NewStats: 3,263,952 , 8,182,061 topics. Date: Monday, 09 June 2025 at 01:33 AM 316a23

Mbahchiboy: plz need ur assistance guys::
(1).y"+2y'+y=0.
(2).2004^2004.
plz wen solvin use simple 4 my better understanding.
TNX IN ADVANCE

Solution.
1). Y''+2Y'+Y=0. Taking an auxiliary equation.
m^2+2m+1=0.
(m+1)(m+1)=0.
m=-1 twice.
The general soln is. Y=Ae^-x+Bxe^-x.

2). 2004^2004. You can't av an analytic solution for this. Buh here is the answer.
2004^2004=1.0069902351614371713287127625540878181640439207...*10^6617. The number length is 6618 decimal digits. All is well.

benbuks: Xy'=2x +y
solve d DE

Solution.
xy'=2x+y.
divide thru by x.
y'=2+y/x.
y'-y/x=2. Nw recall that dy/dx+P(x)y=Q.
P(x)=-1/x and Q=2.
I.F means integrating factor.
I.F= e^§P(x)dx.
I.F=e^§-1/x dx.
I.F=e^-lnx. I.F=x^-1=1/x.
hence I.F=1/x.
y.I.F= §Q.I.Fdx
y.1/x=§2.1/xdx
y/x=2§1/xdx
y/x=2lnx+C.
y=x(2lnx+C). All is well.

wisemania:
d gurus in d house will provide d working 4u,...

Ans:
2/(3)^(1/2)arctan(((3)^1/2)/(2x-1))-ln(x+1)+c...i stand 2b corrected..thank u...

Ur ans is wrong.
Here is some steps.
§x/(x^3+1).
Soln.
Using Partial fractn
§ x+1/3(x^2-x+1) dx - § 1/3(x+1)dx.
1/3§ (x+1)/(x^2-x+1) -1/3§1/(x+1). I'm stopping here. I'm sure u can continue frm here. Cos i'm kind of busy.
Here is the ans. If u move further.
1/6[ln(x^2-x+1)-2ln(x+1)+2rut3 tan^-1(2x-1/rut3)]+ C. Lol. OYO.

Laplacian:
i ve very high esteem 4 ur questns
If x+a^(x^a)=b, from d structure of x we can infer dat d above eqn is a vector eqn, applyin vector tripl produt to d 2nd term on d L.H.S,
x+x(a.a)-a(a.x)=b....eqn1, multiply d given eqn by a. to get,
a.x+a.[a^(x^a)]=a.b, scalar tripl prodct givs
a.x-(x^a).[a^a]=a.b, recall dat
a^a=0, so a.x=a.b, substitt in eqn1, to get; x(1+|a|^2)-a(a.b)=b, so
x=[b+a(a.b)]/(1+|a|^2)

@ laplacian. Hmmm u tried sha buh no be the answer be that.

Here is my question.
1. Solve for x. If x+a^(x^a)=b. Where x=(x1,x2,x3). Note that ^ means cap.

2. Let W be a subspace of real space R^3. Give a geometrical describtion of W in of its dimension.

Laplacian: ...G is a group of order 4, show dat G is abelian...

Soln.
Case 1. Suppose G has an element g of order 3. Then d cyclic subgroup generated by g contains three elements.{g,g^2,g^3=e}, where e is d identity. But the order of every subgroup must divide the order of G, nd dis is a contradiction. So G has no element of order 4.
Case 2. G is nt cyclic. Thus G={e,a,b,c} where e is d identity, nd each of a,b nd c has order 2. Nw let us take a look at d multiplicatn table.
i dnt knw hw to construct a table here. Buh here is a way 4ward. Draw a multiplicatn table.The ist row is (e,a,b,c), ist column (e,a,b,c) , 2nd row is (e,a,b,c), 3rd row is (a,e,c,b), 4th row is (b,c,e,a) nd d 5th row is (c,b,a,e). Nd nw by inspection, we see dat G is abelian. All is well.

Laplacian: ...show that d intersection of any subgroups of a group G is a subgroup of G...

solution.
Suppose we have 2 subgroups of G defined as H1 and H2. Nw the questn implies H1 intersectn H2 must also be a subgroup of G.
For any element a in H1 there exists a^-1 and H2 is closed. The same holds for H2. So the intersectn will only contain an element c in H1 intersectn H2. If c and c^-1 are in H1 and H2 they must contain e the identity of G thus H1 intersectn H2 cannot be empty. Hence the intersectn of any subgroups of a group G is a subgroup of G.

@ jackpot i tried to modify my error yesterday nite but d network hook me. Thanks 4 d correctn.
Question.
State the order axiom for IR. For a,b€ IR. And also prove for x,y€ IR 0<x<y =>0<1/y<1/x.

rhydex 247:

soln.
Suppose u=(3,4). If r & s € k. r=1 and s=2.
(r+s)u=ru+su.
(r+s)u=(1+2)(3,4)=(9,4).
ru+su=1(3,4)+2(3,4)
ru+su=(3,4)+(6,4)= (9,. Since (r+s)u is not equal to ru+su. Hence is not a vector space.

rhydex 247: here is the question again
Show that V=IR^2 is not a vector space over IR with respect to operation (a,b)+(c,d)=(a+c, b+d) and k(a,b)=(ka,kb)

soln.
Suppose u=(3,4). If r & s € k. r=1 and s=2.
(r+s)u=ru+su.
(r+s)u=(1+2)(3,4)=(9,4).
ru+su=1(3,4)+2(3,4)
ru+su=(3,4)+(6,4)= (9,. Since (r+s)u is not equal to ru+su. Hence is not a vector space.

lol @ jackpot u said d questn is nt correct. I dnt expect dat 4rm u. Anyway. All is well.
THE BEST DNT ALWAYS DO THE BEST.

here is the question again
Show that V=IR^2 is not a vector space over IR with respect to operation (a,b)+(c,d)=(a+c, b+d) and k(a,b)=(ka,kb)

he only solved the no 2 question

yea @ alpha miximus

@ laplacian, alpha, d citizen, double dx, jackpot nd other maths general i'm still waiting for the solution to my first question.

@ lanrexlan questn.
x+y=5... (1) nd x^x+y^x=13...(2)
Solutn.
frm eqn 1. x=5-y. Put x=5-y in eqn 2. We av
(5-y)^(5-y)+y^(5-y)=13... eqn * Nw we av reduced d problem to what value of y that must be put in eqn * to obtain 13. Hmmm. A thorough look at eqn 1 nd eqn 2 show dat d values of x nd y must b positive integers within the range 0<x<5 nd 0<y<5.
clearly, y=3. i.e
(5-3)^(5-3)+3^(5-3)=13. Nw dat we av y=3. It is more convenient to get x.
recal dat x=5-y. x=5-3. x=2.
hence d soln is x=2 nd y=3. All is well. @ richiez u knw wat i mean.

I can't view the pdf file @ jackpot. Buh nevertheless i will see what i can do to that. Laplacian,d citizen, doubledx,jackpot,benbuks,alphamux nd oda maths general make una help me solve my questn.

1. Show that V=IR^2 is not a vector space over IR with respect to operation (a,b)+(c,d)=(a+c,b+d) and k(a,b)=(ka,kb).

2. Show that the vectors u=(1,2,3), v=(0,1,2) and w=(0,0,1) span R^3.
Hmmm. All is well.

@ cashio questn.
resolve into partial fractn.
1/x^3-1.
solution.
1/(x-1)(x^2+x+1). Representing with A/(x-1) + Bx+C/(x^2+x+1). Nw we av
1=A/(x-1) + Bx+C/(x^2+x+1).
1=A(x^2+x+1)+(Bx+C)(x-1).
1=Ax^2+Ax+A+Bx^2-Bx+Cx-C.
1=x^2(A+B)+x(A-B+C)+A-C.
Equatin d coefficient
A+B=0... Eqn 1
A-B+C=0... Eqn 2
A-C=1... Eqn 3. Nw solvin simultaneously we av A=1/3, B=-1/3 nd C=-2/3. Putting back in A/(x-1) + Bx+C/(x^2+x+1). Finally we av.
1/3(x-1)+ (-x-2)/3(x^2+x+1).

Break fast questions.
1. Show that V=IR^2 is not a vector space over IR with respect to operation (a,b)+(c,d)=(a+c, b+d) and k(a,b)=(ka,kb).
2. Let W be a subspace of real space R^3. Give a geometrical description of W in of its dimension.

Solution to the question.
(x^2-16)(x-3)^2+9x^2=0.
by expanding out the of the L.H.S. We av
x^4-6x^3+2x^2+96x-144=0.
=> (x^2-8x+24)(x^2+2x-6)=0.
x^2-8x+24=0 or x^2+2x-6=0. Lets take it one after d other. i.e
x^2-8x=-24. Add 16 to both sides we av x^2-8x+16=-8.
(x-4)^2=-8..... x-4=2i(sqrt2) or x-4=-2i(sqrt2). Which x=2i(rut2)+4 or x=4-2i(rut2). Taking the second part i.e
x^2+2x-6=0..... x^2+2x=6. Add 1 to bth sides. We av
x^2^2x+1=7.
(x+1)^2= 7.
x+1= rut 7 or x+1=-rut7.
x=rut7-1 or x=-rut7-1. Hence the possible solution are x= 2irut2+4, x=4-2irut2, x=rut7-1 nd x=-rut7-1.

I'm back from ileya festival. Hmmmn. I can see hw u guys are destroying maths questiön. All is well. Lets kip the thread rolling.

I'm veri sowie for my mistakes. Nobody is perfect. I still remain rhydex 247 either u prove me rigorously or by venn diagram. Happy Eid-il-Kabir.
Enjoy.

@ laplacian.
I can still recall when sum1 post dis questn.
5^(x+1)=3^(x^2-1).
which d soln goes dis way.
take log of bth sides.
(x+1)log5=(x^2-1)log3.
(x+1)log5-[(x+1)(x-1)log3]=0
(x+1)[log5-(x-1)log3]=0
x+1=0 or log5-xlog3+log3=0.
x=-1 or log15=xlog3.
x=-1 or x=log15/log3.
x=-1 or x=2.465.
Similarly to sir tunechi questn.
1^x+2^x=3^(x-1).
take log of bth sides.
xlog1+xlog2=(x-1)log3
xlog1+xlog2=xlog3-log3.
0.3010x=0.4771x-0.4771.
x=0.4771/0.17612.
x=2.708929145
x=3.
Hence my soln nd my ans still stand. U can check by puttin x=2.708929145 or x=3.

Breakfast questn
1. Solve. (x^2-16)(x-3)^2+9x^2=0.

@ Sir tunechi questn.
1^x+2^x=3^(x-1).
solution.
take log of both sides.
log1^x+log2^x=log3^(x-1).
xlog1+xlog2=(x-1)log3.
xlog1+xlog2=xlog3-log3.
recall that log1=0, log2=0.3010 nd log3=0.4771.
Nw we have.
0.3010x=0.4771x-0.4771.
0.4771=0.4771x-0.3010x.
0.4771=0.17612x
x=0.17612/0.4771
x=2.708929145.
Approximately
x=3.
I want to ask wetin newton raphsön method do u @ sir tunechi.

@ laplacian question. Solutiön.
Abstract Algebra
The center is a subgroup. We show closure nd inverses. Suppose x,y in Z(G) implies xa=ax and ya=ay for every a in G. Nw xa=ax implies x=axa^-1 so that. xya=(axa^-1)ya=axy. Since nw that (xy)a=a(xy), by definition xy lies in Z(G). [we av used inverses because all of the elements are in G and hence av inverses].
nw we show that if x is in Z(G), then x^-1 lies in Z(G). If x is in Z(G) then xa=ax for all a in G. Then xa=ax implies a=x^-1ax => ax^-1=x^-1a. Which means x^-1 is in Z(G). Nw dat we av showed Z(G) is a subgroup of G. We nw show dat it is normal in G. Let H=Z(G). We must show that H=yHy^-1 for all y in G. Let x be in H. Then we know that yx=xy but yxy^-1=xyy^-1=x. Hence x is in yHy^-1. Suppse x is in yHy^-1. Then x=xyy^-1=yxy^-1 so dat xy=yx and hence x is in H. Hence the center of a group Z(G) is a normal subgroup of G.

find the general solution of
1. y"+2y'+y=4sinhx
2. y"+4y'+5y = sinx.

find the general solution of
1. y"+2y'+y=4sinhx

@ d citizen
4^x=8x
solution
using newton raphson iterative method
f(x)=4^x-8x
let x1=2
f(2)=0 hence x=2 is a factor
f'(x1)=4^xln4-8
f'(x1)=4^2ln4 - 8
f'(x1)=14.181
recall the formula
x=x1-f(x1)/f'(x1)
where x1=2, f(x1)=0 and f'(x1)=14.181
x=2-0/14.181
x=2.

@ benbuks
recall that x^2+3x+2=(x+1)(x+2)
x^2+4x+3=(x+1)(x+3)
x^2+5x+6=(x+2)(x+3).
the sqrt[(x^2+3x+2)(x^2+4x+3)(x^2+5x+6)]= sqrt[(x+1)(x+2)(x+1)(x+3)(x+1)(x+2)(x+3)].
which implies sqrt[(x+1)^2(x+2)^2(x+3)^2]= (x+1)(x+2)(x+3)
=x^3+6x^2+11x+6.

Calculusf(x):
...great master but i guess this is newton-raphson iterative method

yea this is newton raphson iterative method.