1. Find solution of differential equation y’= єy – αy3 , є>0 and α>0 !
Answer:
y’= єy – αy3 , є>0 and α>0
Rewriting equation in the standard from
y’= єy – αy3 ,…………….(1) (left and right side divided by y3)
» y’/ y3= єy/ y3 – αy3 /y3………….(2)
» (1/ y3)dy/dt = є/ y2 – α ……………………..(3)
Suppose: u = 1/ y2 = y-2
du=-2/ y3 so, -1/2 du = dy/ y3
substitution dy/ y3 and y-2 in equation (3), we have
(du/-2dt) – є u = -α…………….(4)
(-1/2) u’ – є u = -α ……………..(5)
To find this solution,we first compute µ(t)
Left and right side of (5) multiply by -2 µ(t), we have
µ(t) u’ + 2єu µ(t) = 2αµ(t) ………..(6)
µ(t) u’ + µ’(t) u = (µ(t)u)’ ………..(7)
so that,equation (6) =(7)
µ’(t)u = 2єu µ(t)
µ’(t) = 2є µ(t)
µ(t) = e ∫2єdt
µ(t) = e 2єt+ c ,e c= k, constant
we have µ(t) = e2єt
substitution µ(t) = in equation (7), we obtain
e2єtu’ + u e2єt = (e2єtu)’ ……………(8)
and therefor (e2єtu)’ = ∫ 2α e2єtdt
so we have u = (α + c є e-2єt) / є
where u = 1/ y2, it follow that
y2 = 1/ u
y = ± √(1/u)
y = ± √(є/(α + c є e-2єt))
The solution becomes y = ± √(є/(α + c є e-2єt))

2. Prove if {an} konvergen and {bn} divergen , { an + bn }divergen!
Answer:
{an} konvergen, suppose : {an} = 1/n
{bn} divergen, suppose {bn} = n, n2
a 1-5 , where n = 1-5
b 1-5 ,where n = 1-5
a 1-5 = 1/1 , ½, 1/3, ¼,1/5,……..,1/n (1)
b 1-5 = 1, 2, 3, 4, 5, ………….,n (2)
____________________________________ _____ +
2, 5/2, 7/3, 17/4, 26/5,……, (n + 1/n)
{a 1-5 + b 1-5} = 2, 5/2, 7/3, 17/4, 26/5
So, , { an + bn }divergen if {an} konvergen and {bn} divergen
3. In triangle the measure of side is 30, 17, and 26. Determine measure of each angles !
Answer:
C



A B
The three of angles which will counted measure is α, β, γ.
AC = 17, AB = 30, BC = 26
α is angle of CAB, β is angle of ABC, γ is angle of BCA
a. For determine measures α, using cosine
BC2 = AC2 + AB2 – 2 AC.BC cos α
262 = 172 + 302 – 2.17.30 cos α
Cos α = 0,5029
Using calculator or table , value of α = 59,800
b. For determine measure of β, using sine
BC /sin α = AC / sin β
26/ sin 59,80 = 17/ sin β
Sin β = (17. Sin 59, 800)/26
Sin β = 0,5623
β = 34,210
c. Counting measure of γ angle
γ = 1800- (α+β)
γ = 1800- (59,800+ 34,210)
γ = 85,990

4. Prove if C bisector of PAQ angle and AP =AQ . So, APC angle congruent with AQC angle!
Answer:
Triangle APQ, AP ≈ (congruent) AQ
AP = line segment af AP
Angle APC ≈ angle AQC
• AP ≈ AQ (measure of AP = AQ)
• Э C € PQ ( exsistence one point in line segment )
• Э AC ( exsistence line segment from two point)
• Э < QAC and < PAC ( existence of angles)
• Э ! AC
< QAC ≈ < PAC ( existence bisector of angle)
• ∆ QAC and ∆ PAC (existence teiangle)

AP ≈ AQ (given)
AC = AC ( reflektif)
< QAC ≈ < PAC
• ∆ QAC ≈ ∆ PAC (existence side – angle - side)
• Because ∆ QAC ≈ ∆ PAC,
So, PC ≈ QC
< PCA ≈ < QCA
< APC ≈ < AQC
So, we can prove that APC angle congruent with AQC angle

Assigment 5

1.Prove if C bisector of PAQ angle and AP =AQ . So, APC angle congruent with AQC angle!
Answer:
Triangle APQ, AP ≈ (congruent) AQ
AP = line segment af AP
Angle APC ≈ angle AQC
• AP ≈ AQ (measure of AP = AQ)
• Э C € PQ ( exsistence one point in line segment )
• Э AC ( exsistence line segment from two point)
• Э < QAC and < PAC ( existence of angles)
• Э ! AC
< QAC ≈ < PAC ( existence bisector of angle)
• ∆ QAC and ∆ PAC (existence teiangle)

AP ≈ AQ (given)
AC = AC ( reflektif)
< QAC ≈ < PAC
• ∆ QAC ≈ ∆ PAC (existence side – angle - side)
• Because ∆ QAC ≈ ∆ PAC,
So, PC ≈ QC
< PCA ≈ < QCA
< APC ≈ < AQC
So, we can prove that APC angle congruent with AQC angle