differential eqn.. help!
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Huh? I dont really understand. Is it possible if you do this by the standard method by using Tn = Sn - Sn-1 and Tn is in the form a + (n-1)dOriginally posted by papercut87:Sn = pn + qn^2
therefore,
S1 = p + q (first term)
S2 = 2p + 4q
S3 = 3p + 9q
...
1st term = p + q
2nd term = S2 - S1 = p + 3q
3rd term = S3 - S2 = p + 5q
notice that common difference is 2q
the first term is p + q -
ok.Originally posted by polarsnake:No it's not a forum for maths geeks but rather, I created a thread titles' 'maths geeks'
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i got a life...Originally posted by papercut87:so you're here to +1 huh?
get a life then.
i came here to +1
+1 -
nope.Originally posted by coldzero:Huh? I dont really understand. Is it possible if you do this by the standard method by using Tn = Sn - Sn-1 and Tn is in the form a + (n-1)d
but u need to know that:
Tn = Sn - Sn-1 -
Is it possible to express the final answer as Tn = a+(n-1)d ?Originally posted by papercut87:nope.
but u need to know that:
Tn = Sn - Sn-1 -
U need to prove it's an AP, not assume it is an AP and solve it using the method of AP.Originally posted by coldzero:Huh? I dont really understand. Is it possible if you do this by the standard method by using Tn = Sn - Sn-1 and Tn is in the form a + (n-1)d
To prove you need to have values of S1, S2, S3 and so on and see what is the pattern like.
As papercut87 has pointed out, it's a sequence, however, may I add that this could be a better way to prove and solve:
Given Sn = pn + qn²,
Sn+1 = p(n+1) + q(n+1)² = pn+qn² + (p + 2qn + q)
Sn+2 = ... = pn+qn² + (2p + 4qn + 4q)
Sn+3 = ... = pn+qn² + (3p + 6qn + 9q)
If u take the difference bet Sn+3 and Sn+2 u get a difference of (p + 2qn + 5q), and for Sn+2 and Sn+1 u get a difference of (p + 2qn + 3q)
Those are the terms of T3 and T2 respectively.
And from observation, first term is 2q and difference is p+q. -
yep i agree. for that qn u arent supposed to use the Tn formula. because u're supposed to PROVE that its an A.P... not work out an AP. hope it helps
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as above
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Because for my lecture notes, it says as long as Tn can be expressed in the form of Tn= a + (n-1)d, it's an AP. So i believe my tutor would want me to expessed it in the form of Tn = a + (n-1)d to prove it's an AP.
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erm are the term's saparable?
looks like can just saparate and then integrate both sides? -
to prove that its an AP, u need to show that the common difference is alwaes constant. thats all. u dont need to prove the general formula.Originally posted by coldzero:Because for my lecture notes, it says as long as Tn can be expressed in the form of Tn= a + (n-1)d, it's an AP. So i believe my tutor would want me to expessed it in the form of Tn = a + (n-1)d to prove it's an AP. -
poly students... tsk tsk!Originally posted by R3SsH|n:i got a life...
i came here to +1
+1 -
no it isntOriginally posted by hisoka:erm are the term's saparable?
looks like can just saparate and then integrate both sides?
senior....
please help me! -
Just in case you're still at it
It's a non-exact equation. So I guess you're left with
(x+y+1)dx + (-x+y-3)dy = 0
And now you need to do the whole integrating factor thing.
- Or maybe you can just separate dx on oneside, dy on the other, then integrate each side with respect to x or y. Then group like terms and combine them for your answer. You'll probably get -(y^2+x^2)/2 + xy + -x + 3y + C.
I'm too lazy to check if that's right but since it's been a while probably not lol. -
Aiyah, papercut jie, just ask your lecturer lah
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i think polarsnake got it...that solution makes sense
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i'll go work on it abit try and recall my maths.
back later
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Let me give it a try
(x+y+1)+(-x+y-3)(dy/dx) = 0
therefore (dy/dx) = (x+y+1)/(x-y+3)
now we cannot do v=y/x or v=x+y as we cant express the right hand side in functions of (y/x) and (x+y).
Neither is this function exact as d/dy(x+y+1) not = d/dx(-x+y-3)
Thus we require another substitution
This DE is non-homogeneous and we shall use the following substitution
x=X+x_0
y=Y+y_0
where x_0 and y_0 are constants.
By chain rule we can prove that dy/dx = dY/dX as we just added constants to x and y.
Therefore (dY/dX) = (X+Y+x_0+y_0+1)/(X-Y+x_0-y_0+3)
Now we shall make it homogeneous
x_0+y_0+1 = 0
x_0-y_0+3 = 0
solving we find x_0 = -2 y_0 = 1
Thus x= X-2 y = Y+1
And dY/dX = (X+Y)/(X-Y)
Now we can use Y=vX substituion to solve this and then sub x=X-2 and y =Y+1
we can get the ans
Hope it is correct... -
OMG brilliant!!!
THANKS!! 
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Actually if we can see ahead we can make a direct substitution of v=(y-1)/(x+2) to make it a linear separable homogeneous function =p. But it is really quite hard to do that so i guess this method can assist in finding the appropriate substitution.
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oops looks like i came back too late. oh well ats long as you managed to get the thing solved
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*bang head on walll*
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rearranging,Originally posted by papercut87:okay here's the problem:
x + y + 1 + ( -x + y - 3 ) (dy/dx) = 0
find the general soln.
such a qn needs to be approached by substituition. but ive tried using v = x+y and v = -x+y but it doesnt work.
help, any maths pro?
thanks!!!
dy/dx=(-x-y-1)/(-x+y-3)=(x+y+1)/(x-y+3)
This is to find what substitution we need:
let a + b +1 = 0
a - b +3 =0
solving you get a= -2, b=1
now let x=z-2, y=w+1
dy/dx=(z-2+w+1+1)/(z-2-w-1+3)=(z+w)/(z-w)
dy/dx=dw/dx=(dw/dx)*(dz/dx)=dw/dz
so dw/dz=(z+w)/(z-w)
Now let w=vz
dw/dz=v+ (dv/dz)*z
Thus DE becomes
v+ (dv/dz)*z=(1+v)/(1-v)
Solving this, you get tan-1(w/z)=1/2 ln(z^2 + w^2) + c
so answer: tan-1[(y-1)/(x+2)]=1/2ln[(x+2)^2 + (y-1)^2] +c -
Same as her. We use the same substitutionOriginally posted by missallsunday:Let me give it a try
(x+y+1)+(-x+y-3)(dy/dx) = 0
therefore (dy/dx) = (x+y+1)/(x-y+3)
now we cannot do v=y/x or v=x+y as we cant express the right hand side in functions of (y/x) and (x+y).
Neither is this function exact as d/dy(x+y+1) not = d/dx(-x+y-3)
Thus we require another substitution
This DE is non-homogeneous and we shall use the following substitution
x=X+x_0
y=Y+y_0
where x_0 and y_0 are constants.
By chain rule we can prove that dy/dx = dY/dX as we just added constants to x and y.
Therefore (dY/dX) = (X+Y+x_0+y_0+1)/(X-Y+x_0-y_0+3)
Now we shall make it homogeneous
x_0+y_0+1 = 0
x_0-y_0+3 = 0
solving we find x_0 = -2 y_0 = 1
Thus x= X-2 y = Y+1
And dY/dX = (X+Y)/(X-Y)
Now we can use Y=vX substituion to solve this and then sub x=X-2 and y =Y+1
we can get the ans
Hope it is correct... -
haha i am a he . Anyway any NUS ppl around?