Çözüldü Diferansiyel Denklem Sistemi - Belirsiz Katsayılar Yöntemi - Laplace Dönüşümü (3 Soru)

Question - 1
Period (.) has been used for the decimal point because of the language.

dx / dt = x' = -0.08·x + 0.02·y
dy / dt = y' = 0.08·x - 0.08·y
x(0) = 5, y(0) = 2
system of initial value problem solution:
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Differentiating the first equation with regard to t (assumed variable);
x'' = -0.08·x' + 0.02·y'
y' is taken from the second equation and used above as follows;
x'' = -0.08·x' + 0.02·(0.08·x - 0.08·y)....(I)
Again from the first equation, 0.02·y = x' + 0.08·x ⇒ y = 50·x' + 4x is obtained and substituted into (I);
x'' = -0.08·x' + 0.02·[ 0.08·x - 0.08·(50x' + 4x) ]
x'' = -0.08·x' + 0.02·(0.08·x - 4x' - 0.32·x)
x'' = -0.08·x' + 1.6·[ 10^(-3) ]·x - 0.08·x' - 6.4·[ 10^(-3) ]·x
x'' + 0.16·x' + 4.8·[ 10^(-3) ]·x = 0....(II)
Auxiliary equation using derivative operator as r = dx / dt is;
r^2 + 0.16·r + 4.8·[ 10^(-3) ] = 0
Roots are r1 = -3 / 25 = -0.12 and r2 = -1 / 25 = -0.12
x(t) = C1·[ e^(-0.12·t) ] + C2·[ e^(-0.04·t) ]....(III)
Differentiating (III) gives, x' = -0.12·C1·[ e^(-0.12·t) ] - 0.04·C2·[ e^(-0.04·t) ]....(IV)
Moving (III) and (IV) to the first equation in the problem;
-0.12·C1·[ e^(-0.12·t) ] - 0.04·C2·[ e^(-0.04·t) ] = -0.08·{ C1·[ e^(-0.12·t) ] + C2·[ e^(-0.04·t) ] } + 0.02·y
and after simplification, obtaining y from the equation above;
y(t) = 2·[ e^(-3t / 25) ]·{ -C1 + C2·[ e^(2t / 25) ] }....(V)
Applying the initial values as x(0) = 5 and y(0) = 2 in (III) and (V);
5 = C1 + C2....(VI)
2 = -2·C1 + 2·C2...(VII)
Solving the equations (VI) and (VII) yields; C1 = 2 and C2 = 3
So the complete solution based on (III) and (V) using the coefficients C1 and C2 is;

x(t) = 2·[ e^(-0.12·t) ] + 3·[ e^(-0.04·t) ]

y(t) = -4·[ e^(-0.12·t) ] + 6·[ e^(-0.04t) ]

Check by WolframAlpha:

https://i72.servimg.com/u/f72/19/97/10/39/3difde10.png
https://www.wolframalpha.com/input/...y / dt = 0.08·x - 0.08·y, x(0) = 5, y(0) = 2
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Not: Diğer 2 soruya da ilk fırsatta bakacağım.

Rica ederim, lafı mı olurmuş. Peki WolframAlpha ile de kontrol ettiniz mi?
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Question - 2
y'' + 4y = (x^2)sin(2x) solution with Undetermined Coefficients Method

Auxiliary equation: r^2 + 4 = 0 ⇒ r = 0 ∓ 2i
Homogeneous solution: yh(x) = [ e^(0·x) ]·[ C1·cos(2x) + C2·sin(2x) ]
yh(x) = C1·cos(2x) + C2·sin(2x)....(I)

Since the part [ e^(0·x) ]·sin(2x) of the right hand side function also coincides with the onefold repeated root (0 ∓ 2i) of the auxiliary equation, the particular solution must be in the form of (assuming a, b, c, d, e, f ∈ R);

yp(x) = (x^1)·(a·x^2 + b·x + c)·[ e^(0·x) ]·cos(2x) + (x^1)·(d·x^2 + e·x + f)·[ e^(0·x) ]·sin(2x)
yp(x) = (ax^3 + bx^2 + cx)·cos(2x) + (dx^3 + ex^2 + fx)·sin(2x)....(II)

First derivative after differentiating and grouping:
y' = (-2ax^3 - 2bx^2 - 2cx + 3dx^2 + 2ex + f)·sin(2x) + (3ax^2 + 2bx + c + 2dx^3 + 2ex^2 + 2fx)·cos(2x)

Second derivative after differentiating and polynomially grouping:
y'' = [ sin(2x) ]·(-12ax^2 - 8bx - 4c - 4dx^3 + 6dx - 4ex^2 + 2e - 4fx) +
[ cos(2x) ]·(-4ax^3 + 6ax - 4bx^2 + 2b - 4cx + 12dx^2 + 8ex + 4f) and simplifying,

y'' = [ sin(2x) ]·[ -4d·x^3 + (-12a - 4e)·x^2 + (-8b + 6d - 4f)·x - 4c + 2e ] +
[ cos(2x) ]·[ -4a·x^3 + (12d - 4b)·x^2 + (6a - 4c + 8e)·x + 2b + 4f ]....(III)

Substituting (II) and (III) in the equation;
[ sin(2x) ]·[ -4d·x^3 + (-12a - 4e)·x^2 + (-8b + 6d - 4f)·x - 4c + 2e ] +
[ cos(2x) ]·[ -4a·x^3 + (12d - 4b)·x^2 + (6a - 4c + 8e)·x + 2b + 4f ] + (4a·x^3 + 4b·x^2 + 4c·x)·cos(2x) +
(4d·x^3 + 4e·x^2 + 4f·x)·sin(2x) = (x^2)·sin(2x)

[ sin(2x) ]·[ -4d·x^3 + (-12a - 4e)·x^2 + (-8b + 6d - 4f)·x - 4c + 2e + 4d·x^3 + 4e·x^2 + 4f·x ] +
[ cos(2x) ]·[ -4a·x^3 + (12d - 4b)·x^2 + (6a - 4c + 8e)·x + 2b + 4f + 4a·x^3 + 4b·x^2 + 4c·x ] = (x^2)·sin(2x)

[ sin(2x) ]·[ -12a·x^2 + (-8b + 6d)·x - 4c + 2e ] + [ cos(2x) ]·[ 12d·x^2 + (6a + 8e)·x + 2b + 4f ] = (x^2)·sin(2x) and applying the Method of Undetermined Coefficients,

-12a = 1 ⇒ a = -1 / 12
-8b + 6d = 0
-4c + 2e = 0
12d = 0 ⇒ d = 0
6a + 8e = 0 ⇒ 6·(-1 / 12) + 8e = 0 ⇒ -1 / 2 + 8e = 0 ⇒ e = 1 / 16
2b + 4f = 0
-4c + 2·(1 / 16) = 0 ⇒ c = 1 / 32
b = d = f = 0

So yp(x) = y(x) = -(1 / 12)·(x^3)·cos(2x) + (1 / 16)·(x^2)·sin(2x) + (1 / 32)·x·cos(2x)....(IV)

And the final general solution by summing (I) and (IV),
y(x) = C1·cos(2x) + C2·sin(2x) - (1 / 12)·(x^3)·cos(2x) + (1 / 16)·(x^2)·sin(2x) + (1 / 32)·x·cos(2x)

Check by WolframAlpha:

https://i72.servimg.com/u/f72/19/97/10/39/3difde11.png
https://www.wolframalpha.com/input/?i=y'' + 4y = (x^2)sin(2x)

Her ihtimale karşı yine de yazalım belki başkaları yararlanır:
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Question - 3
2x''(t) + 32x = 4sin(4t), x(0) = -1 / 2, x'(0) = 0, solution using Laplace Transform

ℒ{ x''(t) } + ℒ{ 16x } = ℒ{ 2sin(4t) }
[ (s^2)·X(s) - s·(-1 / 2) - 0 ] + 16·X(s) = 2·[ 4 / (s^2 + 16) ]
(s^2 + 16)·X(s) + s / 2 = 8 / (s^2 + 16)
X(s) + s / [ 2·(s^2 + 16) ] = 8 / [ (s^2 + 16)^2 ]
X(s) = 8 / [ (s^2 + 16)^2 ] - s / [ 2·(s^2 + 16) ]

And taking inverse Laplace Transform;
ℒ^(-1) { X(s) } = ℒ^(-1) { 8 / [ (s^2 + 16)^2 ] - s / [ 2·(s^2 + 16) ] }, referring to the Convolution Theorem for the first rational term (unnecessary details were not written), and the standard information given at the relevant tables for the second term the general solution;

x(t) = (1 / 16)·[ sin(4t) - 4t·cos(4t) ] - (1 / 2)·cos(4t)

Check by WolframAlpha:

https://i72.servimg.com/u/f72/19/97/10/39/3difde13.png
https://www.wolframalpha.com/input/?i=laplace transform x''+16x=2sin(4t),x(0)=-0.5,x'(0)=0