Find the differential equations of the following family of curves.

Submitted by Bingo on

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.

Tags
Family of Curves

Jhun Vert Sat, 07/08/2017 - 17:53

(1) Upwards and downwards parabolas with latus rectum equal to 4a.
$y - k = \pm 4a(x - h)^2$

$y' = \pm 8a(x - h)$

$y'' = \pm 8a$

Bingo Sat, 07/08/2017 - 19:41

so for number 2,
x-h=±4a(y-k)2
x'=±8a(y-k)
x''=±8a
am i wrong?

Jhun Vert Sun, 07/09/2017 - 09:37

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

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

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

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

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

Jhun Vert Fri, 08/11/2017 - 13:38

They are different. I made a mistake in $y - k$ of the above solution.
$(y − k)^2 = \pm 4a(x − h)$

$2(y - k) \, y' = \pm 4a$

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

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

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

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

$y'' = \dfrac{\mp 2a \, y'}{\dfrac{4a^2}{(y')^2}}$

$y'' = \dfrac{\mp (y')^3}{2a}$

$2a \, y'' = \mp (y')^3$

$\pm (y')^3 + 2a \, y'' = 0$

Ancha (not verified) Sun, 08/28/2022 - 22:29

Sir do you know how to eliminate arbitrary constant here in this equation?
(y-33)²=4a(x-h)?