Find the differential equations of the following family of curves.

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Find the differential equations of the following family of curves.

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

Romel's picture
(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 ?

Romel's picture
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 ?

Romel's picture
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?

Romel's picture
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.
$(y − k)^2 = \pm 4a(x − h)$

$2(y − k)y' = \pm 4a$

$y' = \dfrac{\pm 2a}{y - k}$

Note: $y - k = \dfrac{y'}{\pm 2a}$

$y'' = \dfrac{\mp 2a \, y'}{(y - k)^2}$

$y'' = \dfrac{\mp 2a \, y'}{\left( \dfrac{y'}{\pm 2a} \right)^2}$

$y'' = \dfrac{\mp 8a}{y'}$

$y'' \, y' = \pm 8a$

ChaCha Ingal Cortez's picture
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 ?

Romel's picture
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