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