Differential equation

B5001- Engineering Mathematics DIFFERENTIAL EQUATION

4.1 INTRODUCTION TO DIFFERENTIAL EQUATIONS

A differential equation is an equation which relates an unknown function of a
single variable with one or more of its derivatives.

Example:

dy
x y cos x 0
dx

Independent variable Dependent variable

Derivative of y wrt x

The order of a differential equation is the highest derivative involved in the
equation, and the degree is the power of the highest derivative in the
equation.

Example :

dy 1 x
a. is a DE of order 1 and degree 1.
dx 1 y2

dy
b. x 2 y sin x is a DE of order 1 and degree 1.
dx
d2y dy
c. 4 2y x 2 is a DE of order 2 and degree 1.
dx 2 dx
dy
d. xy xy 2 x is a DE of order 1 and degree 1.
dx
2
2 dy
e. y 1 x is a DE of order 1 and degree 2
dx
2
d2 y dy
f. 2 10y cos 2x is a DE of order 2 and degree 2.
dx2 dx

Sa`adiah Saad JMSK, POLIMAS Page 1


Example 1

State the dependent variable, the independent variable, the order and degree for each
differential equation.

ds s dm
i. ii. 2mn 0
dt t dn

2
dy 4 x ds
iii. iv. t2 sin t 0
dx x3 dt

dy dy
v. t2 1 yt vi. x xy y2 1
dt dx

d2y dy dv
vii. x 4 2 xy 0 viii. u3 1 uv2 u
dx 2 dx du

2
d2y dy d2y dy
ix. 3 2y 0 x. x 4 2 xy 0
dx 2 dx dx2 dx

Answer

Independent
Dependent Variable Order Degree
variable
i. s t 1 1
ii. m n 1 1
iii. y x 1 2
iv. s t 1 1
v. y t 1 1
vi. y x 1 1
vii. y x 2 1
viii. v u 1 1
ix. y x 2 2
x. y x 2 1



4.2 Formation of Differential Equations

If y 10x2 A ; Differential Equation is formed when the arbitrary constant A
is eliminated from this equation.
Example 2

i. If y 10x2 A

Differentiate with respect to x,

dy
20 x
dx

dy
20x 0 Differential equation.
dx

ii. If y x Ax 2 ------
Differentiate with respect to x, gives,
dy
1 2 Ax -----
dx
y x
From , A 2
x
y x
Subtituting A 2
into 
x
dy y x
Then, 1 2 2
x
dx x

y x
1 2
x
y
1 2 2
x
Multiply both sides with x.
dy
So, x x 2 y 2x
dx
dy
x 2y x
dx



iii. If y Ax 2 Bx x -----

Differentiate with respect to x, gives,

dy
2 Ax B 1 -----
dx

dy
Differentiate with respect to x.
dx

d2y
2A -----
dx 2

1 d2y
From  , A -----
2 dx 2

Substituting  into  to get B.

dy 1 d2y
2 x B 1
dx 2 dx2

d2y
x B 1
dx2

d2y dy
B x 1 -----
dx2 dx

Then, substitute  and  into .

1 d2y 2 d2y dy
y x x 2 1 x x
2 dx2 dx dx

1 2 d2y d2y dy
x x2 x x x
2 dx2 dx2 dx

1 2 d2y dy
x x
2 dx2 dx



EXERCISE 1

Form differential equation of the following:

i. y x3 Ax 2 iv. y 4Bx A

ii. y Ax 2 7x v. y A sin 2 x B kos x

iii. y Dx 2 Ex vi. y Ce x
2 De x

Answer

dy dy
i. x2 3x 4 2( y x3 ) iv x 2 y 7x
dx dx
dy x2 d 2 y d2y
ii. y x v. 0
dx 2 dx2 dx2

d2y d2y
iii. 4y 0 vi y 0
dx 2 dx 2

4.3 SOLUTIONS OF FIRST ORDER DIFFERENTIAL EQUATIONS (DE)

We already know that DE is one that contains differential coefficient.

dy
Example: i. 4x - 1st order DE
dx

d2 y dy
ii.
2
2 4y 0 - 2nd order DE
dx dx



dy
4.3.1 f(x) - Solve By Direct Integration; i.e. y f(x)dx
dx

dy
Example 3: Find the general solution of the DE 3x 2 sin2x
dx
Answer
dy cos2x
dx (3x 2 sin2x)dx so y = x3 c
dx 2

dy
Example 4: Find the particular solution of de 5 2x 3 , given the boundary
dx
2
conditions y 1 when x = 2 .
5
Answer

dy dy dy 3 2x 3 2x
5 2x 3 5 3 2x
dx dx dx 5 5 5

3 x2
Hence y x c
5 5

Substituting the boundary conditions;
7 3 22 7 6 4 7 6 4 5
(2) c c c= 1
5 5 5 5 5 5 5 5 5 5

3 x2
The particular solution, y x 1
5 5



4.3.2 Solve By Variable Separable;
dy dy
4.3.2.1 f(y) - rearranged to give dx and then solve by direct
dx f(y)
dy
integration; i.e. dx
f(y)

dy
Example 5: Find the general solution of de 5 sin2 3y .
dx
Answer

dy dy
5 2
dx dx = 5
sin 3y sin2 3y

cot3y
x = 5 cosec 2 3y dy = 5[- ] +c
3

5
i.e. x = - cot3y +c
3

dy
Example 6: Find the particular solution of de (y 2 1) 3y given that y = 1 when
dx

x 13
6
Answer

(y 2 1) 1 y2 1 1 1
dy dx dx = dy dx = (y ) dy
3y 3 y 3 y

1 y2
x= ( - ln y) + c
3 2

Substituting y=1 and x 13 ;
6


13 1 1 13 1
= ( - ln 1) + c c= - =2
6 3 2 6 6

1 2 1
The particular solution is x = y ln y 2
6 3

dy dy
4.3.2.2 f(x)g(y) - rearranged to give f(x)dx and then solve
dx g(y)

by direct integration

dy 2x3 1
Example 7: Solve
dx 3 2y
Answer
Separating the variables gives: (3 -2y)dy = (2x3 – 1)dx

x4
(3 2y)dy (2x 3 1)dx 3y - y 2 = x c
2

Example 8: The current in an electric circuit containing resistance R and inductance L
di
in series with a constant voltage source E is given by the de E L Ri .
dt
Solve the equation and find i in terms of time t given that when t = 0 and i
= 0.
Answer
di di 1
L E Ri dt
dt E-Ri L

Let u = E-Ri , du = -Rdi
1 du 1 1 1
- = dt - ln (E - Ri) = t c
R u L R L



When t = 0, i= 0;

1 1
- ln E = c ; c=- ln E
R R

1 1 1
- ln (E - Ri) = t ln E
R L R
1 1 t
= ln E - ln (E - Ri) =
R R L
Rt Rt Rt
E Rt E E-Ri
ln ( )= eL e L E-Ri = Ee L
E-Ri L E-Ri E
Rt
E -
i = ( 1-e L )
R

Example 9: Solve the Des:

dy 2x xy dy y(2 3x)
a. 2
d.
dx y 1 dx x(1 3y)

ds s2 6s 9 dy 6t 2 2t 1
b. e.
dt t 2 dt cos y ey
dy sec h y
c.
dx 2 x

4.3.3 Homogeneneous First Order DEs

dy
An equation of the form P = Q , where P and Q are function of both x and y of the
dx
same degree – said to be homogenous in y and x.

Example;



f(x,y) Homogeneous degree
1 x2 + 3xy + y2 yes 2
x 3y
2 yes 1
2x y

4 x 2 3y 2
3 yes 2
2 xy

x2 y y–1
4 no
2 x2 y2 x-2

dy
Procedure to solve DE of the form P =Q
dx
dy dy Q
i. Rearrange P = Q into the form =
dx dx P
dy dv
ii. Make the substitution y =vx, from = v (1) + x , by the product rule.
dx dx
dy dy Q
Substitute for both y and in the equation = . Simplify, by
iii. dx dx P
cancelling, and on equation result in which the variables are separable.
iv. Separate the variable and solve as direct integrating.
y
v. Substitute v = to solve in terms of the original variables.
x

dy
Example 10: Solve the DE y - x = x , given x = 1 when y = 2.
dx
Answer
dy y - x
i. Rearranging; =
dx x
dy dv
ii. Let y = vx, =v+x
dx dx

dy dv v x - x
iii. Substitute for both y and gives: v + x =
dx dx x



dv x(v - 1) dv
v+x = x v-1-v
dx x dx
dv dx
x -1 dv=- v = - ln x +c
dx x
y
= - ln x +c Subtitute x =1 and y = 2
x
2
c c=2
1
y
= - ln x + 2 or y = - x ( ln x - 2 )
x

dy x2 y2
Example 11: Find the particular solution of DE; x given the boundary
dx y
conditions that y = 4 when x = 1.
Answer

dy x2 y2 dy x2 y2
i. Rearranging; x
dx y dx yx
dy dv
ii. Let y = vx, =v+x
dx dx

dv x2 + v 2 x2 x2 (1 + v 2 ) (1 + v 2 )
iii. v+x =
dx vx2 vx2 v

dv (1 + v 2 ) 1 + v2 v2 1
x v
dx v v v

1
vdv = dx
x

v2
= ln x + c
2



y2
2
= ln x + c y2 2x 2 (ln x c)
2x
16
c c=8 y 2 = 2x 2
(lnx + 8)
2

4.3.4 Linear First Order DEs

dy
If P = P(x) and Q = Q(x) are functions of x only, then + Py = Q is called a linear
dx
differential equation order 1. We can solve these linear DEs using an integrating
factor. For linear DEs of order 1, the integrating factor is: e∫Pdx

The solution for the DE is given by multiplying y by the integrating factor (on the left)
and multiplying Q by the integrating factor (on the right) and integrating the right side
with respect to x, as follows:

y eò Q eò
Pdx Pdx
=
ò +K

dy 3
Example 12: Solve for - y = 7.
dx x

Answer

dy 3 3
- y = 7. then P(x) = - and Q(x) = 7
dx x x
Now for the integrating factor:
3 3
ò Pdx = e ò - x dx = e ò - x dx = e- 3 ln x = x- 3
IF= e

For the left hand side of the formula



ye ò Qe ò
Pdx Pdx
=
ò dx

we have ye∫Pdx = yx-3

For the right hand of the formula, Q = 7 and the IF = x-3, so:

Qeò
Pdx - 3
= 7x

Qe ò
Pdx 7 -2
ò ò 7x
- 3
Applying the outer integral: dx = dx = - x +K
2

ye ò Qe ò
Pdx Pdx
Now, applying the whole formula; =
ò dx

- 3 7 -2
we have ; yx =- x +K
2
7 3
Multiplying throughout by x3 gives: y = - x + Kx
2

dy
Example 13: Solve + (cot x)y = cos x
dx
Answer
dy
+ (cot x)y = cos x
dx
Here, then P(x) = cot x and Q(x) = cos x

Determine
ò Pdx = ò cot xdx = lnsin x
IF = e
ò Pdx = eln sin x = sin x

ò Pdx = cos x sin x
Now Qe
Apply the formula:

ye ò Qe ò
Pdx cot xdx
=
ò dx



y sin x =
ò cos x sin xdx

The integral needs a simple substitution: u = sin x, du = cos x dx
2
sin x
y sin x = +K
2
Divide throughout by sin x:
sinx K sinx
y= + = + K cosecx
2 sinx 2

- 3x
Example 14: Solve dy + 3ydx = e dx

Answer
Dividing throughout by dx to get the equation in the required form, we get:
dy - 3x
+ 3y = e
dx
In this example, P(x) = 3 and Q(x) = e-3x.
Now e∫Pdx = e∫3dx = e3x
and

Qe ò
Pdx
òe ò 1dx = x
- 3x 3x
= e dx =

ò Pdx
Qe ò
Pdx
Using ye =
ò dx + K , we have:

ye3x = x + K
or we could write it as:
x +K
y=
3x
e

Example 15: Solve 2(y - 4x2)dx + x dy = 0

Answer



We need to get the equation in the form of a linear DE of order 1.
Expand the bracket and divide throughout by dx:
2 dy
2y - 8x + x = 0
dx
Rearrange:
dy 2
x + 2y = 8x
dx
Divide throughout by x:
dy 2
+ y = 8x
dx x
2
Here, P(x) = and Q(x) = 8x
x
2
ò Pdx ò x dx 2 ln x ln x
2
2
IF = e = e = e = e = x

Now Qe ò
Pdx 2 3
= 8xx = 8x
Applying the formula:

ye ò Qeò
Pdx Pdx
=
ò dx + K

ò
2 3 4
gives: yx = 8x dx + K = 2x + K

2 K
Divide throughout by x2: y = 2x +
2
x

dy 6 x
Example 16: Solve x - 4y = x e
dx
Answer
Divide throughout by x:
dy 4 5 x
- y= x e
dx x
4 5 x
Here, P(x) = - and Q(x) = x e
x



4
ò - dx - 4
IF = e ò
P(x)dx ln x - 4
= e x = e = x

Now Qe
ò P(x)dx = x5 ex x- 4 = xex

ye ò Qeò
Pdx Pdx
Applying the formula: =
ò dx + K gives

ò
- 4 x
yx = xe dx + K

This requires integration by parts, with
x
u= x dv=e
x
du = dx v=e

- 4 x x
So yx = xe - e + K

5 x 4 x 4
Multiplying throughout by x4 gives: y = x e - x e +Kx

dy x
Example 17: Solve e 2y , x 0 , subject to the initial condition y = 2
dx
when x = 0

Answer

The differential equation can be expressed in the proper form by adding
2y to both sides:

dy x
2y e for x 0
dx

x
We have P(x) = 2 and Q(x) = e

P(x)dx 2dx
An integrating factor is given by IF = e e e2x for x 0



Next, use the first-order linear differential equation theorem, where IF =
x
e 2 x and Q(x) = e , to find y:

1
y= 2x
e2x e x dx C
e

1
= 2x
ex C e x
e 2x
C for x 0
e

To find C, insert y = 2 when x = 0, to obtain C = 1

x 2x
Thus y = e e , x 0

dy 2xy
Example 18: Solve sin x with y = 1 when x = 0
dx 1 x2

Answer

2x
P(x) = ; Q(x) = sin x
1 x2

2x
dx
2 1)
IF = e 1 x2 = e ln(x x2 1

1
y= 2
(x 2 1)sin xdx C
1 x

1
= 2
x 2 sin xdx sin xdx C
1 x

1
= 2
2x sin x (2 x 2 )cos x cos x C
1 x



1
= 2
2x sin x (1 x 2 )cos x C
1 x

Since y = 1 when x = 0,

1
1= 0 1 C implies C = 0
1

1
y= 2
2xsinx +(1- x 2 )cosx
1+ x

Example 19:

dy
a. Solve the DE = sec θ + y tan θ , given the boundary conditions y=1 when θ = 0.
dθ

[Ans: y = (θ + 1) sec θ]

dy 5 c
b. Solve the DE t -5 t = -y . [ ans: y t ]
dt 2 t
c. Consider a simple electric circuit with the resistance of 3 an inductance of 2H. If a
battery gives a constant voltage of 24V and the switch is closed when t = 0, the
current, I(t), after t seconds is given by

dI 4
+ t = 15, I(0) = 0
dt 3

4
45 t
i) Obtain I(t) [ans: I(t) (1 e 3 )]
4
ii) Determine the difference in the amount of current flowing through the circuit from
the fourth to eight seconds. Give your answer to 3 d.p.

[ans: 0.05 A]

iii) If the current is allowed to flow through the circuit for a very long period of time,
estimate I(t).



45
[ans: A]
4

4.4 SECOND ORDER DIFFERENTIAL EQUATION

The general form of the second order differential equation with constant coefficients is

d2y dy
a +b + cy = Q( x )
dx 2 dx

where a, b, c are constants with a > 0 and Q(x) is a function of x only.

4.4.1 Homogeneous Equation

In this section, most of our examples are homogeneous 2nd order linear DEs (that is, with
Q(x) = 0):

d 2y dy
a +b + cy = 0 , where a, b, c are constants.
dx 2 dx

Method of Solution

The equation am2 + bm + c = 0 is called the Auxiliary Equation (A.E.) (or
Characteristic Equation)

The general solution of the differential equation depends on the solution of the A.E. To
find the general solution, we must determine the roots of the A.E. The roots of the A.E.
are given by the well-known quadratic formula:

m= -b ± b2 - 4ac
2a



Summary:

If Differential Equation: ay'' + by' + c = 0 and Associated auxiliary equation is:
am2 + bm + c = 0

Nature of roots Condition General Solution

4.4.1. Real and distinct roots,
b2 − 4ac > 0 y = Aem1x + Bem2x
m1, m2

4.4.2. Real and equal roots, m b2 − 4ac = 0 y = emx(A + Bx)

4.4.3. Complex roots

m1 = α + jω b2 − 4ac < 0 y = eαx(A cos ωx + B sin ωx)

m1 = α − jω

Example 20

2
d i di
The current i flowing through a circuit is given by the equation + 60 + 500i = 0 ,
dt dt
Solve for the current i at time t > 0.

Answer

The auxiliary equation arising from the given differential equations is:

2
A.E.: m + 60m + 500 = 0 = (m + 50)(m + 10) = 0
So m = - 50 and m = - 10 and
1 2

We have 2 distinct real roots, so we need to use the first solution from the table above
(y = Aem1x + Bem2x), but we use i instead of y, and t instead of x.



- 10t - 50t
So i = Ae + Be

- 10t - 50t
We could have written this as: i= C e + C e
1 2

Since we have 2 constants of integration. We would be able to find these constants if
we were given some initial conditions.

Example 21

Solve the following equation in which s is the displacement of a object at time t.

2
d s ds ds
- 4 + 4s = 0 , given that s = 1, = 3 when t = 0
2 dt dt
dt

(That is, the object's position is 1 unit and its velocity is 3 units at the beginning of the
motion.)

Answer

The auxiliary equation for our differential equation is:
2 2
A.E. m - 4m + 4 = 0 = (m - 2) = 0
In this case, we have: m = 2 (repeated root or real equal roots)
We need to use the second form from the table above (y = emx(A + Bx)), and once again
use the correct variables (t and i, instead of x and y).
2t
So S(t) = ( A + Bt )e .
Now to find the values of the constants:
s(0) = 1 Þ A = 1
2t
So we can write S(t) = (1+ Bt )e

' 2t 2t
S (t) = 2(1+Bt)e +Be

s '(0) = 3 Þ 2 + B = 3 Þ B = 1



2t
So s t)
( =(1+ t )e

The graph of our solution is as follows:

Example 22

d 2y dy
Solve the equation - 2 + 4y = 0
2 dx
dx

Answer

2
This time the auxiliary equation is: m - 2m + 4 = 0

Solving for m, we find that the solutions are a complex conjugate pair:
m = 1- j 3 and m = 1 + j 3
1 2

The solution for our DE, using the 3rd type from the table above:
y = eαx(A cos ωx + B sin ωx)
x
we get: y(x) = e (Acos 3 x +Bsin 3 x)



Example 23

In a RCL series circuit, R = 10 Ω, C = 0.02 F, L = 1 H and the voltage source is E = 100
V. Solve for the current i(t) in the circuit given that at time t = 0, the current in the circuit
is zero and the charge in the capacitor is 0.1 C.

[Note: Damping and the Natural Response in RLC Circuits

Consider a series RLC circuit (one that has a resistor, an inductor and a capacitor) with
a constant driving electro-motive force (emf) E. The current equation for the circuit is
di 1 di 1
L + Ri + ò idt = E or L + Ri + q = E
dt C dt C
Differentiating, we have
2
d i di 1
L +R + i = 0 ; This is a second order linear homogeneous equation.]
dt
2 dt C

Answer

d 2i + 10 di + 50i = 0
dt 2 dt

AE : m2 + 10m + 50 = 0 ,



The factors are: m = - 5 - j5 and m = - 5 + j5
1 2

So, i = e- 5t ( Acos5t + B sin5t ) ; éù
i
êú
ëû
= A = 0 ; (This means at t = 0, i = A = 0 in this
t= 0
case.)

Then i = e- 5t B sin5t , We need to find the value of B.
Differentiating gives:
di - 5t - 5t - 5t
= e (5B cos5t ) + (B sin5t )(- 5e ) = 5Be (cos5t - sin5t )
dt
di
At t = 0, = 5B
dt
di
Returning to equation + 10i + 50q = 100
dt
di di
Now, at time t = 0, + 10(0) + 50(0.1) = 100 ; So = 95 = 5B , so B = 19.
dt t = 0 dt t = 0

-5t
Therefore, i = 19e sin5t


Differential equation

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Differential equation