Find General Solution to Equation

[JoshMaths]

Homework Statement

y'' - 3y' + 2y = e^2xcos x

The Attempt at a Solution

So this is a second order inhomogeneous equation.

gives λ = {2,1}
get y_CF giving Ae^2x + Be^x
so y_P = e^2x(Ccosx + Dsinx)

Here is where is gets a bit fuzzy.

Subbing this into the top equation gives me 2e^2x((-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.

 

[DryRun]

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.

 

[ehild]

So this is a second order inhomogeneous equation.

gives λ = {2,1}
get y_CF giving Ae^2x + Be^x
so y_P = e^2x(Acosx + Bsinx)

Correct so far.

Here is where is gets a bit fuzzy.

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

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

ehild

 

[SammyS]

Homework Statement

y'' - 3y' + 2y = e^2xcos x

The Attempt at a Solution

So this is a second order inhomogeneous equation.

gives λ = {2,1}
get y_CF giving Ae^2x + Be^x
so y_P = e^2x(Ccosx + Dsinx)

Here is where is gets a bit fuzzy.

Subbing this into the top equation gives me 2e^2x((-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 e^2x((-C+D)cosx + (-C-D)sinx) for the left hand side.

 

[ehild]

Subbing into the top equation, I get e^2x((-C+D)cosx + (-C-D)sinx) for the left hand side.

That is correct, Sammy.

ehild

 

[JoshMaths]

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.

Subbing into the top equation, I get e^2x((-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.

 

[SammyS]

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'' ?

 

[JoshMaths]

What do you get for y' and y'' ?

I have tried with y'' + 3y' + 2y = e^2xcos x

and get e^2x((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.

 

[SammyS]

To repeat my question(s) ...

What do you get for y_P' ?

What do you get for y_P'' ?