Introduction

If given y  e
0.1x 2 dy
, we can get dx
.
dy
But if given  0.2 xe 0.1 x 2
, how to get back
dx
y?

So we need to study differential equation.

Sol:
dy
 0.2 xe 0.1 x 2

dx
dy  0.2 xe 0.1x 2
dx

 dy   0.2 xe
0.1x 2
dx u = 0.1x2
ye 0.1x 2
du= 0.2x dx

Differential Equations
Differential equation can be divided into
 Ordinary Differential Equation (ODE)
 Partial Diffential Equation(PD)

1
ODE : involving only ordinary derivatives
with respect to a single independent
variable

PDE : involving partial derivatives with

respect to more than one independent
variable

Revise
y = f(x)
y : dependent variable
x : independent variable
Eg1
d 2x dx
2
 a  kx  0
dt dt
x : dependent variable
t : independent variable

u u
  x  2y
x y
2
u : dependent variable
x, y : independent variable

Eg2
a dy  5 y  e x ODE
dx
b d 2 y dy
2
  6y  0 ODE
dx dx
c dx dy
  2x  y ODE
dt dt
d (y”) +(y’) +3y=x2
2 3
ODE
e xy’ + y = 3 ODE
f y’’’ + 2 (y”)2 + y’ = cos x ODE
g dz  z  x dz ODE
dx dy
h d 2z d 2z
2
 2  x2  y PDE
dx dy
i  2u  2u
  2
u PDE
x 2 t 2 t
j u v PDE

y x

3
Classification of Ordinary Differential
Equation (ODE)

ODE can be classified by order, degree,

linear or nonlinear.

Order of ODE is the highest-ordered

derivative appearing

Degree of ODE is degree of the highest

derivaties

Eg3
Order Degree
(y’’)2 + (y’)3 + 3y = x2 2 2

xy’ + y = 3 1 1

Linear /Nonlinear Differential Equation

A differantial equation is linear if it has
the following format :-
dny d n 1 y d2y dy
an ( x) n  an 1 ( x) n 1  ....  a2 ( x) 2  a1 ( x)  a0 ( x) y  g ( x)
dx dx dx dx
4
Otherwise it is nonlinear.

Eg4

a (y-x)dx + 4x dy =0 Linear
dy
4x yx
dx
b y’’ – 2y’ + y = 0 Linear
c d y  3x dy  5 y  e Linear
3
x
3
dx dx
d (1-y)y’ + 2y = ex Nonlinear
e d y  sin y  0 Nonlinear
2

dx 2
f d4y
4
 y2  0 Nonlinear
dx
g dy
 y  x2 Linear
dx
h dy
x(1  y)( ) 2  (1  x) y  0 Nonlinear
dx
i d3y
3
 e x dy
 y  ln x Linear
dx dx
j d2y dy
2 x 2  sin x  y  cos x Linear
dx dx
k d2y
 (1  )
dy 2
3
Nonlinear
dx 2 dx

5
Formation of Differential Equation
Eg5
By deleting the constant A from
y = x + A x , form one differential equation.
Sol
y = x + A x ---------(1)

xy = x2 + A

A = xy – x2 ----- (2)

dy A
dx
 1 2
x -----------(3)
(2) in (3)

dy ( xy  x 2 )
 1
dx x2
dy y
 2
dx x
dy
x  y  2x
dx

6
Eg6
By deleting the constants A and B from
the equation y= A cos x + B sin x, form a
differential equation.
Sol
y= A cos x + B sin x -----(1)
dy
  A sin x  B cos x -------------(2)
dx
d2y
2
  A cos x  B sin x
dx
 ( A cos x  B sin x)
= -y
d2y
2
y0
dx

Eg7
By deleting the constant A and B from the
equation y = e-2t(A cos 3t + B sin 3t), form
a differential equation.
Sol
y = e-2t(A cos 3t + B sin 3t) ---------(1)
e2t y = A cos 3t + B sin 3t -----------(2)

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2e2t y +e2t y’ = -3A sin 3t + 3B cos 3t
--(3)

4e2ty + 2e2ty’ +2e2ty’ + e2ty’’ =

-9A cos 3t –9B sin 3t (4)

Simplify it
e2t (4y + 4y’ + y’’) = -9(Acos 3t +B sin 3t)
= - 9e2ty

y’’ + 4y’ + 13y = 0

Solutions of Differential Equation

A function y = (x) is called explicit
solution to the differential equation if it
satisfies the differential equation.

Eg8
Show that y = Aex + (x+2)e2x is an explicit
dy
solution to  y  ( x  3)e 2 x .
dx
Sol
y = Aex + (x+2)e2x
8
dy
 Ae x  e2 x  2( x  2)e2 x
dx
dy
 y = Ae x  e2 x  2( x  2)e2 x  ( Ae x  ( x  2)e2 x )
dx
 ( x  3)e2 x
so y = Aex + (x+2)e2x is the explicit
solution.

Eg9
B
Show y  Ax 3  is an explicit solution to
x3
d2y dy
x2
2
 x  9y  0 .
dx dx
Sol
B
y  Ax 3 
x3
dy 3B
 3 Ax 2  4
dx x
d2y 12 B
2
 6 Ax  5
dx x
d2y dy 12 B 3B B
x2
2
 x  9 y  x 2
( 6 Ax  5
)  x (3 Ax 2
 4
)  9( Ax 3
 3
)0
dx dx x x x
B
so y  Ax 3  is the explicit solution
x3

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