First Order Linear Differential Equations

Any equation containing a derivative is called a differential equation.

A function which satisfies the equation is called a solution to the differential equation.

The order of the differential equation is the order of the highest derivative involved.

The degree of the differential equation is the degree of the power of the highest derivative involved.

)(yfdx

dy form the of is equation aldifferenti a When

dxdyyf

)(

1We can then integrate both sides.

dxyf

dy

)(This will obtain the general solution.

When the equation is of the form ( ) ( ), thendy

f x g ydx

1( )

( )dy f x dx

g y

( )( )dy

f x dxg y

A first order linear differential equation is an equation of the form

( ) ( )dy

P x y Q xdx

1

To find a method for solving this equation, lets consider the simpler equation

( ) 0dy

P x ydx

Which can be solved by separating the variables.

(Remember: a solution is of the form y = f(x) )

( ) 0dy

P x ydx

( )dy

P x ydx

( )dy

P x dxy

ln ( )y P x dx c

( )P x dx cy e

( )P x dx cy e e

( )P x dx

y Ce

or( )P x dx

ye C

Using the product rule to differentiate the LHS we get:

( )P x dxdye

dx

( ) ( )( )

P x dx P x dxdye yP x e

dx

( )( )

P x dxdyP x y e

dx

Returning to equation 1, ( ) ( )dy

P x y Q xdx

If we multiply both sides by ( )P x dx

e

( ) ( )( ) ( )

P x dx P x dxdyP x y e Q x e

dx

( ) ( )( )

P x dx P x dxdye Q x e

dx Now integrate both sides.

( ) ( )( )

P x dx P x dxye Q x e dx

For this to work we need to be able to find ( )

( ) and ( )P x dx

P x dx Q x e dx

2Solve the differential equation

dyy x

dx x

2Step 1: Comparing with equation 1, we have ( )P x

x

2( )P x dx dx

x 2ln x 2

ln x 2ln x

2( ) lnStep 2: P x dx xe e 2x ( )

is called the integrating factorP x dx

e

Step 3: Multiply both sides by the integrating factor.

2 22dyx y x xdx x

2 3dyx x

dx

2 3dyx x

dx

2 3yx x dx4

2

4x

yx c

2 214

y x cx

Note the solution is composed of two parts.

21The particular solution: and

4y x 2the complementary function: y cx

The particular solution satisfies the equation:2

( )dy

y Q xdx x

The complementary function satisfies the equation:2

0dy

ydx x

Solve the differential equation 2 3dy

xy xdx

Step 1: Comparing with equation 1, we have ( ) 2 and ( ) 3P x x Q x x

( ) 2P x dx xdx 2x

2( )Step 2: Integrating Factor

P x dx xe e

Step 3: Multiply both sides by the integrating factor.

2 2

2 3x xdye xy e x

dx

2 2

3x xdye xe

dx

2 2

3x xdye xe

dx

2 2

3x xye xe dx To solve this integration we need to use substitution.

2Let t x 2dt xdx

2 33

2x txe dx e dt

3

2te

23

2xe

2 232

x xye e c

232

xy ce

If you look at page 113 you will see a note regarding the constant when finding the integrating factor. We need not find it as it cancels out when we multiply both sides of the equation.

2 3Solve the differential equation 0dy

x x xydx

1Step 1: Comparing with equation 1, we have ( ) and ( )P x Q x x

x

( )dx

P x dxx

ln x

( ) lnStep 2: Integrating Factor P x dx xe e x

Step 3: Multiply both sides by the integrating factor.

2dyx x

dx

1rewriting in standard form:

dyy x

dx x

(note the shortcut I have taken here)

2yx x dx2 11

3y x cx

The modulus vanishes as we will have either both positive on either side or both negative. Their effect is cancelled.

2Solve the differential equation cos sin cosdyx y x xdx

Step 1: Comparing with equation 1, we have ( ) tan and ( ) cosP x x Q x x

( ) tanP x dx xdx ln sec x( ) ln secStep 2: Integrating Factor secP x dx xe e x

Step 3: Multiply both sides by the integrating factor.

sec 1dy x

dx

rewriting in standard form: tan cosdy

y x xdx

(note the shortcut I have taken here)

sec 1y x dxcos cosy x x c x

The modulus vanishes as we will have either both positive on either side or both negative. Their effect is cancelled.

So far we have looked at the general solutions only. They represent a whole family of curves. If one point can be found which lies on the desired curve, then a unique solution can be identified.

Such a point is often given and its coordinates are referred to as the initial conditions or values. The unique solution is called a particular solution.

2Solve the differential equation given 1 when 0dyx y x x ydx

1Step 1: Comparing with equation 1, we have ( )P x

x

1( )P x dx dx

x ln x

( ) ln 1Step 2: Integrating Factor

P x dx xe ex

Step 3: Multiply both sides by the integrating factor.

1d ydx x

1rewriting in standard form:

dyy x

dx x

1y

dxx

2y x cx

2y x cx given 1 when 0x y

0 1 c 1c

Hence the particular solution is 2y x x

Page 114 and 116 Exercise 1 and 2

TJ Exercise 1

Second-order linear differential equations

2

2Differential equations of the form ( )

are called second order linear differential equations.

d y dya b cy Q xdx dx

When ( ) 0 then the equations are referred to as homogeneous,Q x

When ( ) 0 then the equations are non-homogeneous.Q x

Note that the general solution to such an equation must include two arbitrary constants to be completely general.

Theorem

If ( ) and ( ) are two solutions then so is ( ) ( )y f x y g x y f x g x

Proof

2 2

2 2 0d f df d g dga b cf a b cgdx dx dx dx

2

2and 0d g dga b cgdx dx

Adding:

2

2we have 0d f dfa b cfdx dx

2 2

2 2 0d f d g df dg

a b c f gdx dx dx dx

And so ( ) ( ) is a solution to the differential equation. y f x g x

, for and , is a solution to the equation 0mx dyy Ae A m b cy

dx

It is reasonable to consider it as a possible solution for2

2 0d y dya b cydx dx

mxy Ae mxdyAme

dx

22

2mxd y

Am edx

If is a solution it must satisfymxy Ae 2 0 mx mx mxaAm e bAme cAe

assuming 0, then by division we getmxAe 2 0am bm c

The solutions to this quadratic will provide two values of m which will make y = Aemx a solution.

If we call these two values m1 and m2, then we have two solutions.

1m xy Ae and 2m xy Be

A and B are used to distinguish the two arbitrary constants.

From the theorem given previously;

1 2m x m xy Ae Be Is a solution.

The two arbitrary constants needed for second order differential equations ensure all solutions are covered.

2The equation 0 is called the auxiliary equation.am bm c

The type of solution we get depends on the nature of the roots of this equation.

When roots are real and distinct

2

2Find the general solution of 5 6 0.d y dy

ydx dx

The auxiliary equation is 2 5 6 0m m

( 2)( 3) 0m m 2, 3m or m

2 3Thus the general solution is .x xy Ae Be

To find a particular solution we must be given enough information.

2

2Find the particular solution to 2 7 4 0,

given 1 and 2 when 0.

d y dyy

dx dxdy

y xdx

The auxiliary equation is 22 7 4 0m m

(2 1)( 4) 0m m 1

, 42

m or m 1

42Thus the general solution is .x xy Ae Be

Using the initial conditions.0 01 Ae Be 1 A B

1421

Now 42

x xdyAe Be

dx 0 01

2 42Ae Be

12 4

2A B

1 A B

12 4

2A B

Solving gives4 1

,3 3

A B

Thus the particular solution is

1424 1

3 3

x xy e e

Page 119 Exercise 3 Questions 1(a), (b) 2(a), (b)

But we will now look at the solution when the roots are real and coincident.

Roots are real and coincident

When the roots of the auxiliary equation are both real and equal to m, then the solution would appear to be y = Aemx + Bemx = (A+B)emx

A + B however is equivalent to a single constant and second order equations need two.

With a little further searching we find that y = Bxemx is a solution. Proof on p119

So a general solution is

mx mxy Ae Bxe

Find the general solution of y'' 6 ' 9 0.y y

The auxiliary equation is 2 6 9 0m m

( 3)( 3) 0m m 3 (twice)m

3 3Thus the general solution is .x xy Ae Bxe

Find the particular solution to 9 ''- 6 ' 0,

given 1 when 0 and 4 when 3.

y y y

y x y e x

The auxiliary equation is 29 6 1 0m m (3 1)(3 1) 0m m

1(twice)

3m

1 13 3Thus the general solution is .x x

y Ae Bxe

Using the initial conditions.0 01 0Ae B e 1 A

4 3e e Be 1 B 1 1Now 4 3e Ae Be

1 13 3x x

y e xe 13 1x

e x

Page 120 Exercise 4 Questions 1(a), (b) 2(a), (b)

But we will now look at the solution when the roots are complex conjugates.

Roots are complex conjugatesWhen the roots of the auxiliary equation are complex, they will be of the formm1 = p + iq and m2 = p – iq. Hence the general equation will be

( ) ( )p iq x p iq xy Ae Be px iqx px iqxAe e Be e

px iqx iqxe Ae Be We know that cos sinie i

cos sin cos( ) sin( )pxe A qx i qx B qx i qx

cos sin cos sinpxe A qx i qx B qx i qx

cos sinpxe A B qx A B i qx

cos sinpxe C qx D qx

Where and ( )C A B D A B i

2

2Find the solution to 6 13 0.d y dy

ydx dx

The auxiliary equation is 2 6 13 0m m

3 2m i

3Thus the general solution is cos 2 sin 2xy e C x D x

6 36 522

m

6 162

2

2Find the particular solution to 4 13 0,

given that 2 and 0 when 0.

d y dyy

dx dxdy

y xdx

The auxiliary equation is 2 4 13 0m m

2 3m i

2Thus the general solution is cos3 sin3xy e C x D x

4 16 522

m

4 362

Substituting gives: 2 cos0 sin 0C D 2C

2 22 cos3 sin3 3 cos3 3 sin3x xdye C x D x e D x C x

dx

Substituting gives: 0 2 2cos0 sin 0 3 cos0 6sin 0D D

0 4 3D 43

D

2 22 cos3 sin3 3 cos3 3 sin3x xdye C x D x e D x C x

dx

The particular solution is2 4

2cos3 sin33

xy e x x

Page 122 Exercise 5A Questions 1(a), (b) 2(a), (b)

TJ Exercise 2

Non homogeneoussecond order differential equations

2

2 ( )d g dga b cg Q xdx dx

Non homogeneous equations take the form

2

2 ( )d y dya b cy Q xdx dx

Suppose g(x) is a particular solution to this equation. Then

Now suppose that g(x) + k(x) is another solution. Then

2

2

( ) ( )( ) ( )

d g k d g ka b c g k Q x

dx dx

Giving

2 2

2 2 ( )d g d k dg dka a b b cg ck Q xdx dx dx dx

2 2

2 2 ( )d g dg d k dka b cg a b ck Q xdx dx dx dx

2

2( ) ( )d k dk

Q x a b ck Q xdx dx

2

2 0d k dka b ckdx dx

From the work in previous exercises we know how to find k(x).

This function is referred to as the Complimentary Function. (CF)

The function g(x) is referred to as the Particular Integral. (PI)

General Solution = CF + PI

2

2Find the general solution to 5 6 15 7,

given that the PI is of the form ( )

d y dyy x

dx dxk x Px Q

Finding the (CF): the auxiliary equation is 2 5 6 0m m

2 3m or m 2 3Thus the CF is x xy Ae Be

( 3)( 2) 0m m

Finding the PI: y Px Q dyP

dx

2

2 0d ydx

Substituting into the original equation 0 5 6 15 7P Px Q x 6 6 5 15 7Px Q P x

6 15P 52

p

1112

Q

5 11Thus the PI

2 12x

Hence the general solution is 2 3 5 11

2 12x xy Ae Be x

2

2

2

Find the general solution to 4 6 2,

given that the PI is of the form ( )

d y dyx

dx dx

k x Px Qx

Finding the (CF): the auxiliary equation is 2 4 0m m

0 4m or m 4Thus the CF is xy A Be

( 4) 0m m

2Finding the PI: y Px Qx 2dy

Px Qdx

2

2 2d y

Pdx

Substituting into the original equation 2 4(2 ) 6 2P Px Q x 8 2 4 6 2Px P Q x

8 6P 34

p 78

Q

23 7Thus the PI

4 8x x

Hence the general solution is 4 23 7

4 8xy A Be x x

Finding the form of the PI2

2 ( ) is non homogeneous because ( ) 0d y dya b cy Q x Q xdx dx

The individual terms of the CF make the LHS zero

The PI makes the LHS equal to Q(x). Since Q(x) ≠ 0, it stands to reason that the PI cannot have the same form as the CF.

When choosing the form of the PI, we usually select the same form as Q(x).

This reasoning leads us to select the PI according to the following steps.

•try the same form as Q(x)•If this is the same form as a term of the CF, then try xQ(x)•If this is the same form as a term of the CF, then try x2Q(x)

3 2CF: ; ( ) 2 , for PI try a linear functionx xy Ae Be Q x x y Cx D

3 2 4 4CF: ; ( ) 2 , for PI try 2x x x xy Ae Be Q x e y e

3 2 3 3CF: ; ( ) 2 , for PI try (note extra factor)x x x xy Ae Be Q x e y Cxe x

4CF: ; ( ) 6 2, for PI try ( )

(note extra factor since A is a linear function)

xy A Be Q x x y x Cx D

x

2

2

CF: sin3 cos3 ; ( ) 3 1, for PI try

a quadratic function

y A x B x Q x x

y Cx Dx E

3 3CF: ; ( ) 3cos , for PI try sin cos , a wave functionx xy Ae Be Q x x y C x D x

CF: sin3 cos3 ; ( ) 3cos , for PI try sin cosy A x B x Q x x y Cx x Dx x

22

2Find the general solution to the equation 5 6 8 xd y dyy e

dx dx

Finding the (CF): the auxiliary equation is 2 5 6 0m m

2 3m or m 2 3Thus the CF is x xy Ae Be

( 2)( 3) 0m m

2 2Finding the PI. ( ) 8 so we might try x xQ x e Ce

22 2 2 2 2

22 4 4x x x x xdy d yy Cxe Ce Cxe Ce Cxe

dx dx

Substituting into the original equation

2 2 2 2 2 24 4 5( 2 ) 6 8x x x x x xCe Cxe Ce Cxe Cxe e

Equating Coefficients, we get 8C

2 2but this is the same form as the in the CF. So we try x xAe Cxe

2Hence the PI is 8 xy xe

2 3 2Thus the general solution is 8x x xy Ae Be xe

If the wrong selection of PI is made, you will generally be alerted to this by the occurrence of some contradiction in later work.

Page 126 – Exercise 7A as many as possible.

TJ Exercise 3. This assesses beyond poly and trig work.