h)2 ?
I made a mistake in there, it
I made a mistake in there, it should be (x - h)2 = ±4a(y - k). The solution for (1) should go this way:
$(x - h)^2 = \pm 4a(y - k)$
$2(x - h) = \pm 4ay'$
$2 = \pm 4ay''$
$y'' = \pm \frac{1}{2a}$
y−k=±4a(x−h)2 is this the
y−k=±4a(x−h)2 is this the equation for parabolas?
isn't it 4a(y-k)=(x-h)2 ?
Yes you are right, please
Yes you are right, please refer to my reply above
so for number 2,
so for number 2,
x-h=±4a(y-k)2
x'=±8a(y-k)
x''=±8a
am i wrong?
It is better to express your
It is better to express your answer in terms of y' rather than x'. Although x' will do and simpler.
$(y − k)^2 = \pm 4a(x − h)$
$2(y − k)y' = \pm 4a$
$y' = \dfrac{\pm 2a}{y - k}$
$y'' = \dfrac{\mp 2a \, y'}{(y - k)^2}$
$y'' = \dfrac{\mp 2a \, y'}{\left( \dfrac{y'}{\pm 2a} \right)^2}$
$y'' = \dfrac{\mp 8a}{y'}$
$y'' \, y' = \pm 8a$
our prof. gave the same
our prof. gave the same question but the ans. he gave is different from your ans. in no 2, his answer is (y')^3 + 2ay"=0 are they the same with your answer sir ?
In the above solution, I made
They are different. I made a mistake in $y - k$ of the above solution.
$(y − k)^2 = \pm 4a(x − h)$
$2(y - k) \, y' = \pm 4a$
$y' = \dfrac{\pm 2a}{y - k}$
$y - k = \dfrac{\pm 2a}{y'}$
$y'' = \dfrac{\mp 2a \, y'}{(y - k)^2}$
$y'' = \dfrac{\mp 2a \, y'}{\left( \dfrac{\pm 2a}{y'} \right)^2}$
$y'' = \dfrac{\mp 2a \, y'}{\dfrac{4a^2}{(y')^2}}$
$y'' = \dfrac{\mp (y')^3}{2a}$
$2a \, y'' = \mp (y')^3$
$\pm (y')^3 + 2a \, y'' = 0$
Thank You so much sir.
Thank You so much sir.
sir, do you know the standard
sir, do you know the standard or general equations for the last 3 problems?
what equations should i use?
I think you are from ADZU
I think you are from ADZU haha
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