Find General Solution to Equation

In summary: I see you edited your first post. You must have seen my question in the meantime.ehildTo repeat my question(s) ...What do you get for yP' ?What do you get for yP'' ?I see you edited your first post. You must have seen my question in the meantime.ehildI did see it and edited it before you posted the question, sorry about that. I don't have the time to write out the detail of the algebra but i goty' = e2x(-Csinx + Dcosx)y'' = e2x(-Ccosx - Dsinx)which give (-3D-C)cosx + (3C-D)
  • #1
JoshMaths
26
0

Homework Statement



y'' - 3y' + 2y = e2xcos x

The Attempt at a Solution



So this is a second order inhomogeneous equation.

gives λ = {2,1}
get yCF giving Ae2x + Bex
so yP = e2x(Ccosx + Dsinx)

Here is where is gets a bit fuzzy.

Subbing this into the top equation gives me 2e2x((-2D-C)cosx + (2C-D)sinx)

which is wrong, i have gone through it and am missing something can you help me find this correct expression? Thanks.
 
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  • #2
Well, it depends on what method you used to find ##y_p##. Personally, i use the inverted operator technique (but there is apparently also the variation of parameters method).

Here is what i would do: Use the shift theorem to separate ##e^{2x}## (moves to the left) from ##\cos x##. $$\frac{ke^{ax}.\phi (x)}{L(D)}\to ke^{ax}.\frac{\phi (x)}{L(D+a)}$$where k=1, a=2.
From there, use the appropriate inverted operator technique on ##\cos x## which is: $$\frac{k\cos (ax+b)}{L(D)} \to \frac{k\cos (ax+b)}{L(-a^2)}$$ where k=1, a=1, b =0.
 
  • #3
JoshMaths said:
So this is a second order inhomogeneous equation.

gives λ = {2,1}
get yCF giving Ae2x + Bex
so yP = e2x(Acosx + Bsinx)

Correct so far.


JoshMaths said:
Here is where is gets a bit fuzzy.

Subbing this into the top equation gives me 2e2x((-2D-C)cosx + (2C-D)sinx)

What are C and D? Show your work in detail, please.

ehild
 
  • #4
JoshMaths said:

Homework Statement



y'' - 3y' + 2y = e2xcos x

The Attempt at a Solution



So this is a second order inhomogeneous equation.

gives λ = {2,1}
get yCF giving Ae2x + Bex
so yP = e2x(Ccosx + Dsinx)

Here is where is gets a bit fuzzy.

Subbing this into the top equation gives me 2e2x((-2D-C)cosx + (2C-D)sinx)

which is wrong, i have gone through it and am missing something can you help me find this correct expression? Thanks.
Subbing into the top equation, I get e2x((-C+D)cosx + (-C-D)sinx) for the left hand side.
 
  • #5
SammyS said:
Subbing into the top equation, I get e2x((-C+D)cosx + (-C-D)sinx) for the left hand side.

That is correct, Sammy. :cool:

ehild
 
  • #6
ehild said:
Correct so far.

What are C and D? Show your work in detail, please.

ehild

Yep, my mistake, C and D were in y particular but I always mix them up accidentally. It has been fixed.

SammyS said:
Subbing into the top equation, I get e2x((-C+D)cosx + (-C-D)sinx) for the left hand side.

Yep, this is right, i tried and i got (-C-3D)cosx +(3C-D)sinx i guess i am just not seeing it. I understand if you don't have time to write out some of the algebra manipulations but they would be useful if anyone could find a minute. If not, i'll keep working at it.
 
  • #7
JoshMaths said:
Yep, my mistake, C and D were in y particular but I always mix them up accidentally. It has been fixed.

Yep, this is right, i tried and i got (-C-3D)cosx +(3C-D)sinx i guess i am just not seeing it. I understand if you don't have time to write out some of the algebra manipulations but they would be useful if anyone could find a minute. If not, i'll keep working at it.
What do you get for y' and y'' ?
 
  • #8
SammyS said:
What do you get for y' and y'' ?

I have tried with y'' + 3y' + 2y = e2xcos x

and get e2x((11C+7D)cosx + (11D-7C)sinx) which is correct so i think it was just that specific example i was having trouble with, but i understand the principle behind it better now.
 
  • #9
To repeat my question(s) ...

What do you get for yP' ?

What do you get for yP'' ?
 

What is the general solution to an equation?

The general solution to an equation is a formula or expression that can be used to find all possible solutions to that equation. It includes all possible values for the variables in the equation.

How do you find the general solution to an equation?

To find the general solution to an equation, you must first rearrange the equation to isolate the variable on one side. Then, you can use algebraic manipulation to find the general solution. It may involve solving for multiple variables or using inverse operations to isolate the variable.

Why is it important to find the general solution to an equation?

Finding the general solution to an equation is important because it allows us to find all possible solutions to that equation. This can be useful in many areas of science, such as physics and engineering, where equations are used to model real-world situations and the general solution helps us understand all possible outcomes.

Are there any limitations to finding the general solution to an equation?

Yes, there are some limitations to finding the general solution to an equation. In some cases, the general solution may involve imaginary or complex numbers, which may not be applicable to the real-world situation. Additionally, there may be certain values or conditions that are not included in the general solution and require further analysis.

Can the general solution to an equation be used to find a specific solution?

Yes, the general solution to an equation can be used to find a specific solution by plugging in specific values for the variables. This allows us to find a particular solution that satisfies the given equation. However, it is important to note that the general solution may not always give us the most accurate or practical solution, and further analysis may be needed.

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