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8/9/2019 Non Linear First Order Diff Equ
http://slidepdf.com/reader/full/non-linear-first-order-diff-equ 1/20
NON-LINEAR FIRSTORDER DIFERENTIAL
EQUATIONS
PRESENTED BY:AHSAN ALI08-EE-50
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NON-LINEAR FIRST ORDER
DIFERENTIAL EQUATIONSDEFINITION:-
Any differential equation is said to be non-linear if:-
1.The dependent variable y and itsderivative are all not of degree one.
2.Products of y or its derivative appear
3.Transcendental function of y or itsderivative appear
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FOR EXAMPLE
The equation
Is not linear because of the term
Similarly, the equation
3
2
2
3
3
2 xdx
dy ydx
yd edx
yd x=++
dxdy y
23
2
22
+=
ydx
yd
dx
dy
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is also non linear because of degree 3
Other examples are
2
22
3
2
2
2
2
1
0
dx
yd
dx
dy
dx
dy
xydx
yd
=
+
=
+
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NON –LINEAR
DIFFERENTIAL EQUATIONS
A first order non-linear differential
equation in dependent variable y andindependent variable t can be written inthe form:
00 )(),,( yt y yt f dx
dy==
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DIFFERENCES BETWEENLINEAR AND NON-LINEARDIFFERENTIAL EQUATIONSFORM
Linear differential equation has the form
While non linear has the form
( ) ( ) ( ) 00, yt yt g yt p y ==+′
00 )(),,( yt y yt f
dx
dy==
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Methods of Solution:-
In case of linear equation themethod of solution is by integratingfactor. If non linear differentialequations is separable separate
Explicit solution:-There is always a way to
solve for y explicitly in case of linearequations. In non linear equations itmay not be possible to solve for yexplicitly
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Completeness of solution:-In linear equations, all solutions can begenerated by varying a constant in asingle expression. In non-linear equation,a general solution may be obtained in theform of a algebraic expression, but eventhen it is possible that not all thesolutions may be produced by varying theconstant
Discontinuity:-In linear equations solution can only bediscontinuous if p and g are
discontinuous.
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Even then it may exist even at points
where either p or g are discontinuous.Non-linear equations can bediscontinuous at points other than where f & are discontinuous. Moreover,
there may be nothing in the differentialequation that indicates where theseadditional discontinuities exist
Weaker results:-
In non-linear differentialequations if f is continuous then asolution exists, but I may not be unique
y
f
∂
∂
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Theorems:-For linear equations
“If p and g are continuous on an open intervalI:α<t<β containing the point t0 then there
exists a unique function y=φ(t) that satisfiesthe differential equation for each t in I andthat also satisfies the initial condition”
For non-linear equation
“If f and are containing in the same
Rectangle α<t<β,γ<y<δ containing the point (t o, y 0 ) the some interval t o-h<t<t 0+h
contained in the α<t<β there is a uniquesolution y=φ(t) of the initial value problem”
y
f
∂
∂
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EXAMPLES AND
SOLUTIONSExample no. 1
Consider the equation:
The functions f and are given as
( )1)0(,
12243
2
−=
−
++= y
y x x
dxdy
y
f
∂
∂
( ) ( ),
12
243),(,
12
243),(
2
22
−
++−=
∂
∂
−
++=
y
x x y x
y
f
y
x x y x f
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And are continuous except on the line y=1thus we can draw an open rectangleabout (0,-1) on which f and are
continuous. Graphically shown below
y
f
∂
∂
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Suppose we change the initial conditiony(0)=1 solving the new initial value
problem we obtain
Thus the solution is not unique but exists
EXAMPLE NO. 2
Consider the equation
0,221 23>++±= x x x x y
( )00)0(,3/1 ≥==′ t y y y
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The function f and are given as:
Hence f is continuous everywhere butdoes not exist at y=o thus solution existsbut is not unique separating variables andthen solving
If the initial condition is not on t-axis then wecan ‘t guarantee the uniqueness andexistence
y
f
∂
∂
3/23/1
3
1
),(,),(
−
=∂
∂
= y yt y
f
y yt f
y
f
∂
∂
0,3
2
2
32/3
3/23/1 ≥
±=⇒+=⇒=− t t yct yd t d y y
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APPLICATIONSApplications of non-linear homogenous
equations are :-
1.Determining the motion of projectile,rocket, satellite or planet
2.Finding the charge or current in anelectric circuit
3.Study of chemical reactions4.Determination of curves with given
properties
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Question no 15 Book page no. 47 EX#1.7
Show that
Can also be written as
Solve the equation when E(t)=0 assumingQ(0)=Q0
( )1
RI Idt E t C
+ =∫
( )1dQ
R Q E t dt C
+ =
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Solution:-
Since
Therefore ,
……….(proved)
Now,
dQ I dt
=
( ) ( )
( )
1
1
dQ R I t E t dt C
dQ R Q E t
dt C
+ × =
+ =
( ) 0............... E t Given=
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1
0
1
1
1
dQ
R Qdt C
dQ R Q
dt C dQ
Q
dt RC
dQ Qdt
RC
+ =
=−
=−
=−
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( )
1 1
1 1
ln
t c
RC
t
RC
dQ dt Q RC
dQ dt
Q RC t
Q c RC
Q e
Q ce
− +
−
=−
=−
=− +
=
=
∫ ∫
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When t=0 ,Q=Q0
0
0
t
RC
Q c
Q Q e−
=
=