Solve the simultaneous equations dx/dt+ 2x – 3y = 5t & dy/dt – 3x +2y = 2e^2t
Answers
Step-by-step explanation:
Solution:
Given equations as in question =>
\frac{dx}{dt} + 2x -3y = t
dt
dx
+2x−3y=t
\frac{dy}{dt} - 3x + 2y = e^{2t}
dt
dy
−3x+2y=e
2t
Let D = d/dt
Then,
(D + 2)x -3y = t(D+2)x−3y=t [Equation 1]
-3x + (D + 2)y = e^{2t}−3x+(D+2)y=e
2t
[Equation 2]
Now, we have to eliminate y.
Multiply by (D + 2) on Equation 1
$$\begin{lgathered}(D + 2)^{2} x - 3(D + 2)y = (D + 2)t \\(D + 2)^{2} x - 3(D + 2)y = Dt + 2t\\(D + 2)^{2} x - 3(D + 2)y = \frac{d}{dt} (t) + 2t\\\end{lgathered}$$
$$(D + 2)^{2} x + 3(D + 2)y = 1 + 2t$$ [Equation 3]
Multiply Equation 2 with 3
$$-9x + 3(D + 2)y = 3e^{2t}$$ [Equation 4]
Add Equation 3 and Equation 4
$$[(D + 2)^{2} x - 3(D + 2)y] + [-9x + 3(D + 2)y] = 1 + 2t + 3e^{2t}$$
$$(D + 2)^{2} x - 3(D + 2)y -9x + 3(D + 2)y = 1 + 2t + 3e^{2t}$$
$$(D + 2)^{2} x -9x = 1 + 2t + 3e^{2t}$$
$$(D^{2} + 4D + 4 )x -9x = 1 + 2t + 3e^{2t}$$
$$(D^{2} + 4D + 4 - 9)x = 1 + 2t + 3e^{2t}$$
$$(D^{2} + 4D -5)x = 1 + 2t + 3e^{2t}$$
This differential equation is a second order linear differential equation having constant coefficients.
We need to solve (D² + 4D - 5)x = 0 for complementary function and the auxiliary equation from this is,
(Put D as m)
m² + 4m - 5 = 0 [Auxiliary Equation]
m² + 5m -m - 5 = 0
m(m + 5) -1(m + 5) = 0
(m + 5)(m - 1) = 0
Roots will be =>
m + 5 = 0 => m = -5
m - 1 = 0 => m = 1
m = -5, 1
Now, we know that the roots are real and different.
So,
$$Complementary \ Function(C.F) = C_{1} e^{-5t} + C_{2} e^{t}$$
$$\begin{lgathered}Particular \ Integral(1) [P.I(1)] =>\\ \frac{1}{D^{2} + 4D - 5 } (1 + 2t)\\\end{lgathered}$$
$$\frac{-1}{5} \frac{1}{[1 - \frac{(4D + D^{2}) }{5}] } (1 + 2t)$$
$$\frac{-1}{5} [ 1 - \frac{(4D + D^{2}) }{5}]^{-1} (1 + 2t)$$
$$\frac{-1}{5} [ 1 + \frac{(4D + D^{2}) }{5}] (1 + 2t)$$
$$\frac{-1}{5} [ 1 + \frac{4D }{5}] (1 + 2t)$$
$$\frac{-1}{5} [ 1 + 2t + \frac{4 }{5}D(1 + 2t)]$$
$$\frac{-1}{5} [ 1 + 2t + \frac{4 }{5}\frac{d}{dt} (1 + 2t)]$$
$$\frac{-1}{5} [ 1 + 2t + \frac{4 }{5}*2]$$
$$\frac{-1}{5} [ 2t + \frac{13}{5}]$$
$$\begin{lgathered}Particular \ Integral(2) [P.I(2)] =>\\ \frac{1}{D^{2} + 4D - 5 } 3e^{2t} \\\end{lgathered}$$
$$3\frac{e^{2t} }{4 + 4*2-5}$$
$$\frac{3e^{2t} }{7}$$
x = C.F + P.I(1) + P.I(2)
$$x = C_{1} e^{-5t} + C_{2} e^{t} - \frac{1}{5} [ 2t + \frac{13}{5}] + \frac{3e^{2t} }{7}$$
Now, we know Equation 1
(D + 2)x -3y = t
3y = (D + 2)x - t
$$3y = \frac{dx}{dt} + 2x -t$$
Put value of x
$$3y = \frac{d}{dt} [C_{1} e^{-5t} + C_{2} e^{t} - \frac{1}{5} [ 2t + \frac{13}{5}] + \frac{3e^{2t} }{7}] + 2[C_{1} e^{-5t} + C_{2} e^{t} - \frac{1}{5} [ 2t + \frac{13}{5}] + \frac{3e^{2t} }{7}] -t$$
$$3y = \frac{d}{dt} [C_{1} e^{-5t} (-5)+ C_{2} e^{t} - \frac{2}{5} + \frac{6e^{2t} }{7} + 2C_{1} e^{-5t} + 2C_{2} e^{t} - \frac{4t}{5} - \frac{26}{25} + \frac{6e^{2t} }{7} - t$$
$$3y = -3C_{1} e^{-5t} + 3C_{2} e^{t} - \frac{9t}{5} + \frac{12e^{2t} }{7} - \frac{36}{25}$$
$$y = -C_{1} e^{-5t} + C_{2} e^{t} - \frac{3t}{5} + \frac{4e^{2t} }{7} - \frac{12}{25}$$
The solution is =>
$$x = C_{1} e^{-5t} + C_{2} e^{t} - \frac{1}{5} [ 2t + \frac{13}{5}] + \frac{3e^{2t} }{7}$$
$$y = -C_{1} e^{-5t} + C_{2} e^{t} - \frac{3t}{5} + \frac{4e^{2t} }{7} - \frac{12}{25}$$
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