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# Is this right? watch

1. This differential equation:
where

Characteristic Equation:

Thus:

General Solution:

I don't have the marking scheme and I wanted to double-check if this is right.
2. (Original post by Alpha-Omega)
This differential equation:
where

Characteristic Equation:

Thus:

General Solution:

I don't have the marking scheme and I wanted to double-check if this is right.
I don't think this right
Consider that the equation will be 0 when
1.

2.
3. (Original post by ztibor)
I don't think this right
Consider that the equation will be 0 when
1.

2.
Hmm... do you know how to get the correct solution?
4. (Original post by Alpha-Omega)
Hmm... do you know how to get the correct solution?
It depends on L
A simple multiplication of two operators
as you wrote in the first equation
or one linear operator meaning

as you wrote with characteristic polinomial
I think to this
Then your method is correct with that

For simple multiplication you will get same base solutions
but then the real and complex solutions will be separate solutions so you can not get the linear combination of them (only for the two complex conjugate to get real functions) because the initial equation is not linear.
1.

2.
THe base solutions

so for real

So another solution for x(t)
5. (Original post by ztibor)
It depends on L
A simple multiplication of two operators
as you wrote in the first equation
or one linear operator meaning

as you wrote with characteristic polinomial
I think to this
Then your method is correct with that

For simple multiplication you will get same base solutions
but then the real and complex solutions will be separate solutions so you can not get the linear combination of them (only for the two complex conjugate to get real functions) because the initial equation is not linear.
1.

2.
THe base solutions

so for real

So another solution for x(t)

Although, I don't fully understand why do the real and complex solutions have to be separated if the initial equation is not linear. Could explain this a bit further or give me a link to your reference. Cheers
6. (Original post by Alpha-Omega)

Although, I don't fully understand why do the real and complex solutions have to be separated if the initial equation is not linear. Could explain this a bit further or give me a link to your reference. Cheers

In first half of my answer I assume L is one linear operator of

so

does not mean the following product as you wrote

but
equation
I wrore that you have the correct solution for this except two m values;

So the base solutions are
from this the general solution in Complex is the linear combination of base solutions (only if the equation is linear and it is)

or functions with real variables (with further linear combinations as i wrote)

At the second half of my answer I assumed that
where

and

so
which means

That is

OR

Which means to solve the two equations separately and not simoultaneously
because
involves L(x)=0
and involves that too.
So there is two solutions for x(t)
from the 1st equation

from the 2nd

or for real functions

http://en.wikipedia.org/wiki/Charact...ion_(calculus) (Characteristic equation)
http://en.wikipedia.org/wiki/Superposition_principle (Superposition)
7. (Original post by ztibor)
In first half of my answer I assume L is one linear operator of

so

does not mean the following product as you wrote

but
equation
I wrore that you have the correct solution for this except two m values;

So the base solutions are
from this the general solution in Complex is the linear combination of base solutions (only if the equation is linear and it is)

or in real (with further linear combinations as i wrote)

At the second half of my answer I assumed that
where

and

so
which means

That is

OR

Which means to solve the two equations separately and not simoultaneously
because
involves L(x)=0
and involves that too.
So there is two solutions for x(t)
from the 1st equation

from the 2nd

or for real functions

I have no queation or answer books or links just my knowledge and skills
Substituting back the solution into the original equation it works, and it is enough to prove that this was a good solution.
Cheers, that makes sense.

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Updated: April 7, 2013
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