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Assignment No.01
MTH401 (Fall 2012)
Total marks:30
Question#1
(i)
Marks 10
Solve the following integrals using partial fraction technique:
(2)
dy
a)
  y  1 y  1
b)
  x  1 x  1
dx
Solution:
(a)
We resolve

dy
  y  1 y  1 integrals, into partial fraction;
1  1
1 


 dy
2    y  1  y  1 
1
1
1
 y  1 dy    y  1 dy 


2
1
 ln y  1  ln y  1   C
2
1 y 1
 ln
C
2 y 1

(b)
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We resolve

dx
  x  1 x  1
integrals, into partial fraction.
1  1
1 


 dx

2   x  1  x  1 
1
1
1
x  1 dy    x  1 dy 



2
1
 ln x  1  ln x  1   C
2
1 x 1
 ln
C
2 x 1

(ii) Simplify the following equation using logarithm rules:
1
1
1
1
ln y  1  ln y  1  ln C  ln x  1  ln x  1
2
2
2
2
(2)
Solution:
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1
1
1
1
ln y  1  ln y  1  ln C  ln x  1  ln x  1
2
2
2
2
1
1
1
1



 



ln  y  1 2   ln C  ln  y  1 2   ln x  1 2   ln  x  1 2 



 



According to Multiplication Law of logarithm;
1
1
1



 
ln  y  1 2 * C   ln  y  1 2   ln x  1 2



 
1
1




2* C
2
y

1
ln
x

1



 Ans.
ln
 ln
1
1




 y  1 2 
 x  1 2 
1



2

ln
x

1






(iii) Find the value of "C " from the following equation, when x  2 and y  2 :
ln
y 1
x 1
 C  ln
y 1
x 1
(2)
Solution:
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ln
y 1
x 1
 C  ln
y 1
x 1
Putting the value of x and y,which is 2 & 2.
ln
2 1
2 1
 C  ln
2 1
2 1
ln
3
3
 C  ln
1
1
ln 3  C  ln 3
ln 3 C
e
e
3 C  3
ln 3
3
3
C  1Ans.
C
(iv) Solve the following differential equation subject to the indicated initial condition:
dy y 2  1

dx x 2  1
; y(2)  2
(4)
Solution:
We can also write
dy
dx
 2
2
y 1 x 1
Resolve into partial fraction,
1  1
1 
1  1
1 



 dy   
 dx

2   y  1  y  1 
2   x  1  x  1 
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Rational fraction of integration,
1 y 1 1 x 1
ln
 ln
C
2 y 1 2 x 1
According to condition that y=2 when x=2 we get
1 2 1 1 2 1
ln
 ln
C
2 2 1 2 2 1
1 1 1 1
ln  ln  C
2 3 2 3
1 1 1 1
ln  ln  C
2 3 2 3
Question#2
Marks 10
Reduce the following differential equation into a separable form by taking appropriate
substitution:
dy
4x  7 y  2

dx
4x  7 y  3
Solution:
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Since
a1
b
1 1 ,
a2
b2
taking subsitute of z =4x+7y,
So,
dy 1  dz

   4
dx 7  dx

It becomes
1  dz
 z2
  4 
7  dx
 z3
 dz

 z 2
  4  7

 dx

 z 3
dz
7 z  14
4
dx
z3
dz 7 z  14

4
dx
z3
dz 7 z  14  4 z  12

dx
z3
dz 11z  26

Ans.
dx
z3
Question#3
Marks 10
(i) Determine whether the following differential equation is exact or not?
 6 xy  dx   4 y  9 x 2  dy
(2)
Solution:
 6 xy  dx   4 y  9 x 2  dy
 6 xy  dx   4 y  9 x 2  dy  0
Now,
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M   6 xy  dx
N   4 y  9 x 2  dy
M
 6x
y
N
 18 x
x
M N

y
x
So, we can say, this differential equation is not exact
(ii) If the differential equation in (i) is not exact then reduce it into an exact form by
multiplying it with an appropriate integrating factor:
(4)
Solution:
For finding the integrating factor,
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M N

y
x 6 x  18 x

M
6 xy
12 x

6 xy
2

y
function of y only,then
2
u(y)=e
  y dy
So, y 2 is integrating factor which multiply it and we find this equation,
 6 x y  dx   4 y
3
3
 9 x 2 y 2  dy
(iii) Determine the implicit solution of the following differential equation:
 6 x y  dx   4 y
3
3
 9 x 2 y 2  dy
(4)
Solution:
 6 x y  dx   4 y
3
3
 9 x 2 y 2  dy  0
As we know
M  6x y 3
N  4 y 3  9x 2 y 2
M
N
 18 xy 2 
y
x
So, exact equation is
 6 x y  dx   4 y
3
3
 9 x 2 y 2  dy  0
Taking integration
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  6 x y  dx    4 y  9 x y  dy  0
  6 x y  dx   4 y dy  9 x y dy  0
3
3
3
3
2
2
2
2
 6x2 y 3   4 y 4   9 x 2 y 3 



0
 2   4   3 
3x 2 y 3  y 4  3x 2 y 3  c
y 4  c
c   y 4 Ans.
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