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Method for Solving these Systems of Differential Equations
edit
Write the system of differential equations that must be solved to find the desired flow line, and solve the sytem of differential equations using the methods…
Let:
a
{\displaystyle a}
b
{\displaystyle b}
c
{\displaystyle c}
a
≠
0
{\displaystyle a\neq 0}
Linear Second-Order Differential Equation with constant coefficients is:
a
x
″
(
t
)
+
b
x
′
(
t
)
+
c
x
(
t
)
=
0
{\displaystyle ax^{''}\left(t\right)+bx^{'}\left(t\right)+cx\left(t\right)=0}
Let:
x
(
t
)
=
e
λ
t
{\displaystyle x\left(t\right)=e^{\lambda t}}
x
′
(
t
)
=
λ
e
λ
t
x
″
(
t
)
=
λ
2
e
λ
t
{\displaystyle {\begin{aligned}x^{'}\left(t\right)=\lambda e^{\lambda t}\\x^{''}\left(t\right)=\lambda ^{2}e^{\lambda t}\end{aligned}}}
Substitute the functions from #2 into the differential equation from #1…
a
x
″
(
t
)
+
b
x
′
(
t
)
+
c
x
(
t
)
=
0
a
(
λ
2
e
λ
t
)
+
b
(
λ
e
λ
t
)
+
c
(
e
λ
t
)
=
0
e
λ
t
(
a
λ
2
+
b
λ
+
c
)
=
0
{\displaystyle {\begin{aligned}ax^{''}\left(t\right)+bx^{'}\left(t\right)+cx\left(t\right)&=0\\a\left(\lambda ^{2}e^{\lambda t}\right)+b\left(\lambda e^{\lambda t}\right)+c\left(e^{\lambda t}\right)&=0\\e^{\lambda t}\left(a\lambda ^{2}+b\lambda +c\right)&=0\end{aligned}}}
Since
e
λ
t
≠
0
{\displaystyle e^{\lambda t}\neq 0}
characteristic equation must be
a
λ
2
+
b
λ
+
c
=
0
{\displaystyle a\lambda ^{2}+b\lambda +c=0}
λ
1
,
λ
2
=
−
b
±
b
2
−
4
a
c
2
a
{\displaystyle \lambda _{1},\lambda _{2}={\dfrac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}}
It can now be shown that the general solution to the differential equation in #1 is given by one of the following cases:
Let:
c
1
{\displaystyle c_{1}}
c
2
{\displaystyle c_{2}}
If
λ
1
,
λ
2
{\displaystyle \lambda _{1},\lambda _{2}}
λ
1
≠
λ
2
{\displaystyle \lambda _{1}\neq \lambda _{2}}
x
(
t
)
=
c
1
e
λ
1
t
+
c
2
e
λ
2
t
{\displaystyle x\left(t\right)=c_{1}e^{\lambda _{1}t}+c_{2}e^{\lambda _{2}t}}
If
λ
1
=
λ
2
=
λ
{\displaystyle \lambda _{1}=\lambda _{2}=\lambda }
x
(
t
)
=
c
1
e
λ
t
+
c
2
t
e
λ
t
{\displaystyle x\left(t\right)=c_{1}e^{\lambda t}+c_{2}te^{\lambda t}}
If
λ
1
,
λ
2
=
α
±
i
β
{\displaystyle \lambda _{1},\lambda _{2}=\alpha \pm i\beta }
x
(
t
)
=
e
α
t
(
c
1
cos
β
t
+
c
2
sin
β
t
)
{\displaystyle x\left(t\right)=e^{\alpha t}\left(c_{1}\cos {\beta t}+c_{2}\sin {\beta t}\right)}
Find the flow line
r
→
(
t
)
{\displaystyle {\vec {r}}\left(t\right)}
F
→
=
⟨
3
x
−
y
,
6
x
−
4
y
⟩
{\displaystyle {\vec {F}}=\left\langle 3x-y,\ 6x-4y\right\rangle }
(
2
,
7
)
{\displaystyle (2,7)}
t
=
0
{\displaystyle t=0}
0
≤
t
≤
2
{\displaystyle 0\leq t\leq 2}
Begin by writing out the system of equations, based on the given information about
F
→
{\displaystyle {\vec {F}}}
X
{\displaystyle X}
Y
{\displaystyle Y}
x
(
t
)
{\displaystyle x\left(t\right)}
y
(
t
)
{\displaystyle y\left(t\right)}
x
′
(
t
)
=
3
X
−
Y
⟹
Y
=
3
X
−
X
′
y
′
(
t
)
=
6
X
−
4
Y
{\displaystyle {\begin{aligned}x'\left(t\right)&=3X-Y&&\Longrightarrow &&Y=3X-X'\\y'\left(t\right)&=6X-4Y&&\\\end{aligned}}}
Then, combine the equations such that
Y
′
{\displaystyle Y'}
X
{\displaystyle X}
X
′
{\displaystyle X'}
Y
′
=
6
X
−
4
Y
=
6
X
−
4
(
3
X
−
X
′
)
=
6
X
−
12
X
+
4
X
Y
′
=
−
6
X
+
4
X
′
{\displaystyle {\begin{aligned}Y'&=6X-4Y\\&=6X-4\left(3X-X'\right)\\&=6X-12X+4X\\Y'&=-6X+4X'\end{aligned}}}
Now, find
X
″
{\displaystyle X''}
X
′
{\displaystyle X'}
Y
′
{\displaystyle Y'}
X
′
=
3
X
−
Y
X
″
=
3
X
′
−
Y
′
0
=
−
X
″
+
3
X
′
−
Y
′
=
−
X
″
+
3
X
′
−
(
−
6
X
+
4
X
′
)
=
−
X
″
−
X
′
+
−
6
X
=
X
″
+
X
′
−
6
X
{\displaystyle {\begin{aligned}X'&=3X-Y\\X''&=3X'-Y'\\0&=-X''+3X'-Y'\\&=-X''+3X'-\left(-6X+4X'\right)\\&=-X''-X'+-6X\\&=X''+X'-6X\end{aligned}}}
From here, we can begin to use the method (mentioned earlier) for solving differential equations by substituting
λ
{\displaystyle \lambda }
0
=
X
″
+
X
′
−
6
X
=
λ
2
+
λ
−
6
=
(
λ
+
3
)
(
λ
−
2
)
{\displaystyle {\begin{aligned}0&=X''+X'-6X\\&=\lambda ^{2}+\lambda -6\\&=\left(\lambda +3\right)\left(\lambda -2\right)\end{aligned}}}
λ
=
−
3
,
+
2
{\displaystyle \lambda =-3,+2}
λ
1
≠
λ
2
{\displaystyle \lambda _{1}\neq \lambda _{2}}
x
(
t
)
=
c
1
e
λ
1
t
+
c
2
e
λ
2
t
x
(
t
)
=
c
1
e
−
3
t
+
c
2
e
2
t
x
′
(
t
)
=
−
3
c
1
e
−
3
t
+
2
c
2
e
2
t
{\displaystyle {\begin{aligned}x\left(t\right)&=c_{1}e^{\lambda _{1}t}+c_{2}e^{\lambda _{2}t}\\x\left(t\right)&=c_{1}e^{-3t}+c_{2}e^{2t}\\x'(t)&=-3c_{1}e^{-3t}+2c_{2}e^{2t}\\\end{aligned}}}
Begin with
y
(
t
)
{\displaystyle y(t)}
Fun with Variables ): To avoid repetitiveness, and add clarity in algebraic manipulation, I temporarily use
A
{\displaystyle A}
B
{\displaystyle B}
=
3
(
c
1
⋅
e
−
3
t
)
⏟
A
+
(
c
2
⋅
e
2
t
)
⏟
B
−
(
−
3
c
1
e
−
3
t
+
2
c
2
e
2
t
)
{\displaystyle {\begin{aligned}&=3\underbrace {\left(c_{1}\cdot e^{-3t}\right)} _{A}+\underbrace {\left(c_{2}\cdot e^{2t}\right)} _{B}-\left(-3c_{1}e^{-3t}+2c_{2}e^{2t}\right)\\\end{aligned}}}
Now that we have an equation for
x
(
t
)
{\displaystyle x(t)}
y
(
t
)
{\displaystyle y(t)}
Flow Line by using the given point
(
2
,
7
)
{\displaystyle (2,7)}
t
=
0
{\displaystyle t=0}
x
(
0
)
=
2
y
(
0
)
=
7
{\displaystyle {\begin{aligned}x\left(0\right)=2\quad &&\quad y\left(0\right)=7\end{aligned}}}
y
(
0
)
=
6
(
c
1
⋅
e
−
3
⋅
[
0
]
)
+
(
c
2
⋅
e
2
⋅
[
0
]
)
7
=
6
⋅
c
1
+
c
2
c
2
=
7
−
6
c
1
{\displaystyle {\begin{aligned}y(0)&=6\left(c_{1}\cdot e^{-3\cdot [0]}\right)+\left(c_{2}\cdot e^{2\cdot [0]}\right)\\7&=6\cdot c_{1}+c_{2}\\c_{2}&=7-6c_{1}\end{aligned}}}
x
(
0
)
=
2
=
c
1
e
−
3
⋅
[
0
]
+
c
2
e
2
⋅
[
0
]
=
c
1
+
c
2
2
=
c
1
+
(
7
−
6
c
1
)
1
=
c
1
{\displaystyle {\begin{aligned}x{(0)}=2&=c_{1}e^{-3\cdot [0]}+c_{2}e^{2\cdot [0]}\\&=c_{1}+c_{2}\\2&=c_{1}+\left(7-6c_{1}\right)\\1&=c_{1}\end{aligned}}}
And now that
c
1
{\displaystyle c_{1}}
c
2
{\displaystyle c_{2}}
x
(
t
)
=
c
1
⋅
e
λ
1
t
+
c
2
⋅
e
λ
2
t
y
(
t
)
=
6
(
c
1
⋅
e
−
3
⋅
t
)
+
(
c
2
⋅
e
2
⋅
t
)
x
(
t
)
=
e
−
3
t
+
e
2
t
y
(
t
)
=
6
e
−
3
t
+
e
2
t
{\displaystyle {\begin{aligned}x(t)&=c_{1}\cdot e^{\lambda _{1}t}+c_{2}\cdot e^{\lambda _{2}t}&y(t)&=6\left(c_{1}\cdot e^{-3\cdot t}\right)+\left(c_{2}\cdot e^{2\cdot t}\right)&&\\x(t)&=e^{-3t}+e^{2t}&y(t)&=6e^{-3t}+e^{2t}\\\end{aligned}}}
Find the flow line
r
→
(
t
)
{\displaystyle {\vec {r}}\left(t\right)}
F
→
=
⟨
x
−
y
,
x
+
y
⟩
{\displaystyle {\vec {F}}=\left\langle x-y,\ x+y\right\rangle }
(
2
,
−
1
)
{\displaystyle (2,-1)}
t
=
0
{\displaystyle t=0}
0
≤
t
≤
3.8
{\displaystyle 0\leq t\leq 3.8}
For simplicity of notation, Let
x
=
x
(
t
)
{\displaystyle x=x(t)}
y
=
y
(
t
)
{\displaystyle y=y(t)}
x
′
=
x
−
y
⟶
y
=
x
−
x
′
y
′
=
x
+
y
⟶
y
′
=
x
+
(
x
−
x
′
)
⟶
y
′
=
2
x
−
x
′
{\displaystyle {\begin{aligned}x'=x-y&&\longrightarrow &&y&=x-x'\\y'=x+y&&\longrightarrow &&y'&=x+(x-x')&&\longrightarrow &&y'=2x-x'\\\end{aligned}}}
Solve for
λ
{\displaystyle \lambda }
x
′
=
x
−
y
x
″
=
x
′
−
y
′
0
=
−
x
″
+
x
′
−
y
′
=
x
″
−
x
′
+
y
′
=
x
″
−
x
′
+
(
2
x
−
x
′
)
=
x
″
−
2
x
′
+
2
x
=
λ
2
−
2
λ
+
2
{\displaystyle {\begin{aligned}x{'}&=x-y\\x{''}&=x'-y'\\0&=-x''+x'-y'\\&=x''-x'+y'\\&=x''-x'+(2x-x')\\&=x''-2x'+2x\\&=\lambda ^{2}-2\lambda +2\end{aligned}}}
To solve for
λ
{\displaystyle \lambda }
a
=
1
{\displaystyle a=1}
b
=
−
2
{\displaystyle b=-2}
c
=
2
{\displaystyle c=2}
λ
=
−
b
±
b
2
−
4
a
c
2
a
=
−
(
−
2
)
±
(
−
2
)
2
−
4
(
1
)
(
2
)
2
(
1
)
=
2
±
−
4
2
{\displaystyle {\begin{aligned}\lambda &=&&{\dfrac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}&&=&&{\dfrac {-(-2)\pm {\sqrt {(-2)^{2}-4(1)(2)}}}{2(1)}}&&=&&{\dfrac {2\pm {\sqrt {-4}}}{2}}&&\end{aligned}}}
λ
=
1
±
i
{\displaystyle \lambda =1\pm i}
Since
λ
{\displaystyle \lambda }
complex non real-valued solutions , with the form:
λ
=
α
±
β
i
{\displaystyle \lambda =\alpha \pm \beta i}
α
=
−
2
,
β
=
1
{\displaystyle \alpha =-2,\beta =1}
The general solution to the differential equation is:
x
(
t
)
=
e
α
t
(
c
1
cos
β
t
+
c
2
sin
β
t
)
=
e
t
(
c
1
cos
t
+
c
2
sin
t
)
{\displaystyle {\begin{aligned}x(t)&=e^{\alpha t}\left(c_{1}\cos {\beta t}+c_{2}\sin {\beta t}\right)\\&=e^{t}\left(c_{1}\cos {t}+c_{2}\sin {t}\right)\end{aligned}}}
Find the derivative of
x
(
t
)
{\displaystyle x(t)}
x
(
t
)
=
e
t
(
c
1
cos
t
+
c
2
sin
t
)
x
′
(
t
)
=
e
t
(
c
1
cos
t
+
c
2
sin
t
)
+
e
t
(
−
c
1
sin
t
+
c
2
cos
t
)
=
x
(
t
)
+
e
t
(
−
c
1
sin
t
+
c
2
cos
t
)
{\displaystyle {\begin{aligned}x(t)&=e^{t}\left(c_{1}\cos {t}+c_{2}\sin {t}\right)\\x'(t)&=e^{t}\left(c_{1}\cos {t}+c_{2}\sin {t}\right)+e^{t}\left(-c_{1}\sin {t}+c_{2}\cos {t}\right)\\&=x(t)+e^{t}\left(-c_{1}\sin {t}+c_{2}\cos {t}\right)\end{aligned}}}
Now that we have an equation for
x
′
(
t
)
{\displaystyle x'(t)}
y
(
t
)
{\displaystyle y(t)}
y
(
t
)
=
x
(
t
)
−
x
′
(
t
)
=
x
(
t
)
−
[
x
(
t
)
+
e
t
(
−
c
1
sin
t
+
c
2
cos
t
)
]
=
−
e
t
(
−
c
1
sin
t
+
c
2
cos
t
)
=
e
t
(
c
1
sin
t
−
c
2
cos
t
)
{\displaystyle {\begin{aligned}y(t)&=x(t)-x'(t)\\&=x(t)-\left[x(t)+e^{t}\left(-c_{1}\sin {t}+c_{2}\cos {t}\right)\right]\\&=-e^{t}\left(-c_{1}\sin {t}+c_{2}\cos {t}\right)\\&=e^{t}\left(c_{1}\sin {t}-c_{2}\cos {t}\right)\end{aligned}}}
Next, we find
c
1
{\displaystyle c_{1}}
c
2
{\displaystyle c_{2}}
(
2
,
−
1
)
{\displaystyle (2,-1)}
t
=
0
{\displaystyle t=0}
x
(
0
)
=
2
=
e
0
(
c
1
cos
0
+
c
2
sin
0
)
2
=
1
⋅
(
c
1
⋅
1
+
0
)
2
=
c
1
y
(
0
)
=
−
1
=
e
0
(
c
1
sin
0
−
c
2
cos
0
)
−
1
=
1
⋅
(
0
−
1
⋅
c
2
)
1
=
c
2
{\displaystyle {\begin{aligned}x(0)=2&=e^{0}\left(c_{1}\cos {0}+c_{2}\sin {0}\right)\\2&=1\cdot (c_{1}\cdot 1+0)\\2&=c_{1}\\\\y(0)=-1&=e^{0}\left(c_{1}\sin {0}-c_{2}\cos {0}\right)\\-1&=1\cdot \left(0-1\cdot c_{2}\right)\\1&=c_{2}\end{aligned}}}
Now that we have found
c
1
=
2
{\displaystyle c_{1}=2}
c
2
=
1
{\displaystyle c_{2}=1}
x
(
t
)
=
e
t
(
2
cos
t
−
sin
t
)
y
(
t
)
=
e
t
(
2
sin
t
−
cos
t
)
{\displaystyle {\begin{aligned}x(t)&=e^{t}\left(2\cos {t}-\sin {t}\right)\\y(t)&=e^{t}\left(2\sin {t}-\cos {t}\right)\end{aligned}}}
![{\displaystyle {\begin{aligned}y(0)&=6\left(c_{1}\cdot e^{-3\cdot [0]}\right)+\left(c_{2}\cdot e^{2\cdot [0]}\right)\\7&=6\cdot c_{1}+c_{2}\\c_{2}&=7-6c_{1}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9a41ceaefd3560c82c4339702c5f566cd9834bf8)
![{\displaystyle {\begin{aligned}x{(0)}=2&=c_{1}e^{-3\cdot [0]}+c_{2}e^{2\cdot [0]}\\&=c_{1}+c_{2}\\2&=c_{1}+\left(7-6c_{1}\right)\\1&=c_{1}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b3deac143747ece684c18ec3ca3d771fcacfb37d)
![{\displaystyle {\begin{aligned}y(t)&=x(t)-x'(t)\\&=x(t)-\left[x(t)+e^{t}\left(-c_{1}\sin {t}+c_{2}\cos {t}\right)\right]\\&=-e^{t}\left(-c_{1}\sin {t}+c_{2}\cos {t}\right)\\&=e^{t}\left(c_{1}\sin {t}-c_{2}\cos {t}\right)\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3586932cec2e5aebebaa2287787b2033d89eea55)
