Download - First Order Differential Equations
Chapter 1
First-Order Differential Order (1st-order DE)
Plan:
- Modeling: derivation of DEs from Physical or other problems
- Methods of Solutions of these DEs
- Interpretation of the results
- Discuss existence and uniqueness of solutions
Section 1.1: Basic Concepts and Ideas
Def 1: An ordinary differential equation (ODE) is an equation containing one or
several derivatives of an unknown variable or function (which we want to
determine from the equation) with respect to only one independent variables.
For Example:
orderrdlinearyyeyxyx
orderndnonlinearyyy
orderstlinearxyy
x
3ln2
20cos44
1cos
22
Def 2: A partial differential equation (PDE) is an equation containing an unknown
function and its Partial derivatives with respect to two or more independent variables.
For Example:
EquationHeatuau
EquationWaveuau
EqnsLaplaceuu
txx
xxtt
yyxx
,
,
',0
2
2
In this course we are concerned with ODEs only, no PDEs.
The order of an ODE is the highest derivative that appears in the ODE.
The 1st-order ODE can be represented by the equation: 0),,( yyxF or ),( yxfy
The differential equation is linear if in each terms, the unknown variable and its
derivatives appear linear ( yxa )( or ))( )(nyxb .
Classifications of Differential Equations:
1- According to Type: ODE or PDE.
2- According to Order: 1st, 2
nd or Higher orders.
3- According to Linearity: Linear or Nonlinear.
An explicit solution to the 1st-order DE given above is a function bxaxhy ),(
that satisfies the equation.
An implicit solution is an equation dycbxayxH &,0),( satisfieing the DE.
Example 1: By direct substitution check that 2xy is a solution to yyx 2 for all x.
Solution: 22)2(2 xxxxy (Done)
Example 2: Check that 0122 yx is a solution to xyy on the interval: 11 x .
Solution: By implicit differentiation of the equation 0122 yx , we obtain
xyyyyx 022 (Done)
Classifications of Solutions of Differential Equations :
1- According to Type: Explicit or Implicit.
2- According to Number: Particular or General (family).
3- According to Singularity: Singular or Nonsingular.
General, particular and singular solutions:
(1) General solution: 2),( cxcxfyg is a general solution of yyx 2 for any
number c.
(2) Particular solution: 23xy p is a particular solution to yyx 2 .
Similarly, 0),,( 222 cyxcyxH is a family of solutions (general) to the DE:
xyy , while 0422 yx is a particular solution.
(3) Singular solution: a solution of the DE but not a memebr of the family of solutions.
For example, the DE 0)( 2 yyxy has the general solution 2ccxyg .
(a family of linear equations for each value of c). Now 4
2xy is a solution but not
a member of the family of solutions. So it is singular.
(4) No general solution: the DE: 0 yy has no general solution but has 0y as
the only (particular) solution.
(5) No solution: the DE 1)( 2 y has no solution.
Initial-Value Problem (IVP)
A differential Equation of the form
00 )(),,( yxyyxfy or 00 )(,0),,( yxyyyxF
is called initial-value Problem (IVP) and 00 )( yxy is called initial condition.
A solution of the IVP (if exists) must satisfy the initial condition.
For example, the (IVP) 3)1(,2 yxy has 22 xy as a solution and it is the only
solution. Also, the (IVP) 4)0(, yxyy has 01622 yx as the only solution.
Models and Applications of 1st-Order DE’s
We now give some real life problems whose mathematical model is represented by a
1st-order DE’s.
(1) Exponential Decay
A radioactive substance decomposes at a rate proportional to the amount present.
i.e., let )(ty represent the amount at time t. Then kydt
dy
for some constant k, called
the constant of proportion.
Example 3: A radium 22488Ra decomposes to its half in 3.6 days. What will be the amount
of this radium if the initial amount was 1 gram?
(2) Geometric:
A curve that passes through a pt ),( 00 yx in the xy-plane has a slope ),( yxf at each of
its points. In this case: 00 )(),,( yxyyxfdx
dy .
Example 4: A curve that passes through (1,1) in the xy-plane has, at each of its points, a
slope yx / . Find this curve. .
Section 1.3: Separable DEs
This is the first method or technique to solve 1st- order DE’s.
A Separable 1st –order DE is usually written or simplified to one of the forms:
)()( xfyyg or 0)()( dxxfdyyg .
Technique of solution: Integrate each term with respect to its variable:
cdxxfdyygorcdxxfdyyg )()()()(
Example 1: Solve the DE: 049 xyy .
Solution: 049 xdxydy
By integration, we obtain: *22
2
4
2
9c
xy or 2
22
94c
xy .
Example 2: Solve the IVP: 1)0(,1 2 yyy
Solution: By separating the variables, we obtain dxy
dy
21
By integration, cxy )(tan 1 . To find c, substitute x=0 and y=1 we obtain
4)1(tan 1 c . So, there is a unique solution:
22,
4tan
xxy .
We now return to the models in Section 1.1:
(1) The Exponential Decay
Example 3: A radioactive substance decomposes at a rate proportional to the amount
present )(ty : 0)0(, yykydt
dy
Solution: kdty
dy . By Integrating both sides, *)ln( ckty or ktcey , where
*cec .
Substituting the initial condition, we obtain kteyy 0 .
For the case with the radium 22488Ra that decomposes to its half in 3.6 days. What
will be the amount of this radium if the initial amount was 0y gram?
As seen above, the solution to the decaying substance is kteyy 0 .
By substituting t=3.6 days and 2
0yy we obtain
2
1ln
6.3
1k and the solution will
be 6.3/
02
1t
yy
(after some math manipulations).
Note that: 3.6 =half life of the substance=2
1t , so, 2
1/
02
1tt
yy
.
(2) The Geometric Problem:
Example 4: A curve that passes through (1,1) in the xy-plane has, at each of its points, a
slope yx / . Find the curve, i.e, solve the IVP: 1)1(, yy
x
dx
dy.
Solution: Separating the variables, we obtain 0 xdxydy .
By integration we obtain cyx
22
22
and the substituting the initial condition, we
get 222 yx
(3) Equations Reducible to Separable Equation:
a- (1st – order Homogeneous Equation): Equations written in the form
x
yfy
Method of Solution:
Substitute x
yu , then we have uxy .
By implicit differentiation with respect to x we obtain )(ufuxuy
From this, we have the separable DE: x
dx
uuf
du
)(
After solving this equation, we substitute x
yu back.
Example 5: Solve 222 xyyxy
Step 1: Simplify the DE we get )()(21
x
y
yx
x
yfy
Step 2: Substitute x
yu , then we have
x
dx
uu
du
u
)( 1
21
.
Step 3: Simplify and integrate we obtain )ln()ln()1ln(1
222
cxduux
dx
u
u
Step 4: Substitute x
yu and simplify we obtain the solution 2
2
22
2)()( cc yx
which is a circle of radius 2
||c and center )0,(2c , where c is to be determined
by any initial condition.
b- )( cbyaxfy .
Method of Solution:
Substitute cbyaxv , then we have )(vbfaybav
From this, we have the separable DE: dxvbfa
dv
)(.
After solving this equation, we substitute cbyaxv back.
Example 6: Solve )02(21
)2(21
21
421
yxf
yx
yx
xy
xyy .
Step 1: Substitute yxv 2 and we then have dxdv
vv
1
212
Step 2: Simplify and integrate we obtain cxdxdvvv v 33)1(2
2
Step 3: substitute yxv 2 back into the solution.
Section 1.4: Modeling Separable DEs
We now study some examples on real life problems whose mathematical models
are described by Separable DEs.
Example 1: Radiocarbon Dating
Suppose an archaeologist excavates a bone and measures its content of radioactive
carbon 146 C . If the result is 25% of the content present in bones of a living
organism, what can be said about the age of the bone?
Idea of the Solution:
In the atmosphere, the ratio of radioactive carbon 146 C and ordinary carbon 12
6 C is
constant, i.e., cC
C constant
126
146 . The same is true for living organisms. If an organism
dies, the absorption of 146 C by breathing or eating terminates.
So, by comparing the carbon ratio in the fossil and that of the atmosphere, one can
estimate the age of the fossil. (W. Libby’s idea of radiocarbon dating, 1960 Nobel
Prize of chemistry).
Given the above and the half-life (5730 years) of 146 C we estimate the age of the bone.
Recall in Section 1.3 we solved the radioactive problem: 0)0(, yykydt
dy .
We obtain the following solution (in terms of the half-life 2
1t ) 2
1/
02
1tt
yy
.
Substitute 0%25 yy into the solution above and we obtain the age of the bone as
11460*2*21
21
2
1ln
4
1ln
ttt years.
Example 2: Mixing Problem
A tank contains 200 gallon of water in which 40 Ib of salt are dissolved. Five
gallons of brine, each containing 2 1b of dissolved salt, run into the tank per
minute, and the mixture, kept uniform by stirring, runs out of the tank at the same
rate. Find the amount of salt )(ty in the tank at any time.
Step 1: Modeling
The rate of change dt
dyinflow of salt – outflow of salt
Inflow rate= 2*5=10 Ib/minute. For the outflow rate, note the amount of salt at
each time is )(ty in 200 gallons of water.
So, each gallon of water contains 200
)(tyIb. 5 gallons/per minute of this water
flows out and they contain 40200
)(*5
yty Ib. The IVP then becomes
.40)0(,40
10 yy
dt
dy
Step 2: Solving the IVP:
Separating the variables above, we obtain 40400
dt
y
dy
.
The solution of the above DE ist
cey 40
1
400
.
Substituting the initial condition (t=0 , y=40) we obtain 360c . So the
amount of dissolved salt at any time is t
ey 40
1
360400
.
Interpretation of the result:
Note that the amount of dissolved salt is increasing with time as expected. Note
also that the maximum amount of salt is 2*200=400 Ib, where each of the 200
gallons of water contains 2 Ib of salt each.
Example 3: Heating Problem
Suppose you turned off the heat in your room at night 2 hours before you go to
bed. If the temperature at this time is F66 and the temperature has dropped to
F63 at the time you go to bed. What temperature you expect in the morning, say
after 8 hours of sleeping? Assume outside temperature FTA32 .
Idea of the Solution: (Newton’s Law of cooling)
The time rate of change dtdT / of temperature T is proportional to the difference
between the temperature T and the surrounding temperature AT .
From this, we have the separable IVP: .66)0(),32()( TTkTTkdt
dTA
The solution of the above DE is
ktceT 32 .
Substituting the initial condition (t=0, T=66) we obtain c=34. Again substituting
(t=2 , T=63) and with c=34, we obtain 046187.034
3263ln
21
k . The solution is
then
tetT 046187.3432)( .
The temperature in the morning (after 10 hours) is then
FeT 4.533432)10( 46187. .
Section 1.5: Exact DEs
This is a second solution method of 1st- order DEs
Recall from Calculus 3: The (total) differential of 32),( yxxyxu is given by:
dyyxdxxydyudxudu yx223 3)21(
Note that if: constant32),( yxxyxu , then the solution to the DE
03)21( 223 dyyxdxxy is constant32 yxx .
Definition of Exact DE:
A DE of the form 0),(),( dyyxNdxyxM (1)
is said to be exact if dyyxNdxyxM ),(),( is the (total) differential of a function
u(x,y). Hence, the DE (1) above is exact if there exists u(x,y) such that
NdyMdxdyyudx
xudu
The DE in (1) above then becomes 0du , whose solution is then t),( constanyxu .
Idea of solution
1- Determine if the DE is exact.
2- If the DE is exact, find the function u(x,y).
Question (1): How can we determine if the DE 0),(),( dyyxNdxyxM is exact?
Answer: The DE is exact iff xN
yM
To see this, suppose M and N have continuous first partial derivatives.
If the DE is exact there exists u(x,y) such that NyuM
xu
& , then
xN
yxu
yM
xyu
2
&2
.
By continuity of the first partial derivatives we then have xN
yM
.
Question 2: If the DE is exact, how do we find u(x,y)?
Answer: Since the DE is exact, then NyuM
xu
& .
Hence, )(),( yhMdxyxu (2) and )(),( xgNdyyxu (3)
Hence, by equating the RHS of (2) with the RHS of (3) above, we determine )(yh ,
)(xg and u . Hence, the solution is then cyxu ),( .
Example 1: Solve 03)21( 223 dyyxdxxy .
Solution: Note that 223 3&21 yxNxyM .
Step 1: Determine if the equation is exact.
Since xNxy
yM
26 , then the DE is exact.
Step 2: Find u(x,y).
)(3)()21(),( 223 xgdyyxyhdxxyyxu .
We then have: )()( 3232 xgyxyhyxx .
So, xxgyh )(,0)( , 32),( yxxyxu and the solution is cyxx 32
Example 2: Solve 0)3()3( 3223 dyyyxdxxyx
Solution:
Step 1: The DE is exact since xy NxyM 6
Step 2: )(2
3
4)()3(),(
22423 yh
yxxyhdxxyxyxu
To compute h(y), 322 3)(3 yyxNyhyxu y . Hence, 3)( yyh and 4
)(4y
yh .
So we have 42
3
4),(
4224 yyxxyxu and the solution is c
yyxx
42
3
4
4224
.
Example 3: Solve the IVP: )0(,0sincoshcossinh ydyyxdxyx
Solution: The DE is exact: xy NyxM sinsinh
Then )(coscosh)(sincosh),( xgyxxgdyyxyxu
To compute g(x), yxMxgyxux cossinh)(cossinh . So, 0)( xg ,
hence, 0)( xg and the solution is then cyx coscosh . With )0(y , then
1c and the unique solution is then yx seccosh .
Question 3: What if the equation was not exact? Can it be made exact?
Example 4: The DE 0 dyxdxy is not exact since xy NM 11 .
But note that if we multiply the DE by 2
1
x, then the DE becomes
01
2
dy
xdx
x
y which is now exact. The solution of this new Exact DE is the
same as the solution to the non-Exact DE.
Remark: The multiplying function 2
1
xis an integrating factor (I.F) of the above DE.
Question 4: How do we know if an integrating factor exists for any given DE and how to
find such one (if exists)?
Answer: Compute N
NMR
xy 1 or
M
MNR
yx 2 and use Integrating Factor Theorems
Theorem 1: If )(1 xPN
NMR
xy
, a function of x only, then there exists an integrating
factor (I.F) dxxP
exF)(
)( .
Theorem 2: If )(2 yQM
MNR
yx
, a function of y only, then there exists an integrating
factor (I.F) dyyQ
eyG)(
)( .
To solve the DE, multiply it by the I.F )()( yGorxF (whichever exists) and the
equation then becomes exact and can be solved as an exact DE.
For instance, the I.F 2
1
x of the DE: 0 dyxdxy in Example 4 above is calculated
as follow:
Since )(211
xPxxN
NM xy
, a function of x only, then the I.F is
2
)( 1)(
2
xeexF
dxdxxPx
as given above.
By multiplying the DE above by 2
1)(
xxF , the DE becomes 0
12
dyx
dxx
y.
Now, x
Nx
yM
1,
2
, hence xy N
xM
2
1 and the DE is now exact and
)()( xgdyNyhdxMu . From which we get: )()( xgx
yyh
x
yu
Then 0)()( xgyh and the solution is then cx
y or cxy .
Note also )(211
yQyyM
MN yx
a function of y only , so another I.F
2
)( 1)(
2
yeeyG
dydyyQ y
exists.
Multiply the DE 0 dyxdxy by 2
1)(
yyG , then the DE becomes 0
12
dyy
xdx
y
which is now exact. Compute
)()(1
2xgdy
y
xyhdx
yu , we obtain
)()( xgy
xyh
y
xu
, hence 0)()( xgyh and the solution is then
c
y
x
or
cxy .
Remark: The example above showed the DE has two integrating factors. This is not
always true.
Example 5: Solve the DE dyyxdxxy sin)4cos2( 2
Solution: The DE is not exact since xy NyyM sinsin2
Note that ),(4cos2
sin22 yxQ
xy
y
M
MNR
yx
is a function of x & y and not y
only. So no integrating factor (I.F) of the form dyyQ
eyG)(
)( exists.
On the other hand, )(1
sin
sin1 xP
xyx
y
N
NMR
xy
a function of x only and an
integrating factor I.F xexFdx
x 1
)( exists. So, the DE becomes
0sin)4cos2( 23 dyyxdxxyx and it is exact. To solve it, compute:
)(cos)4cos2(),( 423 yhxyxdxxyxyxu and find h(y).
yxNyhyxy
usin)(sin 22
. Then 0)(,0)( yhhenceyh and the
solution is then .cos 42 cxyx
Section 1.6: Linear DE and Bernoulli DE
A DE is linear if it is linear in the dependent variable and its derivatives.
A 1st-order linear DE can be written in the form )()( xfyxp
dx
dy
If 0)( xf , the DE is called homogeneous, otherwise it is nonhomogeneous.
Special Cases of the 1st-order Linear DE
Case 1: 0)( xp
Then )(xfdx
dy is separable and the general solution is then cdxxfy )(
Case 2: 0)( xf (homogeneous linear DE)
Then 0)( yxpdx
dy is separable its general solution is then dxxp
h cey)(
where hy denotes homogeneous solution.
Case 3: The nonhomogeneous case with 0)( xp
The equation is not separable and can be written as 0))()(( dydxxfyxp . This
equation is not exact either since xy NxpM 0)( . But it can be made exact by
multiplying it by the integrating factor dxxp
eFI)(
. (as in Section 1.5) and the
linear DE becomes )()()()()(
xfeyxpedx
dye
dxxpdxxpdxxp and it is simplified to
)()()(
xfeyedx
d dxxpdxxp
. Hence the general solution to )()( xfyxpdx
dy
becomes
cdxxfey
dxxpedxxp
g )()()(
Or
ph
dxxpdxxpg
yy
dxxfeceydxxp
e
)(
)()()(
Example 1: Solve xeydx
dy 22
Solution: This DE is linear with xexfxp 2)(&2)( . The integrating factor is then
xdx
ee 22 , hence the solution is xxxxxx
g xecedxeeecey 222222 .
Example 2: Solve the IVP: 1)0(,2sintan yxyxdx
dy
Solution: The I.F is xxdx
exF sectan
)( .
Hence, xcx
cyh cos
sec , xxdxx
xy p
2cos22sinsecsec
1 and
xxcyg2cos2cos is the general solution. Substituting x=0 and y=1, then c=3
and we have the unique solution )cos23(cos xxy
Example 3: Mixing Problem
A tank contains 1000 gallon of water in which 200 lb of salt are dissolved. Fifty
gallons of brine, each containing (1+cos t) lb of dissolved salt, run into the tank per
minute, and the mixture, kept uniform by stirring, runs out of the tank at the same
rate. Find the amount of salt y(t) in the tank at any time.
Solution: As in Example 2 of Section 1,
Inflow rate= 50*(1+cos t) lb/min, Outflow rate= 50* y(t)/1000 = 0.05 y(t) 1b/min
The IVP becomes 200)0(,05.0)cos1(50 yytdt
dy or
200)0(),cos1(5005.0 ytydt
dy.
With this, we have: p(t)=0.05, f(t)=50(1+cos t)
and the solution is then
)cos50( 05.005.0 tdtecey tt ttce t cos
2)05.0(1
5.2sin
2)05.0(1
50100005.0
With y(0)=200, then 5.8022)05.0(1
5.21000200
c and
tetty 05.05.802cos494.2sin88.491000
Notes: (see the graph of y in the book)
(1) The solution is exponentialy increasing.
(2) termsineesomeyt
cos&sin1000lim
. So, 1000 Ib is the mean value as
time incereases.
(3) This mean value 1000 is obtained if 50(1+cost) is replaced by 50 only. In this case the
equation becomes 200)0(,5005.0 yydt
dyand its solution is then tey 05.08001000
Bernoulli Equation (Reduction to Linear DE)
A 1st –order Bernoulli DE of degree r is ryxfyxpy )()( , where r is any real
number. If r=0 or 1 then this equation is linear or separable, respectively. If
r ≠ 0 & r ≠ 1, then y = 0 is atrivail solution in this case. So suppose r ≠ 0 & r ≠ 1 and
find the non trivial general solution.
Solution Method:
Put: uyyyu rr 1
uyr
yyyru rr
1
1)1(
Substituting in Bernoulli equation we obtain: rrr yxfuyxpuyr
)()(1
1
Simplifying, we obtain the Linear DE in u: )()1()()1( xfruxpru
We then solve this linear DE for u then for y.
Example 4: Solve (Logistic Population Model) 2ByAydt
dy ,
where A & B are positive constants.
Solution: This is Bernoulli DE of degree r=2. So BAuuyyu 121
The solution is then
A
BceBdteeceu AtAtAtAt
and the general solution for y is then .)/(
1
ABcey
At
This eqn is called the
logistic law of population growth. If B= 0, we obtain the exponential growth
Atec
1. The term 2By is “ a breaking term ” preventing the population from
growing without bound. For small initial population (B
Ay 00 ), the
population increases monotone to A/B. For large initial population (B
Ay 0 ),
the population decreases monotone to A/B.
See Fig.18 page 38 for the graph of y(t) .
Input and Output:
The linear equation reprsents a mathematical model for any system whose input and
output are related by a linear 1st-order DE.
The function f is the forcing input to the system, say a voltage generator for instance.
The solution y is the output response, say the current.
Total Output= response to the initial data (yh) + response to the input (yp)
phdxxpdxxp
g yydxxfeceydxxp
e )()()()(
Section 1.7: Modeling: Electric Circuits
Basic Elements of an Electric Circuit
The simplest electric circuit is a series circuit containing a source of electric energy
(electromotive force) such as a generator or a battery and a resistor which
consumes energy, for instance an electric lightbulb.
When the switch closes, a current I will flow through the resistor and this will
cause a voltage drop, i.e., the electric potential at the two ends of the resistor will
be different. The basic electric elements we discuss here are: Resistors, Inductance
and Capacitors.
Voltage Drop Across the Basic Electric Elements
Name Symbol Notation Unit Voltage Drop
Kirchhoff’s Voltage Law
The algebraic sum of all the instantaneous voltage drops around any closed loop is
zero.
Example 1: RL- Series Circuit
Find the current in the given RL- series circuit for
(a) A constant source(electromotive force) E0
(b) A periodic electromotive force wtEE sin0
Solution: By Kirchhoff\s voltage law: 0v
So, 0)( tERIdt
dIL or 0)0(,
)(II
L
tEI
L
R
dt
dI
This is a linear DE withL
tEtf
L
Rtp
)()(&)(
Hence, the solution is dtL
tEeecetI tLRtLRtLR )(
)( )/()/()/(
Define L
R and
R
LL
1 (the inductive time constant)
(a) With E(t)=E0 (constant)
Then R
EcetI t 0)( where
R
EIc 0
0 . If I0=0, then R
Ec 0 and the solution is
)1()( 0 teR
EtI and the current increases exponentially from 0 to
R
E0 .
Note that R
EtI
t
0)(lim
, i.e., as if L=0.
(b) With E(t)=E0 sin wt (periodic)
Note that:
wtwwt
L
R
wLR
eLwtwwt
w
edtwte
ttt cossin)cossinsin
222
2
22
Hence, )cossin()(222
0 wtLwwtRwLR
EcetI t
Define R
wLtan , then
222222222222
cossincos&
sincossin
wLR
wt
wLR
wtwL
wLR
wt
wLR
wtR
Hence )sin()(222
0
wtwLR
EcetI t
Definition: An electrical system is said to be in steady-state if the variables describing
the behaviour are either periodic or constant. Otherwise, transient.
So, in case(a), the solution R
E0 is steady-state.
In case (b), )sin(222
0
wtwLR
E is in steady-state and te is a transient
solution (it lasts for a short time.)
Example 2: RC- Series Circuit
Find the current in the following RC-series circuit for
(a) A constant source(electromotive force) E0
(b) A periodic electromotive force wtEE sin0
Solution: By Kirchhoff\s voltage law: 0v
So, 0)()(1
tERIdttIC
or 0)0(,11
IIdt
dE
RI
RCdt
dI
This is linear DE withdt
dE
Rtf
RCtp
1)(&
1)(
Hence, the solution is dtdt
dE
ReecetI tRCtRCtRC 1
)( )/1()/1()/1(
Define RC
1 and RCC
1 (the capacitive time constant)
(a) With E(t)=E0 (constant)
Then tcetI )( where 0Ic . If I0=0, then the solution is .0)( tI
Note that 0)(lim
tIt
, i.e., as if the circuit is open.
(b) With E(t)=E0 sin wt (periodic)
Then wtwEdt
dEcos0
and the solution is wtdtee
R
wEcetI CCC ttt
cos)(//0/
Note That
wt
RCwtw
RCw
eRCwtwtw
w
edtwte
ttt cos
1sin
)(1
)()cossincos
2
2
22
Hence, )cos1
sin()(1
)(2
20/ wt
RCwtw
RCw
wRCEcetI RCt
Define wRC
1tan , then
2222)(1
cossin
)(1
cos&
)(1
sincos
)(1
sin
wRC
wt
wRC
wt
wRC
wt
wRC
wtwRC
And the general solution is then )sin()(1
)(2
0/
wt
wRC
wCEcetI Ct
= transient solution + steady-state solution