Question

solve this differential equation dx/dt+2x-3y=t,dy/dt-3x+2y=e^2t​

Answer 1

Solution:

Given equations as in question =>

$\frac{dx}{dt} + 2x -3y = t$

$\frac{dy}{dt} - 3x + 2y = e^{2t}$

Let D = d/dt

Then,

$(D + 2)x -3y = t$               [Equation 1]

$-3x + (D + 2)y = e^{2t}$         [Equation 2]

Now, we have to eliminate y.

Multiply by (D + 2) on Equation 1

$(D + 2)^{2} x - 3(D + 2)y = (D + 2)t \\(D + 2)^{2} x - 3(D + 2)y = Dt + 2t\\(D + 2)^{2} x - 3(D + 2)y = \frac{d}{dt} (t) + 2t$

$(D + 2)^{2} x + 3(D + 2)y = 1 + 2t$       [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,

$Complementary \ Function(C.F) = C_{1} e^{-5t} + C_{2} e^{t}$

$Particular \ Integral(1) [P.I(1)] =>\\ \frac{1}{D^{2} + 4D - 5 } (1 + 2t)$

$\frac{-1}{5} \frac{1}{[1 - \frac{(4D + D^{2}) }{5}] } (1 + 2t)$

$\frac{-1}{5} [ 1 - \frac{(4D + D^{2}) }{5}]^{-1} (1 + 2t)$

$\frac{-1}{5} [ 1 + \frac{(4D + D^{2}) }{5}] (1 + 2t)$

$\frac{-1}{5} [ 1 + \frac{4D }{5}] (1 + 2t)$

$\frac{-1}{5} [ 1 + 2t + \frac{4 }{5}D(1 + 2t)]$

$\frac{-1}{5} [ 1 + 2t + \frac{4 }{5}\frac{d}{dt} (1 + 2t)]$

$\frac{-1}{5} [ 1 + 2t + \frac{4 }{5}*2]$

$\frac{-1}{5} [ 2t + \frac{13}{5}]$

$Particular \ Integral(2) [P.I(2)] =>\\ \frac{1}{D^{2} + 4D - 5 } 3e^{2t}$

$3\frac{e^{2t} }{4 + 4*2-5}$

$\frac{3e^{2t} }{7}$

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 =>

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

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