Unit-1
Linear differential equations
A homogeneous linear ordinary differential equation with constant coefficients is an ordinary differential equation in which coefficients are constants, all terms are linear, and the entire differential equation is equal to zero,
The form of second-order linear differential equation with constant coefficients is,
Where a,b,c are the constants.
Let, aD²y+bDy+cy = f(x), where d² = , D =
∅(D)y = f(x) , where ∅(D)y = aD²y+bDy+cy
Here first we solve, ∅(D)y = 0, which is called complementary function(C.F)
Then we find particular integral (P.I)
P.I. = f(x)
General solution = C.F. +P.I.
Linear differential equations are those in which the independent variable and its derivative occur only in the first degree and are not multiplied together.
Thus the general linear differential equation of the nth order is of the form
Where and X is a function of x.
Linear differential equation with constant coefficient is of the form-
Where are constants.
Let’s do some examples to understand the concept,
Example1: Solve (4D² +4D -3)y =
Solution: Auxiliary equation is 4m² +4m – 3 = 0
We get, (2m+3)(2m – 1) = 0
m = ,
complementary function: CF is A+ B
now we will find particular integral,
P.I. = f(x)
= .
= .
= .
= . = .
General solution is y = CF + PI
= A+ B .
Differential operators
D stands for operation of differential i.e.
stands for the operator of integration.
stands for operation of integration twice.
Thus,
Note:-Complete solution = complementary function + Particular integral
i.e. y=CF + PI
Rules to find the complementary function-
To solve the equation-
This can be written as in symbolic form-
Or-
It is called the auxiliary equation.
Let be the roots-
Case-1: If all the roots are real and distinct, then equation (2) becomes,
Now this equation will be satisfied by the solution of
This is a Leibnitz’s linear and I.F. =
Its solution is-
The complete solution will be-
Case-2: If two roots are equal
Then the complete solution is given by-
Case-3: If one pair of roots be imaginary, i.e. then the complete solution is-
Where and
Case-4: If two points of imaginary roots are equal-
Then the complete solution is-
Example-Solve
Sol.
Its auxiliary equation is-
Where-
Therefore the complete solution is-
Example: Solve
Ans. Given,
Here the Auxiliary equation is
Example: Solve
Or,
Ans. Auxiliary equation is
Note: If roots are in complex form i.e.
Example: Solve
Ans. Auxiliary equation is
Rules to find Particular Integral
Case 1:
If,
If,
Example: Solve
Ans. Given,
Auxiliary equation is
Case2:
Expand by the binomial theorem in ascending powers of D as far as the result of an operation on is zero.
Example.
Given,
For CF,
Auxiliary equation is
For PI
Case 3:
Or,
Example:
Ans. Auxiliary equation is
Case 4:
Example: Solve
Ans. AE=
The complete solution is
Example: Solve
Ans. The AE is
Complete solution y= CF + PI
Example: Solve
Ans. The AE is
Complete solution = CF + PI
Example: Solve
Ans. The AE is
Complete solutio0n is y= CF + PI
Example: Find the PI of
Ans.
Example: solve
Ans. Given equation in symbolic form is
Its Auxiliary equation is
Complete solution is y= CF + PI
Example: Solve
Ans. The AE is
We know,
Complete solution is y= CF + PI
Example: Find the PI of(D2-4D+3)y=ex cos2x
Ans.
Example. Solve(D3-7D-6) y=e2x (1+x)
Ans. The auxiliary equation i9s
Hence complete solution is y= CF + PI
A general method of finding the PI of any function
Or
Or
It is the LDE.
The solution will be-
Example:
Solve
Solution:
Auxiliary equation
Complementary function
Complete Solution is
Example:
Solve
Solution:
Auxiliary equation
C.F is
[]
The Complete Solution is
Example
Solve
Solution:
The Auxiliary equation is
The C.F is
P.I
The Complete Solution is
Example
Solve
Solution:
The Auxiliary equation is
The C.F is
P.I
Now,
The Complete Solution is
Example
Solve
Solution:
The auxiliary equation is
The C.F is
[Put ]
The Complete Solution is
Example
Solve
Solution:
The auxiliary equation is
The C.F is
But
The Complete Solution is
Example
Solve
Solution:
The auxiliary equation is
The C.F is
[By parts]
The Complete Solution is
Example
Solve
Solution:
The auxiliary equation is
The C.F is
Now,
And
The Complete Solution
Consider a second-order LDE with constant coefficients given by
Then let the complimentary function is given by
Then the particular integral is
Where u and v are unknown and to be calculated using the formula
u=
Example-1: Solve the following DE by using a variation of parameters-
Sol. We can write the given equation in symbolic form as-
To find CF-
It’s A.E. is
So that CF is-
To find PI-
Here
Now
Thus PI =
=
=
=
=
So that the complete solution is-
Example-2: Solve the following by using the method of variation of parameters.
Sol. This can be written as-
C.F.-
The auxiliary equation is-
So that the C.F. will be-
P.I.-
Here
Now
Thus PI =
=
=
So that the complete solution is-
Where, are constant is called Cauchy’s homogenous linear equation.
Put,
Example.
Ans. Putting,
AE is
CS = CF + PI
Example: Solve
Ans. Let,
AE is
y= CF + PI
Example: Solve
Ans. Let, so that z = log x
AE is
Legendre’s differential equation-
A differential equation of the form-
Where k is the constants and X is a function of x, is known as Legendre’s linear differential equation.
We can reduce such type of equations to linear equations with constant coefficient by the substitution as-
i.e.
Example: Solve
Sol. As we see that this is Legendre’s linear equation.
Now put
So that-
And
Then the equation becomes- D (D – 1)y+ Dy + y = 2 sin t
Its auxiliary equation is-
And particular integral-
P.I. =
Note -
Hence the solution is -
The equation involving the derivatives of two or more dependent variables, which are the functions of a single independent variable, is called a simultaneous equation.
Such as-
Example 1: Solve the following simultaneous differential equations-
....(2)
Solution:
Consider the given equations
....(2)
Consider eq(1),(2)
Dx+2y = et....(1)
Dx +2x =e-t....(2)
Eliminating ‘x’ from both the equations we get,
12 2Dx + 2y = 2et
D 2Dx +D2y = e-t
y = Ae2t + Be-2t +
Example 2: Solve the following simultaneous differential equations-
Given that x(0)=1 and y(0)= 0
Solution:
Consider the given equations,
Dy +2x = sin2t
Dx -2y = cos2t
By solving the above equations we get,
(D2 +4)Y =0
X(0) = 1, y(0) = 0
A =0, B=-1
Example-3: Solve the following simultaneous differential equations-
It is given that x = 0 and y = 1 when t = 0.
Sol. Given equations can be written as-
Dx + 2y = - sin t ………. (1) and -2x + Dy = cos t ……… (2)
Eliminate x by multiplying (1) by 2 and (2) by D then add-
Here A.E =
So that C.F. =
And P.I. =
So that- …………. (3)
And ………….. (4)
Substitute (3) in (2), we get-
2x = Dy – cos t =
………… (5)
When t = 0, x = 0, y = 1. (3) and (5) gives-
Hence
References
1. Erwin Kreyszig, Advanced Engineering Mathematics, 9thEdition, John Wiley & Sons, 2006.
2. N.P. Bali and Manish Goyal, A textbook of Engineering Mathematics, Laxmi Publications.
3. Higher engineering mathematic, Dr. B.S. Grewal, Khanna publishers