solve
dx/dt-3x-6y=t^2,
dy/dt+dx/dt-3y=e^t
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{gathered}(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{gathered}
(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=
dt
d
(t)+2t
(D + 2)^{2} x + 3(D + 2)y = 1 + 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}−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 - 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)
2
x−9x=1+2t+3e
2t
(D^{2} + 4D + 4 )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+4−9)x=1+2t+3e
2t
(D^{2} + 4D -5)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}Complementary Function(C.F)=C
1
e
−5t
+C
2
e
t
\begin{gathered}Particular \ Integral(1) [P.I(1)] = > \\ \frac{1}{D^{2} + 4D - 5 } (1 + 2t)\\\end{gathered}
Particular Integral(1)[P.I(1)]=>
D
2
+4D−5
1
(1+2t)
\frac{-1}{5} \frac{1}{[1 - \frac{(4D + D^{2}) }{5}] } (1 + 2t)
5
−1
[1−
5
(4D+D
2
)
]
1
(1+2t)
\frac{-1}{5} [ 1 - \frac{(4D + D^{2}) }{5}]^{-1} (1 + 2t)
5
−1
[1−
5
(4D+D
2
)
]
−1
(1+2t)
\frac{-1}{5} [ 1 + \frac{(4D + D^{2}) }{5}] (1 + 2t)
5
−1
[1+
5
(4D+D
2
)
](1+2t)
\frac{-1}{5} [ 1 + \frac{4D }{5}] (1 + 2t)
5
−1
[1+
5
4D
](1+2t)
\frac{-1}{5} [ 1 + 2t + \frac{4 }{5}D(1 + 2t)]
5
−1
[1+2t+
5
4
D(1+2t)]
\frac{-1}{5} [ 1 + 2t + \frac{4 }{5}\frac{d}{dt} (1 + 2t)]
5
−1
[1+2t+
5
4
dt
d
(1+2t)]
\frac{-1}{5} [ 1 + 2t + \frac{4 }{5}*2]
5
−1
[1+2t+
5
4
∗2]
\frac{-1}{5} [ 2t + \frac{13}{5}]
5
−1
[2t+
5
13
]
\begin{gathered}Particular \ Integral(2) [P.I(2)] = > \\ \frac{1}{D^{2} + 4D - 5 } 3e^{2t} \\\end{gathered}
Particular Integral(2)[P.I(2)]=>
D
2
+4D−5
1
3e
2t
3\frac{e^{2t} }{4 + 4*2-5}3
4+4∗2−5
e
2t
\frac{3e^{2t} }{7}
7
3e
2t
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}x=C
1
e
−5t
+C
2
e
t
−
5
1
[2t+
5
13
]+
7
3e
2t
Now, we know Equation 1
(D + 2)x -3y = t
3y = (D + 2)x - t
3y = \frac{dx}{dt} + 2x -t3y=
dt
dx
+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}] -t3y=
dt
d
[C
1
e
−5t
+C
2
e
t
−
5
1
[2t+
5
13
]+
7
3e
2t
]+2[C
1
e
−5t
+C
2
e
t
−
5
1
[2t+
5
13
]+
7
3e
2t
]−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} - t3y=
dt
d
[C
1
e
−5t
(−5)+C
2
e
t
−
5
2
+
7
6e
2t
+2C
1
e
−5t
+2C
2
e
t
−
5
4t
−
25
26
+
7
6e
2t
−t
3y = -3C_{1} e^{-5t} + 3C_{2} e^{t} - \frac{9t}{5} + \frac{12e^{2t} }{7} - \frac{36}{25}3y=−3C
1
e
−5t
+3C
2
e
t
−
5
9t
+
7
12e
2t
−
25
36
y = -C_{1} e^{-5t} + C_{2} e^{t} - \frac{3t}{5} + \frac{4e^{2t} }{7} - \frac{12}{25}y=−C
1
e
−5t
+C
2
e
t
−
5
3t
+
7
4e
2t
−
25
12
The solution is =>
x = C_{1} e^{-5t} + C_{2} e^{t} - \frac{1}{5} [ 2t + \frac{13}{5}] + \frac{3e^{2t} }{7}x=C
1
e
−5t
+C
2
e
t
−
5
1
[2t+
5
13
]+
7
3e
2t
y = -C_{1} e^{-5t} + C_{2} e^{t} - \frac{3t}{5} + \frac{4e^{2t} }{7} - \frac{12}{25}y=−C
1
e
−5t
+C
2
e
t
−
5
3t
+
7
4e
2t
−
25
12