jc maths on DE- urgent

absol SG

hi, i need pretty urgrent help in jc maths (: tml's the exam lol.

Question:

(i)

By differentiating or otherwise, solve the differential equation (dy/dx)^2 - (x)(dy/dx) + y =0,

showing that the set of solution curves includes precisely one parabola and any tangent to this parabola.

(ii)

Is the curve defined by y= { (1/4)(x^2) , if x lesser or equal to 1,
{ (1/2)x - 1/4 , if x greater than 1,

a solution curve to the differential equation above? Justify your answer.

part (i) i solve already, but i dont know if my method of showing works. there are 2 sets of solution curves which i obtain from d2y/dx2 = 0 and dy/dx = x/2 which gives y=mx + k and y=(x^2)/4 + D respectively upon integration. my answer to (i) is that since m is a constant which can take on any values, when m=1/2, gradient to parabola "y=(x^2)/4 + D" is obtained. hence the set of solution curves does include only 1 parabola and a tangent to it.

^dont know if my method of explaining is correct.

(ii) totally no clue how to start-.-

any help is greatly appreciated, hopefully it can be solved by today (: and i apologize for typing the whole question out(i think its harder to see this way) cuz my scanner wont work with my reformatted com for some reason....

eagle SG

Hi, I don't understand what you mean by the following sentence

there are 2 sets of solution curves which i obtain from d2y/dx2 = 0

Was it given in the question that d²y/dx² = 0?

Also, were any other points given? Because if it ends with an unknown D, there's still multiple number of parabolas

absol SG

you differentiate that equation given in the question, manipulate a bit, will get either d2y/dx2 =0 or dy/dx=x/2.

i show my working here then.

(dy/dx)^2 - x(dy/dx) + y =0

differentiating wrt x,

2(dy/dx)(d2y/dx2) - ( (1)(dy/dx) + (x)(d2y/dx2)) + dy/dx = 0

because -dy/dx+dy/dx = 0,

therefore (d2y/dx2)( (2)(dy/dx) - x ) = 0

therefore, d2y/dx2 =0 or dy/dx = x/2

integrating wrt x,

y = [(x^2) / 4] + D or y= mx + k, where D, m, k are arbitrary constants

...until here lor. haha. thanks for any help, will check back in 2-3 hrs time again.

absol SG

er..the solution key says this about D:

that if u substitute y = [(x^2)/4] + D into the original given differential equation, (bearing in mind its dy/dx = x/2),

(x/2)^2 - (x)(x/2) + (x^2)/4 + D =0

then the answer simply says D = 0 when ...err...D is = (x^2)/2?

i dont know what is wrong here. because D is supposed to be a constant? so it cannot take (x^2)/2 which is a variable. or is it okay to give it a value of 0 just because there are no constants which it can take from the equation D = (x^2)/2?

wee_ws SG

Hi,

I agree with eagle that perhaps some pieces of information are missing.

y = 1/4 x^2 + constant is certainly not ONE parabola, unless constant = 0.

Also, y = mx + c is certainly not TANGENT to the above parabola, unless m = 1/2.

Thanks!

Cheers,
Wen Shih

wee_ws SG

Hi,

I have another view.

The solution curves are of the form y = 1/4 x^2 + d or y = mx + c.

Let d take some fixed value and let m take 1/2. Then indeed, we have ONE parabola and any tangent to it.

Thanks!

Cheers,
Wen Shih

wee_ws SG

Hi,

For (ii), the easiest way is to substitute back into the LHS of DE and verify if RHS is 0.

For x <= 1, y = 1/4 x^2, so dy/dx = x/2.

LHS of DE = (x/2)^2 - x(x/2) + 1/4 x^2 = 0 = RHS of DE.

For x > 1, y = 1/2 x - 1/4, so dy/dx = 1/2.

LHS of DE = (1/2)^2 - x(1/2) + 1/2 x - 1/4 = 0 = RHS of DE.

Also, y = 1/2 x - 1/4 is tangent to y = 1/4 x^2 at x = 1.

Hence, the set of two equations is solution to DE.

Thanks!

Cheers,
Wen Shih

absol SG

ah thank you, although they give the limit for what (x lesser or equal to 1 or x greater than 1)? does it serve any use here? as in, why is there the need to define the domain of those 2 equations? i think i only saw this thing once before in another question.....

wee_ws SG

Hi,

It is called a piece-wise function. This type of function behaves differently for different intervals of x. In our question, the function has the behaviour of a parabola when x <= 1 and it then becomes linear when x > 1.

Piece-wise functions are not common in this syllabus, but they used to get featured in the old syllabus when students studied discrete and continuous random variables in statistics.

Thanks!

Cheers,
Wen Shih

absol SG

ah thank you(: