HIGHER DIFFERENTIAL EQUATION

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Group : B

DIFFERENTIAL EQUATIONDIFFERENTIAL EQUATIONnth order differential equation

General form of equation is ó General form of equation is

dny/dxn + a1(x)dn−1y/dxn−1 ・・・ + an−1(x)dy/dx+ an(x)y = f(x) .

Here,

a1(x), a2(x), a3(x) ………….an(x)= Constants

f(x)=Driving or Forcing

If ,

f(x)=0 ………….Homogenous equation

f(x)≠0 ………….Non-Homogenous equation

HIGHER ORDER HOMOGENOUS HIGHER ORDER HOMOGENOUS DIFFERENTIAL EQUATIONDIFFERENTIAL EQUATION• General form ;

dny/dxn + a1(x)dn−1y/dxn−1 ・・・ + an−1(x)dy/dx+ an(x)y = 0

• General solution Method :

v Make operator form.

f(x)

v Construct an Auxiliary equation.

v Solve the equation & find the Roots of equation.

vFinal sol’n is called Complimentary/General sol’n.

Operator form

g(x)

HIGHER ORDER HOMOGENOUS HIGHER ORDER HOMOGENOUS DIFFERENTIAL EQUATIONDIFFERENTIAL EQUATION

v Theorem for Linear Combination :

y= c1y1+c2y2+c3y3+……………cnyn

ó Corollary : If y1 is the sol’n of a homogenous differential equation then c1y1 will be its general sol’n. & c(-∞, +∞)

HIGHER ORDER HOMOGENOUS HIGHER ORDER HOMOGENOUS DIFFERENTIAL EQUATIONDIFFERENTIAL EQUATION

• Steps for solution :4y’’+4y’+y=0

• Operator form ;

(4D^2+4D+1)y=0

• Auxiliary equation :

4D^2+4D+1=0

• Roots :

D= -1/2 , D= -1/2

• General Solution (Complimentary solution) :

y= c1y1+c2y2 , y=(c1+c2x)e^-1/2x

HIGHER ORDER HOMOGENOUS HIGHER ORDER HOMOGENOUS DIFFERENTIAL EQUATIONDIFFERENTIAL EQUATIONóó Nature of roots ;Nature of roots ;

1. Real & Distinct :

y= c1y1+c2y2+……………cnyn

Since :

y1=e^m1x , y2=e^m2x , y3=e^m3x , …… yn=e^mnx

Plugging in the values, we get :

y= c1e^m1x+c2e^m2x+c3e^m3x……………..cne^mnx

2. Real & Repeated :y= (c1+c2x+c3x^2+……………cnx^n-1)e^mx

Solution :The characteristic equation is,

Hence, roots are REAL & DISTINCTREAL & DISTINCT so the general solution is,

Example :

ó Example :

Solution ;

The characteristic equation is,

So, we have two roots here,

r3= -2 r4= -2

[----y1------ ] [--------y2-----]

Note : no of sol’n= no.of roots ………..for Real n distinct roots

3. Complex Roots :

y= (c1Cosbx+c2xCosbx+c3x^2Cosbx+…….cnx^n-1Cosbx)e^ax

+

(c1Sinbx+c2xSinbx+c3x^2Sinbx+…….cnx^n-1Sinbx)e^ax

Note : no of sol’n= 2(no.of roots)…………for Complex roots

Example : Solve the following differential equation .

SolutionThe characteristic equation is ;

The 4 ,4th roots of -16 can be found by evaluating the following

So, we have two sets of complex roots :

HIGHER ORDER NON HOMOGENOUS HIGHER ORDER NON HOMOGENOUS DIFFERENT EQUATIONDIFFERENT EQUATION

• General form :

dny/dxn + a1(x)dn−1y/dxn−1 ・・・ + an−1(x)dy/dx+ an(x)y = f(x)

(We‘ll deal non- homogenous, when f(x) = single function : )

4y”+3y’+1=Sinx

v Solution Of Non-Homogenous Equation :

Relating to solution we use method named as :

üMethod Of “UNDETERMINED CO-EFFICIENTS”

This method is also known as “ Judicious Guessing Method”“ Judicious Guessing Method”

Since co-efficients are unknown as A, B, C etc. in Particular sol’n.

METHOD OF METHOD OF UNDETERMINED COUNDETERMINED CO--EFFICIENTSEFFICIENTS

• Limitations :

This method is limited to following functions :

The particular sol’n (yp) may also be the product of any two or all of these functions.

(Sinx / Cosx)

(Trignometric func:)

Acosx+Bsinx

X^n+X^n-1+…….E(Polynomial func:)

Ax^n-1+Bx^n-1……C

e^x(Exponential func:)

Ae^x

METHOD OF METHOD OF UNDETERMINED COUNDETERMINED CO--EFFICIENTSEFFICIENTS

• How to Solve these :

o Find the Complimentary solution of associated Homogenous equation . e.g.

y''' + y'' = Cos(2t)

(Associated homogeneous differential equation)

r3 + r2 = 0

o Find out the Particular sol’n for function f(x)= Cos(2t).

yp= Acos(2t)+Bsin(2t)

o General n Final sol’n is :

y= yc(Transient sol’n)+yp(Steady state sol’n)

v Solution of nth order non-homogenous equation

“A sol’n of nth order non-homogenous differential equation is a

function y(x) which is differentiable n-times & its derivatives satisfy the equation”

Ø Linear dependent functions (solutions)

The linear equation solution will be linearly dependent if :

y1=cy2 or y2=cy1

Ø Linear Independent functions (solutions)

The linear equation solution will be linearly independent if :

y1≠cy2 or y2≠cy1

Satisfying this condition the solutions will be mutually dependent or independent.

ü For Non-Homogenous equations we check the same case in between the Complimentary sol’n & Particular sol’n.

Solution of nth order non-homogenous equation

v Conditions & Methods for Dependency & Independency

qSimple Method :

To make the functions (sol’ns) Linearly indpendent only one case is possible that c1=c2=c3=cn=0 , otherwise it won’t be solved.

q WRONSKIAN Determinant Method :

It states that if ;

(i) W=0 solutions are LINEARLY DEPENDENT

(ii) W≠0 solu� ons are LINEARLY INDEPENDENT

It is made clear by examples here :

|e^x x e^x x^2 e^x |W=| e^x e^x+ xe^x 2x ex + x2 e^x |

| e^x 2 e^x + x e^x 2 e^x + 4 x e^x + x^2 e^2 |

q (1) s := [ex, x ex, x2 ex ]

Let's calculate the Wronskian.

Result = 2 (e^x )^3Since the Wronskian is not zero, the given set in LINEARLY INDEPENDENT.

Solution of nth order non-homogenous equation

§ Importance of Linear dependency :

Its importance can be depicted from an example quoted here .

Ø Example:

y″ − 2y′ − 3y = 5e^3t

Here,

yc = C1 e^−t + C2 e^3t.

Since, g(t) = 5e^3t, So, Y = A e^3t,

If we substitute Y, Y ′ = 3A e^3t, and Y ″ = 9A e^3t into the differential equation and simplify, we would get the equation

0 = 5e^3t

ü The remedy is surprisingly simple: multiply our usual choice by “x^k” . Y=(x^k)A e^3t.

Only in this case we can get the solution………

v Principle Of Modification :“If there is any common term in Yp & Yc (in Non-Homogenous case) then multiply

the factor x^k with Yp to make the solutions or Yp &Yc Linearly Independent.”

q Non-Homogenous equation & Particular functions f(x) :

General Principle of SuperpositionGeneral Principle of Superposition: : If y1 is a solution of the equation

y″ + p(t) y′ + q(t) y = g1(t),

and y2 is a solution of the equation

y″ + p(t) y′ + q(t) y = g2(t).

Then, for any pair of constants C1 and C2, the function y = C1 y1 + C2 y2

is a solution of the equation

y″ + p(t) y′ + q(t) y = C1 g1(t) + C2 g2(t).

v Example : y''' + y'' = cos(2t)

Solve :

The characteristic equation isr3 + r2 = 0

Factoring givesr2(r + 1) = 0

r = 0 (repeated twice) or r = -1

The homogeneous solution isyc = c1 + c2t + c3e

-t ………………..(i)

The UC-set generated by g(t) = Cos(2t) is

Notice that the UC-set does not intersect the homogeneous solution set. We can write

yp = Acos(2t) + Bsin(2t)

yp' = -2Acos(2t) + 2Bsin(2t)

yp'' = -4Asin(2t) - 4Bcos(2t)

yp''' = -8Acos(2t) + 8Bsin(2t)Plugging back into the original differential equation gives

[-8Acos(2t) + 8Bsin(2t)] + [-4Asin(2t) - 4Bcos(2t)] = cos(2t)

Combining like terms gives

(-8A - 4B)cos(2t) + (8B - 4A)sin(2t) = cos(2t)

Equating coefficients gives

-8A - 4B = 1-4A + 8B = 0

We get :A = -0.1 and B = -0.05

yp = -0.1cos(2t) -0.05sin(2t)…………………………(ii)

Combining (i) & (ii) we get ,

General solution = y = c1 + c2t + c3e-t -0.1cos(2t) - 0.05sin(2t)

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ó Example :y(iv) +4y'' = 16 + 15et

Solution :

r4 + 4r2 = 0 r2(r2 + 4) = 0

The roots are

r = 0 (repeated twice) r = 2i r = -2i

The homogeneous solution is :

yh = c1 + c2t + c3sin(2t) + c4cos(2t)

Since g(t) is a sum of two terms, we can work each term separately. The UC-Set for the function g(t) = 16 is

yp=ASince this is a solution to the homogeneous equation, we multiply by t to get

yp=AtThis is also a solution to the homogeneous equation, so multiply by t again to get

yp=At2

which is not a solution of the homogeneous equation. We write

yp1 = At2 , yp1' = 2At , yp1'' = 2A , yp1''' = 2A , yp1(iv) = 0

Substituting back in, we get0 +4(2A) = 16

A = 2Hence

yp1 = 2t2 ………………….(i)

Now we work on the second piece. The UC-set for g(t) = 15et isyp= Aet

Since this in not a solution to the homogeneous equation, we getyp2 = Aet

yp2' = Aet

yp2'' = Aet

yp2''' = Aet

yp2(iv) = Aet

Plugging back into the original equation givesAet + 4Aet = 15et

5A = 15A = 3

Henceyp2 = 3et ………..(ii)

The general solution to the nonhomogeneous differential equation isy = c1 + c2t + c3sin(2t) + c4cos(2t) + 2t2 + 3et

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