Hi,
Thank you for your question. I have worked out the solution to the two
problems in a step by step manner as you have requested.
Problem #1:
(X^2)y''+xy'-6y=10 ; y(1)=1,y'(1)=-6
This for my convenience, I will reproduce as:
[(x^2)D^2 + xD - 6]y = 10 equation 1
Note: when you multiply y into the '[' you will get the original
expression.
Now consider this,
let z = logx
then,
dz/dx = 1/x
or,
(d/dx)/(d/dz) = 1/x
D/D' = 1/x
or D' = xD
where D' = d/dz
As you can see the problem has now reduced to:
[(D'^2) + D' - 6]y = 10 Equation 2
This can be solved the normal manner in which all differential
equation problems of this format are solved.
First the Complementary Function is found,
The auxilary equation from the above equation 2 is :
m^2 + m - 6 = 0
This gives (m+3)(m-2) = 0
or m = -3 and m = 2
From this,
y = c1exp(2z) + c2exp(-3z)
Note: c1 and c2 are constants
The above equation becomes,
y = c1exp(2logx) + c2exp(-3logx)
or,
y = c1x^2 + c2/(x^3)
To find the Particular Integral,
Consider 10 = 10[exp(0*x)]
For this case,
10[exp(0*x)]/ [D'^2 + D' -6]
Substituting D' = 0 for the above,
we get y = -5/3
hence the complete solution is y = c1x^2 + c2/(x^3) - 5/3
----equation 3
Now we have to consider the initial values for this solution.
if x = 1 y = 1
hence,
1 = c1 + c2 - 5/3
or,
c1 + c2 = 8/3 ----------equation 4
From the second condition,
x = 1 y' = -6
Now from equation 3,
y' = 2*(c1)x - 3*(c2)/(x^4)
putting x = 1
y' = 2*(c1) -3*(c2)
or,
2c1 - 3c2 = -6 ------------equation 5
From equation 4 and 5,
c1 = 2/5 and c2 = 34/15
hence the final answer looks like this:
y = (2/5)x^2 + (34/15)/(x^3) - 5/3
____________________________________________________________________________
Problem #2:
The given problem is:
y''-y'-2y=4x^2
This in Laplace format looks like:
( s^2 - s - 2 ) Y(s) = 2/s ---------Equation 1
What does that mean?
's' is expressed in the following equation which happens to be the
basic definition for a Laplace transform:
L[y(x)] = 1/2*pi(integral)(-oo to +oo ) y* exp(-sx)ds
x^2 reduces to 2/s in this case.
From equation 1 above,
Y(s) = 2/[s(s^2 - s - 2)]
Please note that this problem from this point can be solved in many
wasys. i have chosen the simplest method for myself which is the
Partial Fraction method.
(s^2 - s - 2) = (s+1)(s-2)
Hence,
Y(s) = 2/[s(s+1)(s-2)]
Solving,
A/s + B/(s+1) + C/(s-2)
A = -1 ; B = 2/3 and C = 1/3
Hence,
Y(s) = -1/s + (2/3)/(s+1) + (1/3)/(s-2)
Taking Laplace inverse of the above, w get the final solution to the
problem.
y(x) = -x^2/2 + (2/3)exp(-x) + (1/3)exp(2x)
______________________________________________________________________________
There! I have solved both the problems for you. If you notice any
error that might have crept into the solution or if any part of the
answer remains a mystery to you please dont hesitate to clarify.
Warmest Regards,
Shiv Reddy |
Clarification of Answer by
shivreddy-ga
on
27 Nov 2002 08:28 PST
Hi,
The answer to the first equation changes a bit with that little
correction.
Let us re-examine the problem again with this change:
The steps I have outlined above will remain the same and the
Complementary Function can be found out exactly as I have already
shown.
The difference however will be reflected in the calculation of the
Particular Integral.
Particular Integral(PI) is give by,
10*x^2 / [(D')^2 + D' - 6]
now z = logx
x = exp(z)
Hence it follows that,
x^2 = exp(2*z)
Now PI = 10*exp(2*z) / [(D')^2 + D' - 6]
=> 10*exp(2*z) / [2^2 + 2 - 6] This is got by sustituting D'
= 2 in the equation above.
Note: The denominator vaule becomes zero!
In this case, The denominator is differentiated w.r.t D' again...
10*z*exp(2*z) / [2*D' + 1 ]
=> 10*z*exp(2*z) /[ 4 + 1 ] this is also got by
substituting D' = 2 in the equation given above.
Note: Denominator in this case is NOT zero.
=> 10*z*exp(2*z) / 5
=> 2*z*exp(2*z)
Since z = logx the above equation can be altered in terms of x.
=> 2*logx*x^2
=> 2*(x^2)*logx
this is added to the complementary function which from the main answer
that I have give is equal to:
CF (Complementary function) = c1(x^2) + c2/(x^3)
the total solution is,
y = c1(x^2) + c2/(x^3) + 2*(x^2)*logx
Now from the given conditions,
y(1) = 1
putting x = 1;
y = c1 + c2 = 1 ----------------Equation 6
differentiating y to get y'(1)
y' = 2*c1*(x) - 3*(c2)/(x^4) + 2 [ (x^2)/x + 2xlogx]
or,
y' = 2*c1*x - 3*(c2)/(x^4) + 2x + 4xlogx
y'(1) = 2c1 - 3c2 + 2
-6 = 2c1 - 3c2 + 2 ---------------Equation 7
From the above equations, 6 and 7 we get:
2c1 - 3c2 = -8
3c1 + 3c2 = 3
_______________
5c1 = -5
or,
c1 = -1
Also from equation 6,
c1 + c2 = 1
-1 + c2 = 1
or,
c2 = 2.
Hence the final solution is given thus:
y = (-1)*x^2 + 2/(x^3) + 2(x^2)logx
This is the altered solution to the first problem. If you have any
further
doubts regarding these problems please do not hesitate to clarify
before rating this answer.
Thank you for using this service.
Warmest Regards,
Shiv Reddy
|
