The document discusses higher order differential equations. It defines nth order differential equations and describes their general forms. For homogeneous equations, the general solution method involves making an operator form, constructing an auxiliary equation, solving for roots, and finding the complementary solution. For non-homogeneous equations, the method of undetermined coefficients is used to find a particular solution and the general solution is the sum of the complementary and particular solutions. Examples are provided to illustrate the solution methods.

1. HIGHER DIFFERENTIAL EQUATION
PREPARED BY:
Group : B

2. DIFFERENTIAL EQUATIONDIFFERENTIAL EQUATION
nth order differential equation
General form of equation is— 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 ………….Homogenousequation
f(x)≠0 ………….Non-Homogenous equation

3. HIGHER ORDER HOMOGENOUSHIGHER 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)

4. HIGHER ORDER HOMOGENOUSHIGHER ORDER HOMOGENOUS
DIFFERENTIAL EQUATIONDIFFERENTIAL EQUATION
v Theorem for Linear Combination :
y= c1y1+c2y2+c3y3+……………cnyn
— Corollary : If y1 is the sol’n of a homogenous differential
equation then c1y1 will be its general sol’n. & c(-∞, +∞)

5. HIGHER ORDER HOMOGENOUSHIGHER 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

6. HIGHER ORDER HOMOGENOUSHIGHER ORDER HOMOGENOUS
DIFFERENTIAL EQUATIONDIFFERENTIAL EQUATION
—— 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

7. Solution :
The characteristic equation is,
Hence, roots are REAL & DISTINCTREAL & DISTINCT so the general solution is,
Example :

8. — Example :
Solution ;
The characteristic equation is,
So, we have two roots here,
r3= -2 r4= -2
[----y1------ ] [--------y2-----]

9. 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

10. Example : Solve the following differential equation .
Solution
The characteristic equation is ;
The 4 ,4th roots of -16 can be found by evaluating the following
So, we have two sets of complex roots :

11. HIGHER ORDER NON HOMOGENOUSHIGHER 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 :
ü Method 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.

12. METHOD OFMETHOD 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

13. METHOD OFMETHOD 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)

14. 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.

15. 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 :

16. |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 )^3
Since the Wronskian is not zero, the given set in LINEARLY INDEPENDENT.

17. 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.

18. 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).

19. v Example :
y''' + y'' = cos(2t)
Solve :
The characteristic equation is
r3 + r2 = 0
Factoring gives
r2(r + 1) = 0
r = 0 (repeated twice) or r = -1
The homogeneous solution is
yc = c1 + c2t + c3e-t ………………..(i)

20. 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)

21. 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)
-------------------------- &&&&& ---------------------------

22. — 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)

23. 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=A
Since this is a solution to the homogeneous equation, we multiply
by t to get
yp=At
This is also a solution to the homogeneousequation, 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 get
0 +4(2A) = 16
A = 2
Hence
yp1 = 2t2 ………………….(i)

24. Now we work on the second piece. The UC-set for g(t) = 15et is
yp= Aet
Since this in not a solution to the homogeneous equation, we get
yp2 = Aet
yp2' = Aet
yp2'' = Aet
yp2''' = Aet
yp2
(iv) = Aet
Plugging back into the original equation gives
Aet + 4Aet = 15et
5A = 15
A = 3
Hence
yp2 = 3et ………..(ii)
The general solution to the nonhomogeneous differential equation is
y = c1 + c2t + c3sin(2t) + c4cos(2t) + 2t2 + 3et
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