1.Determine the integration factors of the following differential equation and the find the solution
(3x2y + 2xy + y3) dx + (x2+ Y2) dy =0

2.Known : sin -1 x =
∫_0^x▒1/√((1-t2)) dt

Determine the first four tribes taknol in Mac Laurin series for sin -1

3.Known :
g =((█(x@y@z))= (█(1@-3@5))+ a (█(-6@4@7)) ) a parameter
Determine the line g and h has an intersection

4.Chanted a coin in a row as much as 6 times, how many chances at least one comes face?

5.Consider the following data for 120 mathematics at the college concerning the languages French, German, Russian
65 study French
45 study German
42 study Russian
20 study French and German
15 study German and Russian
8 study all three languages
Let F, G, and R denote the sets of students studying French, German, and Russian, respectively. We wish to find the number of students who study at least one of the three languages, and to fill the correct number of students in each of the eight regions


Answers:
1.(3x2y + 2xy + y3) dx + (x2+ Y2) dy =0
M(x,y) = 3x2y + 2xy + y3
N(x,y) = x2+ Y2
∂M(x,y) / ∂y = 3x2+ 2x + 3y2
∂N(x,y)/ ∂x = 2x
Formula to get the integrations factors is :
(1/N(x,y)) [(∂M(x,y)/∂y)] - [(∂N(x,y)/∂x)]
Substitute the equation 1 and 2 to formulas;
1/(x2 + y2) ((3x2+ 2x + 3y2) – (2x))
1/(x2 + y2) (3x2 + 3y2) =3
You can get μ(x) = e∫3 dx = e3x
Multiply e3x to M (x,y) and M (x,y) be M* (x,y)
M* (x,y) = 3x2ye3x + 2xy e3x + y3e3x
Differential M* (x,y) to y, so:
∂M*(x,y) / ∂y = 3x2e3x + 2xe3x + 3y2e3x
And so N(x,y) multiply with e3x
N*(x,y) = x2e3x + y2e3x
Differential N*(x,y) to x so:
∂N*(x,y)/∂x = 2xe3x + 3e3xx2 + 3y2e3x
Cause the different equation is exact differential equation, so:
∂F/∂x=M* AND ( ∂F)/∂y=N*
To get the solution we must be :
F (x,y) = ∫(3x2ye3x + 2xye3x + y3e3x) dx + Q (y)
= yx2e + 1/3 y3 e3x + Q(y)
∂F(x,y) /∂y = x2e3x + y2 e3x +Q’(y)
∂F/∂y = N*
e2 e3x + y2 e3x = x2e3x + y2e3x + Q’(y)
Q’(y) = 0
Q(y) = c
So the solution from this problem is
F(x,y) = yx2e + 1/3 y3 e3x + c

2. Known : sin -1 x =
∫_0^x▒1/√((1-t2)) dt
We have a formula :
(1 + X)P = 1 + (█(p@1))x + (█(p@2))x2 +…
Where : + (█(p@k)) =( p (p-1) (p-2))/k!
So we can apply that
(1-t2)-1/2 = 1 + (-1/2) (-t2) + (-1/2) (-3/2) (-t2)2 + (-1/2) (-t2) (-5/2) (-t2)3+…
= 1 + t2/2 + (3/8) t4 + (15/48) t6 +…
= t2/2 + (3/8) t4 + (5/16) t6 +…
Sin -1 x = ∫_0^x▒〖1 〗+ t2/2 + (3/8)t4 + (5/16)t6 +…
= x + x3/6 + (3/40)x5 + (5/112)x7


3.Known :
g = (█(x@y@z)) = (█(1@-3@5))+ a(█(-6@4@7)) a parameter
x = -2 -9b
h = y = 8-3b b parameter (2)
z = -6 +4b
to prof gnh = N
N € g => coordinate N must on the g
N € h => coordinate N must on the h
x = 1 – 6a
y = -3 + 4a
z = 5 + 7a xN= 1 – 6a
substitute coordinate N to (1) => yN = -3 +4a (3)
zN = 5 + 7a



substitute coordinate N to (2) => xN = -2 -9b
h = yN = 8-3b b parameter (4)
zN = -6 +4b
left segment of (3) equals left segment of (4)
right segment of (3) equals right segment of (4)
1 – 6a = -2 – 9b  3 = 6a – 9b
1 = 2a – 3b …………. (5)
-3 + 4a = 8-3b  4a + 3b =11……………(6)
5 + 7a = -6 + 4b  7a – 4b = -11……….(7)
So:
2a – 3b = 1
(8) 4a + 3b=11
7a – 4b = -11
Eliminate (5) and (6) => b = 3
Substitute b to (5) => a = 5
Substitute value a and b to (7)
7a – 4b = -11
7.5 – 4.3 = -11
35 - 12 ≠ -11
If that no similar gnh = empty seT
N if (8) have a solution

4.Suppose that E happened at least once come home. Sample space S contains the sample is 26 = 64 points. Because each reflection can produce two kinds of results (fronts or near), we know that P (E) = 1 – P (E’). if E’ represents the event that no face appeared. This will only happen in a way, which is all behind chanting produces. Be P(E’) = 1/64 so P(E)=1 - 1/64 = 63/64

5.(F U G U R) = n(F) + n(G) + n(R) –n(F slices G) – n(F slices R) - n( G slices R)
+ n(F sliceS G slices R)
= 65+45+42-20-25-15+8 = 100
That is, n(F U G U R)=100 students at least one of the three languages
Now, we use this result to fill in the venn diagram. We have 8 study all the three languages
20 – 8 =12 study French and german but not Russian
25 – 8 = 17 study French and Russian but not german
15 – 8 = 7 study german and Russian but not French
65 – 12 – 8 -17 =28 study only French
45 – 12 – 8 – 7 =18 study only german
42 – 17 – 8 -7 = 10 study only Russian
120 – 100 =20 do not study at the languages
Observe that 28 + 18 + 10 = 56 students only one of the languages

Problems solving created by salindri murbarani 08305141034
Solutions the problems created by Narita yuri