UNIT 1
Linear Differential Equations (LDE) and Applications
Q1) Solve the equation y’’’+11y’-5y=0
A1)
The characteristic equation of the give D.E is
Here one of the root is 
then factorising the term 
from the equation we obtain
(
(
(
=0
(
(
=0
Thus the equation has two roots 
Hence the general equation of the D.E is
Y(x)=(
Where 
are arbitrary constants.
Q2) Solve 
A2) The given equation can be written in the symbolic form as-
So that A.E is 
Hence the complete solution is-
Q3) Determine the complementary function for the following D.E
at t=0,1
A3)
We can write the given homogeneous equation as follows
The auxiliary equation is 
By factorising the auxiliary equations we get,
(m-3)(m-4)=0
And the two real solutions are m=3,4
Hence the complementary function is,
Since f(t) = 2 then 
= c
12c=2
c=1/6 and 
= 1/6
Therefore the general solution is
X(t) = 
Q4) Solve 
-
Given y=4 ,
when x=0
A4)
The auxiliary equation is
(m-1)(m-2)=0
m=1,2
Therefore complimentary function is,
The particular integral is given by y=
Substituting the above values in given equation we get,
-
= 0-3
by comparing the co-efficients we get 2
=1 
= ½
-3
+ 2
= 4+x
= 11/4
Therefore P.I is,
11/4+(1/2)x
General solution is, y = A
+ 11/4 +1/2x
Q5) Find P.I of 
A5) P.I. = 
Put 
= 
Multiply and divide by 1+4D
= 
= 
Q6) Find P.I. Of 
A6) P.I = 
Replace D by D+1
Put 
Q7) Solve the following by using the method of variation of parameters.
A7) This can be written as-
C.F.-
Auxiliary equation is- 
So that the C.F. Will be- 
P.I.-
Here 
Now 
Thus PI = 
= 
= 
So that the complete solution is-
Q8) Solve 
A8) On putting 
so that,
and 
The given equation becomes-
Or it can be written as-
So that the auxiliary equation is- 
C.F. = 
Particular integral- 
Where 
It’s a Leibnitz’s linear equation having I.F.= 
Its solution will be-
P.I. = 
= 
So that the complete solution is-
Q9) Solve the following simultaneous differential equations-
Given that x(0)=1 and y(0)= 0
A9)
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
Q10) Finding the optimal current of an electrical circuit (RL circuits) in which the initial
Condition is i=0 at t=0
A10)
By Kirchhoff voltage law (KVL) method, we get
The differential equation for the RL circuit will be 
In which initial conditions are i=0 at t=0
The standard form of the equation is,
Dividing the differential equation by L to obtain
The integrating factor is
Multiplying the above equation with standard form gives rise to
By applying integration on both sides we get
Now applying i=0 at t-0 gives us
0=
C=-
NOW
i=
= t 
Therefore by finding current of the RL circuits is i=
