Math, asked by rekhamccmcc, 9 months ago

Solve the simultaneous equations dx/dt+ 2x – 3y = 5t & dy/dt – 3x +2y = 2e^2t

Answers

Answered by kanakdurge31gmalicom
3

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 =>

$$x = C_{1} e^{-5t} + C_{2} e^{t} - \frac{1}{5} [ 2t + \frac{13}{5}] + \frac{3e^{2t} }{7}$$

$$y = -C_{1} e^{-5t} + C_{2} e^{t} - \frac{3t}{5} + \frac{4e^{2t} }{7} - \frac{12}{25}$$

Marked as brainly....

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