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### Messages - Victorwoshinidie

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1
##### Quiz-6 / Quiz6 LEC5101
« on: November 15, 2019, 05:29:21 PM »
Q: x' = (1 2
-5 -1)x
Express the general solution of the given system of equations in terms of real_value functions.

det of (1−𝜆      2
-5        −1−𝜆)   = 𝜆^2 + 9
𝜆1 = 3i, 𝜆2 = -3i
Consider 𝜆1 = 3i, N(P-3iI) = N(1- 3i      2
-5       -1-3i)  = span[(-2,1-3i)]

Consider e3it(-2, 1-3i) = cos(3t) + isin(3t) (-2,1-3i) = (-2cos(3t),cos(3t) + 3sin(3t)) + i (-2sin(3t), sin(3t) - 3cos(3t))

Thus, the general solution of this system is

x(t) = c1(-2cos(3t),cos(3t) + 3sin(3t)) + i c2(-2sin(3t), sin(3t) - 3cos(3t))

2
##### Quiz-5 / Quiz5 Victor's Section
« on: October 30, 2019, 02:51:36 PM »
Question: (1-t)y'' + ty' - y = 2(t-1)^2 e^-t , 0 < t < 1 , y1(t) = e^t, y2(t) = t.
Answer: y1(t) = e^t, y1'(t) = e^t, y1''(t) = e^t, y2(t) = t , y2'(t) = 1 , y2''(t) = 0
Substitute back in to the homogeneous equation: (1-t)y'' +ty' - y = 0
Verified that y1(t) and y2(t) both satisfy the corresponding homogeneous equation .
And the complementary solution yc(t) = c1e^t + c2
Now divide both sides of the original eqaution by 1-t :
y'' + t/1-t  -  1/1-t = -2(t-1)e^-t
Then:
p(t) = t/1-t  , q(t) = -1/1-t  , g(t) = -2(t-1)e^-t

W[y1,y2](t) = (1-t)e^t

Since the particular solution has the form:
Y(t) = u1(t)y1(t) + u2(t)y2(t)
and
u1(t) = - ∫y2(t)g(t)/W[y1,y2](t) dt = -∫t.(-2(t-1)e^-t)/(1-t)e^t dt = -2∫te^-2tdt = (t+1/2)e^-2t
u2(t) = - ∫e^t.(-2(t-1)e^-t)/(1-t)e^t .dt = 2 ∫e^-t = -2e^ -t

Therefore,
Y(t) = (t+1/2)e^-2t.e^t + ( -2e^-t) t = (1/2 -t)e^-t
Hence, the general solution:
y(t) = yc(t) + Y(t) = c1e^t + c2t + (1/2 - t) e^-t

Therefore , the particular solution of the given nonhomogeneous equation is
Y(t) = (1/2 - t)e^-t

3
##### Quiz-2 / Quiz2 TUT0702
« on: October 19, 2019, 12:26:04 PM »
(3x+6/y) + (x^2/y + 3 y/x) dy/dx = 0
We want to find an integrating factor u as a function of xy st. (uM)y = (uN)x, Let z = xy, Thus, u(xy) = u(z(x,y)). Then

ux(xy) = du/dz dz/dx = y du/dz and uy(xy) = du/dz dz/dy = x du/dz

Therefore,
(uM)y = (uN)x => uMy +xMdu/dz = uNx + yNdu/dz
du/dz = u(Nx - My/xM - yN)

Therefore, u(z) = exp(integral R(z) dz ) where R(z) = R(xy) = Nx - My / xM - yN

Returning to our orginal diffrential equation, let
M(x,y) = 3x + 6/y and N(x,y) = x^2/y +3y/x = 0

Then derive both M and N
we get -6/y^2 and 2x/y - 3y/x^2

u(xy) = exp( integral 1/z dz) e ^ logz = z = xy
(3x^2y +6x) + (x^3+ 3y ^2) dy/dx = 0

fi (x,y) = x^3y + 3x^2 + y ^3

Thus the solutions of the diffential equation are given implicitly by
x^3y + 3x^2 + y ^3 = C

4
##### Quiz-4 / Quiz4 TUT0702
« on: October 19, 2019, 12:09:31 PM »
Q: y'' - 2y' + y = e^t/(1+t^2)

for homogenous equation y'' - 2y +1 = 0
Characteristic equation:r^2 - 2r + 1 = 0
r1 = 1 , r2 = 1
Complementary solution:
yc(t) = c1y1(t) + c2te^t

For non-homogenous equation: y'' - 2y + y = e^t/(1+t^2)
p(t) = -2, q(t) = 1, g(t) = e^t/(1+t^2) are continous everywhere.
Now,
W[y1,y2](t) = e^(2t) != 0

Thus, y1(t) and y2(t) form a fundamental set of solutions.
Therefore,
u1(t) = -1/2ln(1+t^2)
u2(t) = arctant

yp(t) = -1/2e^tln(1+t^2) + te^t arctant

Therefore, the general solution is:
y(t) = yc(t) +yp(t)
= c1e^t +c2e^t -1/2e^tln(1+t^2) + te^t arctant

5
##### Quiz-3 / Quiz3 TUT0702
« on: October 14, 2019, 08:16:43 PM »
Questions:y'' - 2y' - 2y = 0

we assume that y = e^(rt)
r^2 - 2r - = 0
r = -b+-√b^2-4ac/2a

Hence, r1 = (-1+√33)/4  , r2 = (-1-√33)/4
y = c1e^r1t +c2e^r2t
y = c1e^(-1+√33)/4t +c2e^(-1-√33)/4 t

c1 + c2 = 0
c1(-1+√33)/4 +c2(-1-√33)/4 = 1

c1 = 2/√33
c2 = -2√33

6
##### Quiz-1 / y' - 2y = e2t, y(0) = 2
« on: September 29, 2019, 12:12:05 AM »
μ(t) = e^∫-2 dt = e^(-2t)
Multiply both sides by μ(t):
e^(-2t)y' - 2e^(-2)ty = e^(-2t)e^(2t)
d/dt (e^(-2t)y) = I
e^(-2t)y = t+c(By integrating both sides)
y=(t+c)/(e^(-2t))
y = (t+c)e^2t
Substitue y(0) = 2 into the equation above, then:
2 = (0 + c)e^0 = C*1
The solution to the IVP is y = (t+2)e^(2t)

7
##### Quiz-1 / Quiz1 TUT0702
« on: September 28, 2019, 11:54:06 PM »
μ(t) = e^∫-2 dt = e^(-2t)
Multiply both sides by μ(t):
e^(-2t)y' - 2e^(-2)ty = e^(-2t)e^(2t)
d/dt (e^(-2t)y) = I
e^(-2t)y = t+c(By integrating both sides)
y=(t+c)/(e^(-2t))
y = (t+c)e^2t
Substitue y(0) = 2 into the equation above, then:
2 = (0 + c)e^0 = C*1
The solution to the IVP is y = (t+2)e^(2t)

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