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Solution of Assignment # 2
MTH401 (Spring 2012)
Total marks: 30 Lecture # 12- Due date: 06-06-
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Question#1 Marks 10
In the following differential equation the indicated function y 1 x is a solution of the
associated homogeneous equation. Use the method of reduction of order to find a second solution y 2 (^) x (^) of the homogeneous equation and a particular solution of the given
nonhomogeneous equation using method of undetermined coefficients-superposition approach.
1
2 2 2 4 2 ;^1
d y (^) y y e x dx
^
Solution: Let y 2 (^) v x (^) y 1 (^) x is the second solution of the given differential equation. That is we have to determine v x (^) .
Now y 2 and its derivatives are
2
2
2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
x
x x
x x x x
x x x x
x x x x
x x x
y v x e
y v x e e v
y v x e e v e v v e
y v x e e v e v v e
y e v x e v e v e v
y e v x e v e v
Now plugging the above values into associated homogeneous Eq. (^) (^1) so one can solve
for v x (^) ;
^ ^ ^
2 2 2 2
2 2
x x x x
x x
e v x e v e v v x e
e v e v
v v
This is second order linear differential equation, so, substitute w x (^) v ' (^) x into Eq.
2 , it yields
Now, we will determine the particular solution of the given nonhomogeneous equation using method of undetermined coefficients-superposition approach.
Since the function g x 2 , let us assume a particular solution that is in the form of a
constant:
y p A
We seek to determine specific coefficients A for which y (^) p is a solution of (^) 1 ,
substituting y (^) p A and y '' (^) p 0 into (^) 1 , we have
0 4 2 4 2 1 2
A
A
A
Hence,
y p is a particular solution of the given nonhomogeneous differential
equation.
Question#2 Marks 10
Solve the following initial value problem.
0 ;^ 0 ^1
dy p x y y dx
Where
x p x x
^ ^
Hint: Linear differential equations sometimes occur in which the function p (^) x (^) have
jump discontinuities. If x 0 is such a point of discontinuity, then it is necessary to solve
the equation separately for x x 0 and x x 0. Afterwards, the two solutions are
matched so that the function y x is continuous at x (^) 0 ; this is accomplished by a proper
choice of the arbitrary constants. In the given differential equation p (^) x has a jump discontinuity at x 0 (^) 1 , then it is
necessary for all of you to solve the equation separately for x 1 and x 1.
Solution: The given differential equation is
0 ^1
dy p x y dx
Firstly, we will solve it for 0 x 1 , for this we will take p x (^) 2 , then (^) 1 becomes:
(^2 0) 2
dy y dx
The corresponding auxiliary equation is: 2 0 2
m m
Hence the general solution of (^) 2 is
1 1
y c e ^2 x
To find the particular solution of (^) 2 , applying the initial condition
1 1 1
20 0 1
y c e c
So, the particular solution is
1
y e ^2 x
Now, we will solve 1 for x 1 , for this take p x 1 , then 1 becomes:
0 3
dy y dx
The corresponding auxiliary equation is: 1 0 1
m m
Hence the general solution of (^) 3 is
y (^) 2 c e 2 ^^ x a
Hence the solution of (^) 1 is
2
x x
x e y x c e x
^
Above is discontinuous function, and we are interested in continuous solution, because physics demands the continuous solution. So, the above solution should be continuous
at x 1. For this c 2 must be properly choosed.
There may be many techniques to choose c (^) 2 but according to my point of view we should
check the continuity of (^) 4 at x 1. So,
(^) 2
2
lim lim 1 1
y x e^ x x x
e
Moreover,
(^) 2
2
1
lim lim 1 1
y x c e^ x x x c e
^
