EDO 01

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In[2]:= y'(t)=C - k×y(t)
Input:
y′(t)C - k y(t)
Separable equation:
y′(t)
C - k y(t)  1
ODE classification:
first-order linear ordinary differential equation
Alternate form:
C k y(t) + y′(t)
Differential equation solutions: Approximate form
Solve as a separable equation | ▾
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y(t) c1 ⅇ-k t + C
k
Possible intermediate steps:
Solve the separable equationⅆy(t)ⅆt C - k y(t) :
Divide both sides by
C - k y(t) :ⅆy(t)ⅆt
C - k y(t)  1
Integrate both sides with respect to
t :
 ⅆy(t)ⅆt
C - k y(t) ⅆ t   1 ⅆ t
Evaluate the integrals:- log(C - k y(t))
k
 t + c1, where c1 is an arbitrary constant.
Solve for
y(t) :
Answer:
y(t) C - ⅇ-k (t+c1)
k
In[3]:= y'(t)=C - k×y(t)
Input:
y′(t)C - k y(t)
Separable equation:
y′(t)
C - k y(t)  1
ODE classification:
first-order linear ordinary differential equation
Alternate form:
C k y(t) + y′(t)
Differential equation solutions: Approximate form
Solve as a linear equation | ▾ Hide steps
y(t) c1 ⅇ-k t + C
k
Possible intermediate steps:
Solve the linear equationⅆy(t)ⅆt C - k y(t) :
Add k y(t)
to both sides:ⅆy(t)ⅆ t + k y(t)  C
Let μ(t) ⅇ∫ k ⅆt  ⅇk t .
Multiply both sides byμ(t) :ⅇk t ⅆy(t)ⅆ t + k ⅇk t y(t)  C ⅇk t
Substitute
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k ⅇk t  ⅆⅆt ⅇk t :ⅇk t ⅆy(t)ⅆ t + ⅆⅆ t ⅇk t y(t)C ⅇk t
Apply the reverse product rule
f
ⅆgⅆt + g ⅆ fⅆtⅆⅆt ( f g) to
the left-hand side:ⅆⅆ t ⅇk t y(t) C ⅇk t
Integrate both sides with respect to
t : ⅆⅆ t ⅇk t y(t) ⅆ t   C ⅇk t ⅆ t
Evaluate the integrals:ⅇk t y(t)  C ⅇk t
k
+ c1, where c1 is an arbitrary constant.
Divide both sides byμ(t)ⅇk t :
Answer:
y(t)  C
k
+ c1 ⅇ-k t
In[4]:= y'(t)=C - k×y(t)
Input:
y′(t)C - k y(t)
Separable equation:
y′(t)
C - k y(t)  1
ODE classification:
 3
ODE classification:
first-order linear ordinary differential equation
Alternate form:
C k y(t) + y′(t)
Differential equation solutions: Approximate form
Solve with Laplace transform | ▾
Hide steps
y(t) c1 ⅇ-k t + C
k
Possible intermediate steps:
Solve ⅆy(t)ⅆt C - k y(t) :
Apply the Laplace transformationℒt[ f (t)] (s)∫0∞ f (t) ⅇ-s t ⅆ t to
both sides:ℒt ⅆy(t)ⅆ t  (s)ℒt[C - k y(t)] (s)
Find the Laplace transformation term by term and factor out constants:ℒt ⅆy(t)ⅆ t  (s)C (ℒt[1] (s)) - k (ℒt[y(t)] (s))
Apply ℒt[1] (s)  1
s
:
ℒt ⅆy(t)ⅆ t  (s)C 1s - k (ℒt[y(t)] (s))
Apply ℒt ⅆy(t)ⅆ t  (s)  s (ℒt[y(t)] (s)) - y(0):
s (ℒt[y(t)] (s)) - y(0)  C
s
- k (ℒt[y(t)] (s))
Solve forℒt[y(t)] (s) :ℒt[y(t)] (s) C + y(0) s
s (k + s)
Decompose
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Decomposeℒt[y(t)] (s) via
partial fractions:ℒt[y(t)] (s)  C
k s
- C
k (k + s) + y(0)k + s
Compute
y(t)ℒs-1 Ck s - Ck (k+s) + y(0)k+s  (t) :
Find the inverse Laplace transformation term by term:
y(t)  ℒs-1 C
k s
 (t) +ℒs-1- C
k (k + s)  (t) +ℒs-1 y(0)k + s  (t)
Apply ℒs-1 C
k s
 (t) C
k
:
y(t)  C
k
+ℒs-1- C
k (k + s)  (t) +ℒs-1 y(0)k + s  (t)
Apply ℒs-1- C
k (k + s)  (t) -C ⅇ-k tk :
y(t)  C
k
+ -C ⅇ-k t
k
+ℒs-1 y(0)
k + s  (t)
Apply ℒs-1 y(0)
k + s  (t) y(0) ⅇ-k t:
y(t)  C
k
- C ⅇ-k t
k
+ y(0) ⅇ-k t
Substitute
c1  y(0) :
y(t) C
k
- C ⅇ-k t
k
+ c1 ⅇ-k t
Simplify:
y(t)  ⅇ-k t -C +C ⅇk t + k c1
k
Simplify the arbitrary constants:
Answer:
y(t) C + c1 ⅇ-k t
 5
y( )
k
1
In[5]:= y'(t)=C - k×y(t)
Input:
y′(t)C - k y(t)
Separable equation:
y′(t)
C - k y(t)  1
ODE classification:
first-order linear ordinary differential equation
Alternate form:
C k y(t) + y′(t)
Differential equation solutions: Approximate form
Transform into an exact equation | ▾
Hide steps
y(t) c1 ⅇ-k t + C
k
Possible intermediate steps:
Solve ⅆy(t)ⅆt C - k y(t)
:
Rewrite the equation:-C + k y(t) + ⅆy(t)ⅆ t  0
Let R(t, y)-C + k y
and
S(t, y)
1 .
This is not an exact equation, because ∂R(t,y)∂y  k ≠ 0  ∂S(t,y)∂t  .
Find an integrating factor
6 
integratingμ(t)
such thatμ(t) R(t, y) + ⅆy(t)ⅆt μ(t) S(t, y) 0 is
exact.
This means∂∂y (μ(t) R(t, y))∂∂t (μ(t) S(t, y)) :
k μ(t)  ⅆμ(t)ⅆ t
Isolateμ(t) to the
left-hand side:∂μ(t)∂tμ(t)  k
Integrate both sides with respect to
t :
log(μ(t)) k t
Take exponentials of both sides:μ(t)  ⅇk t
Multiply both sides of-C + k y(t) + ⅆy(t)ⅆt  0 byμ(t) :ⅇk t (-C + k y(t)) + ⅇk t ⅆy(t)ⅆ t  0
Let P(t, y) ⅇk t (-C + k y)
and
Q(t, y) ⅇk t .
This is an exact equation, because∂P(t,y)∂y
k ⅇk t∂Q(t,y)∂t .
Define
 7
f (t, y)
such that∂ f (t,y)∂t  P(t, y)
and∂ f (t,y)∂y Q(t, y)
. :
Then, the solution will be given by
f (t, y) c1
, where
c1
is an arbitrary constant.
Integrate∂ f (t,y)∂t
with respect to
t
in order to find
f (t, y) :
f (t, y)∫ (-C + k y) ⅇk t ⅆ t(-C+k y) ⅇk t
k
+ g(y)
where
g(y)
is an arbitrary function of
y .
Differentiate
f (t, y)
with respect to
y
in order to find
g(y) :∂ f (t, y)∂y  ∂∂y (-C + k y) ⅇk tk + g(y)  ⅇk t + ⅆg(y)ⅆy
Substitute into∂ f (t,y)∂y Q(t, y) :ⅇk t + ⅆg(y)ⅆy  ⅇk t
8 
Solve forⅆg(y)ⅆy :ⅆg(y)ⅆy  0
Integrateⅆg(y)ⅆy
with respect to
y :
g(y)∫ 0 ⅆy
0
Substitute
g(y) into
f (t, y) :
f (t, y) (-C + k y) ⅇk t
k
The solution is
f (t, y) c1 :(-C + k y) ⅇk t
k
 c1
Solve for
y :
y(t) ⅇ-k t C ⅇk t + k c1
k
Simplify the arbitrary constants:
Answer:
y(t) C
k
+ c1 ⅇ-k t
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