Find the differential equations of the following family of curves.

14 posts / 0 new
Last post
Bingo
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.

Rate: 
0
No votes yet
Bingo

1

Rate: 
0
Average: 1 (1 vote)
Jhun Vert
Jhun Vert's picture

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

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

$y'' = \pm 8a$

Rate: 
0
No votes yet
Bingo

y−k=±4a(x−h)2 is this the equation for parabolas?
isn't it 4a(y-k)=(x-h)2 ?

Rate: 
0
No votes yet
Jhun Vert
Jhun Vert's picture

I made a mistake in there, it should be (x - h)2 = ±4a(y - k). The solution for (1) should go this way:
$(x - h)^2 = \pm 4a(y - k)$

$2(x - h) = \pm 4ay'$

$2 = \pm 4ay''$

$y'' = \pm \frac{1}{2a}$

Rate: 
0
No votes yet
Bingo

y−k=±4a(x−h)2 is this the equation for parabolas?
isn't it 4a(y-k)=(x-h)2 ?

Rate: 
0
No votes yet
Jhun Vert
Jhun Vert's picture

Yes you are right, please refer to my reply above

Rate: 
0
No votes yet
Bingo

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

Rate: 
0
No votes yet
Jhun Vert
Jhun Vert's picture

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

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

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

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

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

Rate: 
0
No votes yet
ChaCha Ingal Cortez
ChaCha Ingal Cortez's picture

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 ?

Rate: 
0
No votes yet
Jhun Vert
Jhun Vert's picture

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

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

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

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

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

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

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

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

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

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

Rate: 
0
No votes yet
Bingo

Thank You so much sir.

Rate: 
0
No votes yet
Bingo

sir, do you know the standard or general equations for the last 3 problems?
what equations should i use?

Rate: 
0
No votes yet
Shandy

I think you are from ADZU haha

Rate: 
0
No votes yet