Unit – 1
Q 1:Solve the differential equation y’’’+2y’’-y’-2y=0
Solution:
The corresponding characteristic equation is,
Solving it, we find the roots
=0
=-1
The general solution for the differential equation is
Y(x)= 
Where 
are arbitrary constants.
Q 2: Solve the equation y’’’+11y’-5y=0
Solution:
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.
Q 3: Determine the complementary function for the following D.E
at t=0,1
Solution:
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) = 
Q 4: Solve 
-
Given y=4 ,
when x=0
Solution :
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
Q 5: Solve the following linear equations generally
9x+2y=6 9x+2y=6
3x-y=7
6x-2y=14
Solving the above two equations we get,
9x+2y=6
6x-2y=14
15x = 20
x= 20/15 = 4/3
Q 6: Short cut methods:
(1) 2x-a+4=x+3a-1
Solution: 2x-x=3a-1+a-4 (-a+4 moves to right side)
x= 4a – 5
(2) 5x-7=2x+5
3x = 12 
x= 12/3=4
(3) X-m+3=2m+1
Solution: x= 2m+1+m-3= 3m-2 (-m and +3 to right side)
Q 7 : find the particular integral for the following
Solution:
Now the C.F is as follow,
We find u and v accordingly
= -
= -x
= 
= log(sinx)
P.I = 
= -xcosx +log(sinx)
The general solution is y = C.F +P.I
Y = 
-xcosx +log(sinx)
Q 8: Wronskain ,W= 
= 
= 
=1
Particular integral is given by,
secx= 1/cosx
The general solution is,
Y = C.F +P.I
Y = 
Q 9: Check the validity for cauchy’s mean value theorem f(x) = 
and g(x) = 
on the interval [1,2]
Solution:
The derivatives of this function are,
f’(x) = (x4) = 4x3, g’(x)=(x2) = 2x
substituting in the Cauchy formula we get,
We take into account that the boundaries of the segment are a=1 and b=2
C=
c=
=
Q 10: Check the validity for cauchy’s theorem for the function f(x) = x2 and g(x) = arctanx on the interval [0,1]
Solution:
At first we calculate the derivatives of the given function
f’(x) = (x3)’ , g’(x) = (arctanx)’= 
substitute f(x),g(x) and their derivatives in the Cauchy formula
= 
= 
For the values of a=0, b=1 we obtain,
12c2 = 
c = 
Given we consider the segment [0,1],we choose the positive value for c which lies in interval(0,1)
C=
0.60
Legendre’s formul:
= 
Where p is a prime and 
is the exponent of p in the prime factorizatio of n! And sp(n) is the sum of the digits of n when written in bas p
Q 11: Using the first form of Legendre’s equation , substitute the values in the formula
The values are n=27 and p =2
Solution :
Consider the Legendre’s formula,
= 
substituting the given values in the formula we get,
= 
= 
= 
= 27 – s2(27)
the number 27 when expressed in the base 2 gives rise to 11011.this gives us,
s2(27) = 1+1+0+1+1 = 4
= 27 – s2(27) = 27 – 4 =23
Which means that the larget integer k for which 2k divides 27! Is 23.
Q 12:
....(2)
Solve the given simultaneous differential equations
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,
1
2
2Dx + 2y = 2et
D 
2Dx +D2y = e-t
y = Ae2t + Be-2t + 
Q 13:
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
Q 14: Finding the optimal current of an electrical circuit(RL circuits) in which the initial Condition is i=0 at t=0
Solution:
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=
Hence we complete the solution by first order differential equation of first order and even several types of networking circuits and fluid mechanics uses this method.
