Example. Solve the differential equation2(1 )(1 ) 0y x y′− + + =

Example. Solve the differential equation2(1 )(1 ) 0y x y′− + + =

Solution: This equation is separable, and by separating the variables, we get 2(1 )(1 ) so (1 )

21

dy dyx y x dxdx y

= + + = ++

Example. Solve the differential equation2(1 )(1 ) 0y x y′− + + =

Solution: This equation is separable, and by separating the variables, we get

thus

2tan

2xy x C

= + +

so

21(1 ) or tan ( )2 2(1 )

dy xx dx y x Cy

−= + = + +∫ ∫+

2(1 )(1 ) so (1 )21

dy dyx y x dxdx y

= + + = ++

Example. Solve the differential equation

3tan sec 0dyy xdx

− =

Example. Solve the differential equation

3tan sec 0dyy xdx

− =

Solution: 3tan sec 0dyy xdx

− = is equivalent with

sec 3tan or cot 3cosdy x y ydy xdxdx

= =

Example. Solve the differential equation

3tan sec 0dyy xdx

− =

Solution:

thus

so

3tan sec 0dyy xdx

− = is equivalent with

sec 3tan or cot 3cosdy x y ydy xdxdx

= =

ln sin 3siny x C= +

3sinsin xy Ce=

This is an implicit solution.

3sinsin xy e=

A first order equation is linear if it can be expressed in the following way:

where p(x) and q(x) are continuous functions.

( ) ( )dy p x y q xdx

+ =

A first order equation is linear if it can be expressed in the following way:

where p(x) and q(x) are continuous functions.

( ) ( )dy p x y q xdx

+ =

Examples:

2dy xx y edx

+ = 3(sin ) 0dy x y xdx

+ + = 5 2dy ydx

+ =

To solve a linear first order equation, we define a function µ(x) by

( )( ) p x dxx eµ ∫=

To solve a linear first order equation, we define a function µ(x) by

( )( ) p x dxx eµ ∫=

Then ( )( ) ( )p x dxd p x e p xdxµ µ∫= =

By the product formula and chain rule, we have

To solve a linear first order equation, we define a function µ(x) by

( )( ) p x dxx eµ ∫=

Then ( )( ) ( )p x dxd p x e p xdxµ µ∫= =

By the product formula and chain rule, we have

( ) ( ) ( ) (1)dyd y dy d dyy p x y p x ydxdx dx dx dx

µ µµ µ µ µ = + = + = +

To solve a linear first order equation, we define a function µ(x) by

( )( ) p x dxx eµ ∫=

Then ( )( ) ( )p x dxd p x e p xdxµ µ∫= =

By the product formula and chain rule, we have

( ) ( ) ( ) (1)dyd y dy d dyy p x y p x ydxdx dx dx dx

µ µµ µ µ µ = + = + = +

( ) ( )dy p x y q xdx

+ =Now multiply the DE by µ. We get

( ) ( )dy p x y q xdx

µ µ + =

or by (1), ( ) ( )d y q xdxµ µ=

( ) ( )d y q xdxµ µ=Since we can integrate both sides to get:

1( ) ( ) or ( ) ( )y x q x dx C y x q x dx Cµ µ µµ

= + = + ∫∫ To summarize, we have

( ) ( )d y q xdxµ µ=Since we can integrate both sides to get:

1( ) ( ) or ( ) ( )y x q x dx C y x q x dx Cµ µ µµ

= + = + ∫∫ To summarize, we have

The Method of Integrating FactorsStep 1. Calculate

This is called the integrating factor. We can take the constant of integration to be 1.

Step 2. Write

and perform the required integration, including a constant of integration.

Step 3. Solve for y.

( ) ( )y x q x dxµ µ=∫

( )( ) p x dxx eµ ∫=

Example. Solve the DE 4 1dy y

dx− =

Example. Solve the DE 4 1dy y

dx− =

Solution: Here p(x) = − 4 and q(x) = 1.

Then 4 4( ) dx xx e eµ −∫ −= =

Example. Solve the DE 4 1dy y

dx− =

Solution: Here p(x) = − 4 and q(x) = 1.

Then 4 4( ) dx xx e eµ −∫ −= =

14 4( ) ( )4

x xy x q x dx e dx e Cµ µ − −= = =− +∫ ∫

Example. Solve the DE 4 1dy y

dx− =

Solution: Here p(x) = − 4 and q(x) = 1.

Then 4 4( ) dx xx e eµ −∫ −= =

14 4( ) ( )4

x xy x q x dx e dx e Cµ µ − −= = =− +∫ ∫

1 11 14 44 44 4 4

x xx xy e C e e C Ceµ − −= − + = − + =− +

Example. Solve the Initial Value Problem:

2 ; (0) 2dy xy x ydx

+ = =

Example. Solve the Initial Value Problem:

2 ; (0) 2dy xy x ydx

+ = =

Solution: Here p(x) = 2x and q(x) =x.

Then 22( ) xdx xx e eµ ∫= =

Example. Solve the Initial Value Problem:

2 ; (0) 2dy xy x ydx

+ = =

Solution: Here p(x) = 2x and q(x) =x.

Then 22( ) xdx xx e eµ ∫= =2 21( ) ( )

2x xy x q x dx xe dx e Cµ µ= = = +∫ ∫

Example. Solve the Initial Value Problem:

2 ; (0) 2dy xy x ydx

+ = =

Solution: Here p(x) = 2x and q(x) =x.

Then 22( ) xdx xx e eµ ∫= =2 21( ) ( )

2x xy x q x dx xe dx e Cµ µ= = = +∫ ∫

22 221 1 112 2 2

xx xxy e C e e C Ceµ

− − = + = + = +

Since y(0) = 2, we substitute y = 2, x = 0 and get

201 122 2

Ce C − = + = +

Since y(0) = 2, we substitute y = 2, x = 0 and get

201 122 2

Ce C − = + = +

Thus we have C = 3/2, and so

21 32 2

xy e

− = +

Example. Solve the DE: 1 0

1

dy yxdx e

+ − =+

Example. Solve the DE: 1 0

1

dy yxdx e

+ − =+

Solution: Here p(x) = 1 and q(x) =1/(1 + ex).

Then ( ) dx xx e eµ ∫= =

Example. Solve the DE: 1 0

1

dy yxdx e

+ − =+

Solution: Here p(x) = 1 and q(x) =1/(1 + ex).

Then ( ) dx xx e eµ ∫= =

( )( ) ( ) ln 11

xe xy x q x dx dx e Cxeµ µ = = = + +∫ ∫

+

Example. Solve the DE: 1 0

1

dy yxdx e

+ − =+

Solution: Here p(x) = 1 and q(x) =1/(1 + ex).

Then ( ) dx xx e eµ ∫= =

( )( ) ( ) ln 11

xe xy x q x dx dx e Cxeµ µ = = = + +∫ ∫

+

( ) ( ) ( )1 ln 1 ln 1 ln 1x x x x xxy e C e e C e e Ceµ

− − −= + + = + + = + +

Example. A tank initially contains 200 gal of pure water. Then at time t = 0, brine containing 5 lb of salt per gallon of brine is allowed to enter the tank at the rate of 10 gal/min andthe mixed solution is drained from the tank at the same rate.

(a) How much salt is in the tank at an arbitrary time?(b) How much salt is in the tank after 30 minutes?

Brine in at 10 gal/min

Mixture out at 10 gal/min

200 gal

Brine in at 10 gal/min

Mixture out at 10 gal/minLet Q(t) be the number of lbs of salt in the tank at

time t.Then salt runs in at the rate of (5 lb/gal)(10 gal/min) = 50 lb/min.Salt is running out at the rate of (Q(t) lb/200 gal)(10 gal/min) = Q /20 lb/min.at any time t.

200 gal

Brine in at 10 gal/min

Mixture out at 10 gal/minLet Q(t) be the number of lbs of salt in the tank at

time t.Then salt runs in at the rate of (5 lb/gal)(10 gal/min) = 50 lb/min.Salt is running out at the rate of (Q(t) lb/200 gal)(10 gal/min) = Q /20 lb/min.at any time t. This means that Q satisfies the initial value problem:

50 ; (0) 020

dQ Q Qdt

= − =

200 gal

Brine in at 10 gal/min

Mixture out at 10 gal/min

5020

dQ Qdt

= −The equation is linear, so we can use the method

of integrating factors; p(t) = 1/20 and q(t) = 50.Thus

20 20( )dt t

t e eµ ∫= =

200 gal

Brine in at 10 gal/min

Mixture out at 10 gal/min

5020

dQ Qdt

= −The equation is linear, so we can use the method

of integrating factors; p(t) = 1/20 and q(t) = 50.Thus

20 20( )dt t

t e eµ ∫= = 20 20( ) ( ) 50 1000t t

t q t dt e dt e Cµ = = +∫ ∫

200 gal

20( ) ( ) 1000t

t Q t e Cµ = +Thus

20 20 20 201( ) [1000 ] [1000 ] 1000t t t t

Q t e C e e C Ceµ

− −= + = + = +

20( ) ( ) 1000t

t Q t e Cµ = +Thus

20 20 20 201( ) [1000 ] [1000 ] 1000t t t t

Q t e C e e C Ceµ

− −= + = + = +

We also know that Q(0) = 0, so 0 = 1000 + C, and C = −1000.

Thus

20( ) 1000[1 ]t

Q t e−

= −

20( ) ( ) 1000t

t Q t e Cµ = +Thus

20 20 20 201( ) [1000 ] [1000 ] 1000t t t t

Q t e C e e C Ceµ

− −= + = + = +

We also know that Q(0) = 0, so 0 = 1000 + C, and C = −1000.

Thus

20( ) 1000[1 ]t

Q t e−

= −

After 30 minutes, we have

3020(30) 1000[1 ] 776.86Q e

−= − ≈ lbs of salt.

20( ) ( ) 1000t

t Q t e Cµ = +Thus

Example. A tank with 1000 gal capacity initially contains 500 gal of water that is polluted with 50 lb. of particulate matter.At time t = 0, pure water is added at a rate of 20 gal/min and the mixed solution is drained off at a rate of 10 gal/min. How much particulate matter is in the tank when it reaches the pointof overflowing?

Water in at 20 gal/min

Mixture out at 10 gal/min

500 gal

Water in at 20 gal/min

Mixture out at 10 gal/min

500 gal

Let Q(t) be the number of lbs of particulate matter in the tank at time t, and let V(t) be the volume of the solution at time t. Clearly V(t) = 500 + 10t.

Water in at 20 gal/min

Mixture out at 10 gal/min

500 gal

Let Q(t) be the number of lbs of particulate matter in the tank at time t, and let V(t) be the volume of the solution at time t. Clearly V(t) = 500 + 10t.Particulate matter is running out at the rate of (Q(t) lb/V(t) gal)(10 gal/min) = 10Q /V lb/min.at any time t. This means that Q satisfies the initial value problem:

10 ; (0) 5050

dQ Q Q Qdt V t

=− =− =+

50dQ Qdt t

=−+

For the linear equation we have:

1( ) ; ( ) 050

p t q tt

= =+ ( )( ) exp exp ln(50 ) 50

50dtt t t

tµ = = + = +∫ +

50dQ Qdt t

=−+

For the linear equation we have:

1( ) ; ( ) 050

p t q tt

= =+ ( )( ) exp exp ln(50 ) 50

50dtt t t

tµ = = + = +∫ + ( ) ( ) 0t q t dt dt Cµ = =∫ ∫

Thus we have 1( )50

CQ t Ctµ

= =+

50dQ Qdt t

=−+

For the linear equation we have:

1( ) ; ( ) 050

p t q tt

= =+ ( )( ) exp exp ln(50 ) 50

50dtt t t

tµ = = + = +∫ + ( ) ( ) 0t q t dt dt Cµ = =∫ ∫

Thus we have 1( )50

CQ t Ctµ

= =+

Since Q(0) = 50, we have 50 = C/50, or C = 2500. Thus

2500( )50

Q tt

=+

50dQ Qdt t

=−+

For the linear equation we have:

1( ) ; ( ) 050

p t q tt

= =+ ( )( ) exp exp ln(50 ) 50

50dtt t t

tµ = = + = +∫ + ( ) ( ) 0t q t dt dt Cµ = =∫ ∫

Thus we have 1( )50

CQ t Ctµ

= =+

Since Q(0) = 50, we have 50 = C/50, or C = 2500. Thus

2500( )50

Q tt

=+

The tank fills in 50 minutes, at which time the amount of matter is 2500(50) 25 lbs

100Q = =

Example. An object has mass m and is dropped from a plane with an initial velocity of v0. We assume that as it falls through the air, it experiences a drag proportional to its velocity. Describe its motion.

Example. An object has mass m and is dropped from a plane with an initial velocity of v0. We assume that as it falls through the air, it experiences a drag proportional to its velocity. Describe its motion.

Solution. The force on the object at any time t (if we take up as the positive direction) is −mg − cv(t), where g is the acceleration dur to gravity and c is the ‘drag coefficient’. By Newton’s second law, F = ma, that is

( ) dvmg cv t mdt

− − =

Thus we see that the velocity satisfies the linear DE

Example. An object has mass m and is dropped from a plane with an initial velocity of v0. We assume that as it falls through the air, it experiences a drag proportional to its velocity. Describe its motion.

Solution. The force on the object at any time t (if we take up as the positive direction) is −mg − cv(t), where g is the acceleration dur to gravity and c is the ‘drag coefficient’. By Newton’s second law, F = ma, that is

( ) dvmg cv t mdt

− − =

Thus we see that the velocity satisfies the linear DE

( )dv c v t gdt m

+ =−

In this case, we also have the initial condition v(0) = v0.

For the equation we see that ( )dv c v t gdt m

+ =− ( ) , ( ) .cp t q t gm

= =−

In this case, we also have the initial condition v(0) = v0.

For the equation we see that ( )dv c v t gdt m

+ =− ( ) , ( ) .cp t q t gm

= =−

Thus

( ) ; ( ) ( )ctc ct ctdt mm m mgmt e e t q t dt ge dt e C

cµ µ∫ −= = = − = +∫ ∫

In this case, we also have the initial condition v(0) = v0.

For the equation we see that ( )dv c v t gdt m

+ =− ( ) , ( ) .cp t q t gm

= =−

Thus

( ) ; ( ) ( )ctc ct ctdt mm m mgmt e e t q t dt ge dt e C

cµ µ∫ −= = = − = +∫ ∫

ct ctm mgmve e C

c−= + so ( )

ctct ctmm m mgmgv t e e C Ce

cc

−− −− = + = +

In this case, we also have the initial condition v(0) = v0.

For the equation we see that ( )dv c v t gdt m

+ =− ( ) , ( ) .cp t q t gm

= =−

Thus

( ) ; ( ) ( )ctc ct ctdt mm m mgmt e e t q t dt ge dt e C

cµ µ∫ −= = = − = +∫ ∫

ct ctm mgmve e C

c−= + so ( )

ctct ctmm m mgmgv t e e C Ce

cc

−− −− = + = +

0mgv Cc

− = +

so .0

mgC vc

= + Thus

( ) 0

ctmg mgmv t e vc c

− = + −

( ) 0

ctmg mgmv t e vc c

− = + −

Notice that as t increases, the first term tends to 0, and the object tends to ‘terminal velocity’ −mg/c.

( ) 0

ctmg mgmv t e vc c

− = + −

Notice that as t increases, the first term tends to 0, and the object tends to ‘terminal velocity’ −mg/c.

For a 240 lb human, terminal velocity is about 120 ft/sec.