Non Linear First Order Diff Equ

8/9/2019 Non Linear First Order Diff Equ

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NON-LINEAR FIRSTORDER DIFERENTIAL

EQUATIONS

PRESENTED BY:AHSAN ALI08-EE-50



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



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



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

=

+

=

+



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==



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==



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



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.



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

∂

∂



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

∂

∂



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



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

∂

∂



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



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



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



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

+ =



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=



1

0

1

1

1

dQ

R Qdt C

dQ R Q

dt C dQ

Q

dt RC

dQ Qdt

RC

+ =

=−

=−

=−



( )

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

− +

−

=−

=−

=− +

=

=

∫ ∫



When t=0 ,Q=Q0

0

0

t

RC

Q c

Q Q e−

=

=

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Non Linear First Order Diff Equ