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Name: Dave Paul Y. Jandayan Schedule: MWF (4:30 – 5:30)
P.M.
Course/Year: BSECE -
RESEARCH REPORT: SHIELDS
I. SEPARABLE
A first-order differential equation of the form
dy
dx
=g ( x ) h( y ) is said to be separable
or to have separable variables.
Solving a Separable DE: Solve (1 + x) dy - y dx = 0.
ALTERNATIVE SOLUTION
Because each integral result in a logarithm, a judicious choice for the
constant of integration is ln |c|_ rather than c. Rewriting the second line of the
solution as ln |y| = ln |1 + x | + ln|c| enables us to combine the terms on the right-
hand side by the properties of logarithms. From ln|y| = ln|c (1 +x) | we immediately
get y = c (1 + x). Even if the indefinite integrals are not all logarithms, it may still
be advantageous to use ln|c|. However, no firm rule can be given.
SOLUTION: Dividing by (1 + x ) y , we can write dy / y = dx / (1 + x ), from
which it follows that.
∫
dy
y
∫
dx
1 + x
ln|y|=ln| 1 +x|+C
1
y=e
ln| 1 + x|+C
1
=e
ln| 1 + x|
∗e
C
1
¿| 1 + x|e
C 1
¿ ± e
C 1
( 1 + x )
± e
C
1
=C
Laws of
exponent
| 1 + x|= 1 +x x ≥− 1
| 1 + x|=−( 1 + x )
x ≥− 1
II. HOMOGENOUS
A linear nth-order differential equation of the form
a
n
x
d
n
y
d x
n
n− 1
x
d
n− 1
y
d x
n− 1
1
x
dy
dx
0
x
y= 0
is said to be homogeneous, whereas an equation.
a
n
( x )
d
n
y
d x
n
n− 1
( x )
d
n− 1
y
d x
n− 1
1
( x )
dy
dx
0
( x ) y=g ( x ) ; g ( x ) not identicaly zero
is said to be nonhomogeneous.
Solving a Homogenous DE: Solve 2x dy = (x + y) dx
c
2
=c
SOLUTION: Replace y= vx. Differentiate y=vx using product rule; dy=
vdx +xdv
2 x dy =( x+ y ) dx 2 x ( vdx+ xdv ) =( x+ y ) dx 2 x ( vdx+ xdv ) =( x+ vx ) dx
2 ( vdx + xdv )=( 1 + v ) dx
vdx+ 2 xdv=dx 2 xdv=vdx−dx
∫
dv
( 1 −v )
∫
dx
2 x
−ln| 1 −v|=
ln|x|+c
ln ¿ ¿
ln
1 −v
=ln ( c
√
x )
x
[
y
x
=c √
x
]
( x− y )
[
x− y
=c √
x
]
( x− y )
( x )
2
c √
x ( x− y )
2
x
2
=c
2
x ( x− y )
2
x=c ( x− y )
2
|x|=c ( x− y )
2
y=vx ; v=
y
x
IV. EXACT
A differential expression M (x, y) dx + N (x, y) dy is an exact differential in a
region R of the xy -plane if it corresponds to the differential of some function f (x, y)
defined in R. A first-order differential equation of the form
M ( x , y ) dx+ N ( x , y ) = 0
is said to be an exact equation if the expression on the left-hand side is an exact
differential.
Solving an Exact Differential Equation: Solve 2xy dx + (x
2
[A]
SOLUTION: With M ( x , y ) = 2 xy and N ( x , y ) = x2- 1 we have
∂ M
∂ y
= 2 x=
∂ N
∂ x
Thus, the equation is exact, and so by [A] there exists a function f (x, y)
such that
∂ f
∂ x
xy ∧∂ f
∂ y
=x
2
From the first of these equations we obtain, after integrating,
f
x , y
=x
2
y + g( y)
Taking the partial derivative of the last expression with respect to y and
setting the result equal to N (x, y) gives
∂ f
∂ y
=x
2
'
( y )=x
2
It follows that g’(y)= -1 and g(y)= -y. Hence f (x, y) x
2
y – y, so the
solution
of the equation in implicit form is
x
2
y − y=c
The explicit form of the solution is easily seen to be
y=
c
( 1 −x
2
N ( x , y)
V. DIRECT INTEGRATION
A method for equations where the right-hand side does not depend on the
unknown.
General Form:
dy
dt
=f
t
d
n
y
d t
n
=f (t)
Solving by Direct Integration: Solve
dy
dt
= 3 y− 210
SOLUTION:
Factoring the righthand side of this equation to get
dy
dt
= 3 ( y− 70 )
(
dy
dt
)
( y− 70 )
( y− 70 )
dy= 3 dt
∫
( y− 70 )
dy=
∫
3 dt
ln|y − 70 |= 3 t +c
e
ln|y− 70 |
=e
3 t+ c
y − 70
=e
3 t
∗e
c
y=C e
3 t
VII. SUBSTITUTION
Often the first step in solving a differential equation consists of transforming it
into another differential equation by means of a substitution.
For example, suppose we wish to transform the first-order differential equation
dy
dx
=f (x , y ) by the substitution y=g (x , u), where u is regarded as a function of
variable x. If g possesses first-partial derivatives, then the Chain Rule,
dy
dx
∂ g
∂ x
dx
dx
∂ g
∂ u
du
du
gives
dy
dx
=g
x
( x ,u )+ g
u
( x ,u )
du
dx
If we replace
dy
dx
by the foregoing derivative and replace y
in f (x , y )
by g( x , u)
then the DE
dy
dx
=f (x , y ) becomes
g
x
( x ,u )+ g
x
( x , u)
du
dx
=f (x , g ( x , u)) which, solved for
du
dx
, has the form
du
dx
F ( x , u). If we can determine a solution u= ∅ ( x) of this last
equation, then a solution of the original differential equation is y=g
x , ∅ ( x )
.
Solving a Linear Equation: Solve
dy
dx
( 1 +x
3
3 x
2
1 + x
2
y
SOLUTION:
Rewrite as:
dy
dx
3 x
2
1 + x
2
y =
( 1 + x
3
Comparing to
dy
dx
P=
3 x
2
1 +x
2
; Q=
( 1 + x
3
Figure out integrating factor I.F.,
I. F .=e
∫
3 x
2
1 + x
2
dx=e
ln ( 1 +x
3
) ; I. F .= 1 + x
3
Rewrite LHS.
d
y ( I. F. )
dx
; d ( y ( 1 +x
3
) ) dx=
( 1 +x
3
( 1 + x
3
Integrating both sides.
y (
1 + x
3
¿=x
y=
x
( 1 + x
3
y=
x
( 1 + x
3
