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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)
