1. Originally Posted by mr fantastic
NO! Did you substitute it into the DE and check if it worked. You would see in a flash that it doesn't.

The particular solution is of the form $y = \alpha$ where $\alpha$ is a constant whose value you must find.
I didn't. Sorry.

Ok I've subbed the e^(x) into the function, and I find that it equals zero. so, can the '2' part be e^(x)? Or is there a proper method to do this? I don't see any examples or theory in my notes or textbooks about constant...

Thank you for your continued help Mr. Fantastic

P.S. I haven't come across your notation of y = alpha before.... Can you explain?

2. Hey Danny,

Thanks for your continued support and assistance - much appreciated

With regards to your 'guess' equation, assuming i'm wrong and your right, why is your guess equation y = A + Be^(-2x) and not what mine is: y = A - Be^(-2x) ?

Originally Posted by danny arrigo
In this solution we have two parts, the first the complimentary and the second the particular. We notice the extra x so in order to guess a particular solution when rhs contains the complimentary solution we "bump" up the exponential in our first guess and use

$y = A + Bx e^{-2x}$

$y' = B e^{-2x} -2x Bx e^{-2x}$
$y'' = -4B e^{-2x} + 4x Bx e^{-2x}$
I do not comprehend this part...So, you purposely added an extra 'x'? How did you achieve your y' - where did the second 'B' come from? And where did the '-2x' come from? where did the 4x come from? Is that a simple typo?
Sorry to bombard you with question....this topic is frustrating...Also, would you agree on, where you've typed 'complementary equation is y' - y = 0' also can be referred to as the homogeneous equation?

Also,

From which we obtain and the particular solution

this is what the final product is supposed to look like?

Thanks Danny

3. Originally Posted by Prove It
Where did the t's come from?
Yes well, I dont know. But I've mentally rectified this problem...

4. Originally Posted by tsal15
I didn't. Sorry.

Ok I've subbed the e^(x) into the function, and I find that it equals zero. so, can the '2' part be e^(x)? Or is there a proper method to do this? I don't see any examples or theory in my notes or textbooks about constant...

Thank you for your continued help Mr. Fantastic

P.S. I haven't come across your notation of y = alpha before.... Can you explain?
Please read all of the thread carefully. The answers to all these questions are given and, in fact, the solution is all but given too.

5. Originally Posted by tsal15
With regards to your 'guess' equation, assuming i'm wrong and your right, why is your guess equation y = A + Be^(-2x) and not what mine is: y = A - Be^(-2x) ?
They're are both correct. In the end we will get number for the constants. Our numbers for B will differ by a negative but in the end we will get the same $y_p$

Originally Posted by tsal15
I do not comprehend this part...So, you purposely added an extra 'x'? How did you achieve your y' - where did the second 'B' come from? And where did the '-2x' come from? where did the 4x come from? Is that a simple typo?

$y = A + B x e^{-2x}$

The derivatives (using the product rule) are

$y' = B e^{-2x} -2 B x e^{-2x}$
$y'' = -4 B e^{-2x} + 4 B x e^{-2x}$

6. Originally Posted by tsal15
Also, would you agree on, where you've typed 'complementary equation is y' - y = 0' also can be referred to as the homogeneous equation?
Yes, the complimentary equation is always homeogeneous. As the word says, it compliments the original ODE.

For your final anwswer, add $y_p$ to your complimentary solution.

7. Originally Posted by mr fantastic
Please read all of the thread carefully. The answers to all these questions are given and, in fact, the solution is all but given too.
OK I've figured out my problem.... Danny, Mr. Fantastic, Prove it, and everyone else that can help... What is the main process to solving a non-homogenous linear 2nd order differential equation? I think then I'll be able to do this sort of thing...Thanks heaps

8. Originally Posted by danny arrigo
Yes, your complimentary solution is right (via your characteristic equation) except you want x's and not t's. Let me explain a little bit more about the construction of this particular solution. There are primary two methods to do this (1) the method of undetermined coefficients and (2) the variation of parameters. The first method is usually easier to use but involves some educated guessing. The method basic seeks the form of the particular solution by looking at the rhs of the ODE. If it has power's guess power's, exponential's guess exponential's, sin's and cos's guess sin's and cos's. In your example

$y'' + y' -2y = 2 - 2e^{-2x}$

it has powers $x^0$ and exponential's $e^{-2x}.$ Therefore one would try a guess of

$y = A + B e^{-2x}$

Taking derivatives

$y' = -2 B e^{-2x},\;\;\;y'' = 4 B e^{-2x},$

and substituting gives

$y'' + y' - 2y = 4 B e^{-2x} + \left( -2 B e^{-2x}\right) - 2 \left( A + B e^{-2x}\right) = 2 - 2e^{-2x}$

Expanding gives $-2A = 2 - 2e^{-2x}$ (yes, the B's cancel but why?) The reason is that $y = Be^{-2x}$ is part of the complimentary solution so

$y'' + y' - 2y = 0$ identically! So in this case, what do we do for a guess. Let us onsider the following first order example

$(1)\;\;\; y' - y = e^x$

The complimentary equation is

$y' - y = 0$

the characteristic equations is

$m - 1 = 0$

the complimentary solution is

$y = e^x$

However, since eqn. (1) is linear we solve giving

$y = c e^x + x\,e^x$

In this solution we have two parts, the first the complimentary and the second the particular. We notice the extra x so in order to guess a particular solution when rhs contains the complimentary solution we "bump" up the exponential in our first guess and use

$y = A + Bx e^{-2x}$

$y' = B e^{-2x} -2x Bx e^{-2x}$
$y'' = -4B e^{-2x} + 4x Bx e^{-2x}$

substituting gives

$y'' + y' - 2y = -4B e^{-2x} + 4 Bx e^{-2x} + \left( B e^{-2x} -2 Bx e^{-2x} \right) - 2 \left( A + Bx e^{-2x} \right)=$
$= 2 - 2e^{-2x}$

and when we expand we get

$-3B e^{-2x} - 2 A = 2 - 2e^{-2x}$

from which we obtain $A = -1,\;\;B = \frac{2}{3}$ and the particular solution

$y_p = - 1 +\frac{2}{3} \,e^{-2x}$
Ok so becuase the initial trial was identically equal to zero we couldn't use it right?

9. Originally Posted by tsal15
Ok so becuase the initial trial was identically equal to zero we couldn't use it right?
Yes, that's why we need to "bump" up the guess and include an $x$

10. Originally Posted by danny arrigo
$y_p = - 1 +\frac{2}{3} \,e^{-2x}$
Is the $y_p$ supposed to have an extra x? i.e. $y_p = -1 +\frac{2}{3}xe^{-2x}$ ?

In regards to 'bump' up, my text doesn't refer to it at all...its a pretty crappy text...

Thanks again Danny

11. ok so because this (the o.d.e we've been working out) o.d.e is non linear, we use $y_c(x) = c_1e^{r_1 x} + c_2e^{r_2 x}$ and not $y_c(x) = c_1e^{rx} + c_2xe^{rx}$ ?

Thank you all

12. Originally Posted by tsal15
Is the $y_p$ supposed to have an extra x? i.e. $y_p = -1 +\frac{2}{3}xe^{-2x}$ ?

In regards to 'bump' up, my text doesn't refer to it at all...its a pretty crappy text...

Thanks again Danny
Yes, sorry I missed the $x$.

13. Originally Posted by tsal15
ok so because this (the o.d.e we've been working out) o.d.e is non linear, we use $y_c(x) = c_1e^{r_1 x} + c_2e^{r_2 x}$ and not $y_c(x) = c_1e^{rx} + c_2xe^{rx}$ ?

Thank you all
The ODE is linear. The complimently solution is

$y_c = c_1 e^x + c_2 e^{-2x}$

and to this we add the particular solution

$y_p = -1 + \frac{2}{3} x e^{-2x}$

which gives the final solution as

$y = c_1 e^x + c_2 e^{-2x} -1 + \frac{2}{3} x e^{-2x}$

14. Originally Posted by danny arrigo
However, since eqn. (1) is linear we solve giving

$y = c e^x + x\,e^x$
But you said earlier, the above... is this a different case?
I know that eqn (1) was an example but the fact you refer to it as a linear d.e is what i'm picking up on...

Thanks

Page 2 of 2 First 12