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hey guys try this asap, also on it

obtain W(y1 , y2) (t) ...(i.e wronskian )

if y1 and y2 are two solutions of

t^2 y"+2t^3 y' -t^-2 y=0

agentofchange1:
hey guys try this asap, also on it

obtain W(y1 , y2) (t) ...(i.e wronskian )

if y1 and y2 are two solutions of

t^2 y"+2t^3 y' -t^-2 y=0

This is elementary.

Defining the Wronskian as [img]http://tutorial.math.lamar.edu/Classes/DE/Wronskian_files/eq0036M.gif[/img]

The differential equation is:
t²y" +2t³y' - t²y = 0

First we divide all through by the coefficient of the second derivative (y" ) which leaves us with:

y" +2ty' - y = 0

Applying the Wronskian defined above we get:
W = c*e^{-∫2t dt}
W = c* e^{-t^2}
(Because the integral (anti derivative) of 2t = t² + C)
W = c* e^{-t^2}
where c is an arbitrary nonzero constant.
That's all.
Cheers.

Both of una don dey mad,,,, I swear!

Karmanaut:
This is elementary.

Defining the Wronskian as [img]http://tutorial.math.lamar.edu/Classes/DE/Wronskian_files/eq0036M.gif[/img]

The differential equation is:
t²y" +2t³y' - t²y = 0

First we divide all through by the coefficient of the second derivative (y" ) which leaves us with:

y" +2ty' - y = 0

Applying the Wronskian defined above we get:
W = c*e^{-∫2t dt}
W = c* e^{-t^2}
(Because the integral (anti derivative) of 2t = t² + C)
W = c* e^{-t^2}
where c is an arbitrary nonzero constant.
That's all.
Cheers.

good Job Bro. I would have posted the solution before you but i taught it should be more complex lol. it was just two lines. t^2y"+2t^3y'-t^-2y=0...divide 2ru by t^2...y"+2ty'-t^-4y=0...what i need here is just {2t}...W=ce^-$2tdt..-$2t=t^2+c..thus W=ce^-t^2..

^^i read it sometimes back(just roughly during first semester holiday i think). i am very happy that i recalled the formula without much racking of brain. cool

agentofchange1:
seems we gat a female math killer in the house , nice1 dear , welcome on board

Thanks

uwc pretty, so what do have for the boys?

Vixxie:
Thanks

my prof. ok na , i see your hand work sir , what do you have for the boys?

Karmanaut:
This is elementary.

Defining the Wronskian as [img]http://tutorial.math.lamar.edu/Classes/DE/Wronskian_files/eq0036M.gif[/img]

The differential equation is:
t²y" +2t³y' - t²y = 0

First we divide all through by the coefficient of the second derivative (y" ) which leaves us with:

y" +2ty' - y = 0

Applying the Wronskian defined above we get:
W = c*e^{-∫2t dt}
W = c* e^{-t^2}
(Because the integral (anti derivative) of 2t = t² + C)
W = c* e^{-t^2}
where c is an arbitrary nonzero constant.
That's all.
Cheers.

agentofchange1:
my prof. ok na , i see your hand work sir , what do you have for the boys?

Something simple:
Find the limit of tan (x)^cos(x) as x tends to π/2

Vixxie:
Looks pretty straightforward to me.



First, we should note: if, for example, 200ml of 30% conc is added to 200ml of water, it gives 400ml of x% conc (more diluted)
200/400 = x/30; x=15% conc

1 part of 35% conc added to 11 parts of water gives 3% conc.
, i.e. 1 part can be any volume, but we can assume 1 ml for now.
1/(x+1) = 3/35
x = 32
for 11 parts of water; we can have
3/(3 + (32/11).11) = 3/35
where 32/11 = ~3, which makes sense for 1 part of 3ml of solution and 11 parts of 3ml of water

1 part of 12% conc added to 3 parts of water gives 3% conc.
1:4= 3:12
where 4 = 3 parts of water + 1 part of solution

how to get 12% conc from 35% conc?
1 part of 35% conc, how many parts of water to get 12%?

1/(x+1) = 12/35

x = (35-12)/12 = 23/12 = ~2 parts of water

Ans: ~2 parts of water.

Anyways, I love maths and this was a good one to do!

U're pretty good. But the answer to the question is not an approximation. U may have to adopt another approach.

Karmanaut:
Something simple:
Find the limit of tan (x)^cos(x) as x tends to π/2

Is that tan(x^cos(x)) or (tanx)^(cosx)?

Vixxie:
Is that tan(x^cos(x)) or (tanx)^(cosx)?

The second: (tanx)^(cosx)

Karmanaut:
The second: (tanx)^(cosx)

Hmm... tricky.
Answer is most likely 1 though.
There's no answer @π/2 since tanx will be undefined. The tan curve is a crazy one. But as x approaches π/2 from the left and tanx begins to go to +infinity while cosx goes to zero, limit tends to 1.
Also, as x approaches π/2 from the right, and tanx again begins to go to -infinity and cosx to 0, limit goes to 1.
I can show a detailed calculation if you want, but i think this demonstrates the concept.

elmajor:
U're pretty good. But the answer to the question is not an approximation. U may have to adopt another approach.

I only solved from intuition and i think I'm close. Looking at the question though, the answer is most likely 2 parts. I managed to search through Google and got this interesting result

https://www.momentum98.com/peroxide.html

Vixxie:
Hmm... tricky.
Answer is most likely 1 though.
There's no answer @π/2 since tanx will be undefined. The tan curve is a crazy one. But as x approaches π/2 from the left and tanx begins to go to +infinity while cosx goes to zero, limit tends to 1.
Also, as x approaches π/2 from the right, and tanx again begins to go to -infinity and cosx to 0, limit goes to 1.
I can show a detailed calculation if you want, but i think this demonstrates the concept.

The answer is 1.
But I want to see how you solved it.

L hopitals rule man , aloow that damsel to rest watching movies now , maybe later . i will post solution

Karmanaut:
The answer is 1.
But I want to see how you solved it.

agentofchange1:
L hopitals rule man , aloow that damsel to rest watching movies now , maybe later . i will post solution

There's a way to solve it without L'Hopital's rule.

ok boss lets see it,

Karmanaut:
There's a way to solve it without L'Hopital's rule.

agentofchange1:
ok boss lets see it,

Easy.
Set tan(x) as sin(x)/cos(x)
So you'll rewrite the question as:
lim^x->π/2 sinx^cosx/cosx^cosx
Then you substitute π/2 for x.
Since sinx^cosx= 1^0 =1 and
Since cosx^cosx resolves to 0^0 which is mathematically undefined (0 or 1, depending on who you ask) substitute y for cosx.
So you have 0/y^y
Then using the exponential function the denominator becomes e^{lny^y}
which is e^y*lny
Since y = cos π/2 = 0
we have:
e⁰
which is 1.
So we have 1/1
= 1.
Done.

that's right boss, thought of that too , leme finish this movies , will join u guys later

Karmanaut:
Easy.
Set tan(x) as sin(x)/cos(x)
So you'll rewrite the question as:
lim^x->π/2 sinx^cosx/cosx^cosx
Then you substitute π/2 for x.
Since cosx^cosx resolves to 0^0 which is mathematically undefined (0 or 1, depending on who you ask) substitute y for cosx.
So you have 0/y^y
Then using the exponential function the denominator becomes e^{lny^y}
which is e^y*lny
Since y = cos π/2 = 0
we have:
e[sup]0[sup]
which is 1.
So we have 0/1
= 0.
Done.

Vixxie:
I only solved from intuition and i think I'm close. Looking at the question though, the answer is most likely 2 parts. I managed to search through Google and got this interesting result

https://www.momentum98.com/peroxide.html

The answer can be got without first consulting any source.
By the way, u know about hydrogen peroxide? What can u say about the product?

elmajor:
The answer can be got without first consulting any source.
By the way, u know about hydrogen peroxide? What can u say about the product?

I actually got the answer first without consulting any source. The snapshot is after the fact.

Vixxie:
I actually got the answer first without consulting any source. The snapshot is after the fact.

Yes, the answer is a whole number and not approximation. I actually want a step by step process to arrive at the number. The formula u used did not give the exact number. Ur answer is just close.

Divide ( y^a/y^b )^a+b by ( y^a+b/y^{a-b)[sup]a^2/b} or solve it using this image

https://scontent-ams3-1.xx.fbcdn.net/hphotos-xap1/v/t1.0-9/11140269_657765617659139_3050089635620568439_n.jpg?efg=eyJpIjoiYiJ9&oh=7957b2270599fe05565559b347a83df7&oe=56B570E7

Sirneij:
Divide ( y^a/y^b )^a+b by ( y^a+b/y^{a-b)[sup]a^2/b} or solve it using this image

https://scontent-ams3-1.xx.fbcdn.net/hphotos-xap1/v/t1.0-9/11140269_657765617659139_3050089635620568439_n.jpg?efg=eyJpIjoiYiJ9&oh=7957b2270599fe05565559b347a83df7&oe=56B570E7

use laws of indices. House, apart from NMC, Abuja competition, is there any other math competition(s) for undergraduate student?

Given the D. E, xy'-y-x-1 =0 , state the conditions for the series solution to exist.

hence or otherwise solve the DE..

Can you solve this abstract algebra
1. Consider the ring of integers and its ideal 5Z . find the quotient ring Z/5Z.
2. Show that (Z×Z, + ×) is a cumulative ring with unity.

Sirneij:
Divide ( y^a/y^b )^a+b by ( y^a+b/y^{a-b)[sup]a^2/b} or solve it using this image

https://scontent-ams3-1.xx.fbcdn.net/hphotos-xap1/v/t1.0-9/11140269_657765617659139_3050089635620568439_n.jpg?efg=eyJpIjoiYiJ9&oh=7957b2270599fe05565559b347a83df7&oe=56B570E7

Straight forward

MathsChic:
Straight forward

kfb

she av changed her user name

MathsChic:
Straight forward

here,

Sirneij:
Divide ( y^a/y^b )^a+b by ( y^a+b/y^{a-b)[sup]a^2/b} or solve it using this image

https://scontent-ams3-1.xx.fbcdn.net/hphotos-xap1/v/t1.0-9/11140269_657765617659139_3050089635620568439_n.jpg?efg=eyJpIjoiYiJ9&oh=7957b2270599fe05565559b347a83df7&oe=56B570E7

Karmanaut:
Easy.
Set tan(x) as sin(x)/cos(x)
So you'll rewrite the question as:
lim^x->π/2 sinx^cosx/cosx^cosx
Then you substitute π/2 for x.
Since sinx^cosx= 1^0 =1 and
Since cosx^cosx resolves to 0^0 which is mathematically undefined (0 or 1, depending on who you ask) substitute y for cosx.
So you have 0/y^y
Then using the exponential function the denominator becomes e^{lny^y}
which is e^y*lny
Since y = cos π/2 = 0
we have:
e⁰
which is 1.
So we have 1/1
= 1.
Done.

0/1=0 or 1/1=1?