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Mathematics

1 answer:

snow_lady [41]3 years ago

6 0

M<2 = 89
m<7 = m<2 = 89
m<3 = 180 - 89 = 91
m<8 = m<3 = 91
m<10 =23
m<11 = 62
m<9 = 180 - (62 + 91)
m<9 = 27
m<14 = 1/2 (63+18)
m<14 = 40.5
m<13 = 180 -40.5 = 139.5
m<12 = 180 - (57 + 63) = 60
m<17 = 180 - (60 + 40.5) = 79.5
m<27 = 90
m<25 = 63/2 = 31.5
m<24 = 57/2 = 28.5
m<26 = 90 - (31.5 + 28.5) = 30
m<19 = 180 - (30 + 79.5) = 70.5
m<18 = 180 - 70.5 = 108.5
m<15 = (60+23+63)/ 2 = 73
m<20 = 60/2 =30
m<21 = 90 - 30 = 60
m<22 = 180 -(70.5 + 60) = 49.5
m<23 = 180 -(31.5 + 108.5) = 40
62 + 23 + 63 + 57 = 205
60 + 18 = 78
m<16 = 1/2(205 - 78) = 63.5
m<4 = 1/2(62-13) = 24.5
m<5 = 180 - 24.5 = 155.5
m<6 = m<5 = 155.5
m<1 = 180 -(24.5 + 91) = 64.5

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Solve 5y'' + 3y' – 2y = 0, y(0) = 0, y'(0) = 2.8 y(t) = 0 Preview

mario62 [17]

Answer: The required solution is

$y(t)=-\dfrac{7}{3}e^{-t}+\dfrac{7}{3}e^{\frac{1}{5}t}.$

Step-by-step explanation: We are given to solve the following differential equation :

$5y^{\prime\prime}+3y^\prime-2y=0,~~~~~~~y(0)=0,~~y^\prime(0)=2.8~~~~~~~~~~~~~~~~~~~~~~~~(i)$

Let us consider that

$y=e^{mt}$ be an auxiliary solution of equation (i).

Then, we have

$y^prime=me^{mt},~~~~~y^{\prime\prime}=m^2e^{mt}.$

Substituting these values in equation (i), we get

$5m^2e^{mt}+3me^{mt}-2e^{mt}=0\\\\\Rightarrow (5m^2+3y-2)e^{mt}=0\\\\\Rightarrow 5m^2+3m-2=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow 5m^2+5m-2m-2=0\\\\\Rightarrow 5m(m+1)-2(m+1)=0\\\\\Rightarrow (m+1)(5m-1)=0\\\\\Rightarrow m+1=0,~~~~~5m-1=0\\\\\Rightarrow m=-1,~\dfrac{1}{5}.$

So, the general solution of the given equation is

$y(t)=Ae^{-t}+Be^{\frac{1}{5}t}.$

Differentiating with respect to t, we get

$y^\prime(t)=-Ae^{-t}+\dfrac{B}{5}e^{\frac{1}{5}t}.$

According to the given conditions, we have

$y(0)=0\\\\\Rightarrow A+B=0\\\\\Rightarrow B=-A~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)$

and

$y^\prime(0)=2.8\\\\\Rightarrow -A+\dfrac{B}{5}=2.8\\\\\Rightarrow -5A+B=14\\\\\Rightarrow -5A-A=14~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{Uisng equation (ii)}]\\\\\Rightarrow -6A=14\\\\\Rightarrow A=-\dfrac{14}{6}\\\\\Rightarrow A=-\dfrac{7}{3}.$