Differential Equations-Method of Undetermined Coefficients

[cookiemnstr510510]

Homework Statement

Find the particular solution
y''-y'+9y=3sin3t
My question is related to this and the homogeneous sol'n and the general sol'n

Relevant Equations

possibly
[r=+-a]-->cosh(ax),sinh(ax)
or
[r=+-bi]-->cos(bx),sin(bx)

So to answer my question I need to reference another problem, I hope i won't get flagged for this... it is only to make a point about the way I am trying to approach this current problem.
the previous problem stated:
y''+2y'-y=10
so first I am finding Y_(homogeneous) and going straight to the auxiliary equation:
after completing the square→(r+1)^2=2→r=-1±√2 [since when r=±a, we get cosh(ax) and sinh(ax), the solution becomes...]
Y_(H)=e^((-1±√2)t) →e^(-t)⋅e^((±√2)t)→e^-t[C1cosh((√2)t)+C2sinh((√2)t)] etc. etc.

My question: with the current problem y''-y'+9y=3sin3t, when finding the homogeneous equation(solution?) after finding my r values which I will show now:
Aux: r^2-r+9=0 and solving for r I get: r=(1/2)±(√-35)/2 using this to find the homogeneous equation (shown 3 lines up with the other problem) am I using the same property of r, which is when r=±a we get cosh(ax) and sinh(ax)? or am I using the property that says when r=±√-# we get ±i out of it?

Sorry if this question is confusing, I tried to explain my logic as much as possible.
I see similarities in both r equations:
r=-1±√2 I know you use cosh and sinh with this one?
r=(1/2)±(√-35)/2 If i don't use cosh and sinh with this one why not?

I've been looking at my textbook, the internet, and my notes and cannot why you would get "i" for some r's and cosh and sinh for others.

Thanks!

 

[lurflurf]

it is desired to solve
$$(a D^2+b D+c)y=f(t)$$
we complete the square
$$\frac{1}{4a}[ 4a^2 D^2+4a b D+b^2-b^2+4a c ]y= f(t)$$
$$\frac{1}{4a}[ (2a D+b)^2-(b^2-4a c)]y= f(t)$$
the homogeneous solution will have one of three forms
if b^2-4a c=0
$$y_H=C_1+C_2 t$$
if b^2-4a c<0
$$y_H=\left[ C_1 \cos\left(\frac{\sqrt{4a c-b^2}}{2a}\, t\right)+C_2 \sin\left(\frac{\sqrt{4a c-b^2}}{2a}\, t\right) \right]\exp\left(-\frac{b}{2a}\right)$$
if b^2-4a c>0
$$y_H=\left[ C_1 \cosh\left(\frac{\sqrt{b^2-4a c}}{2a}\, t\right)+C_2 \sinh\left(\frac{\sqrt{b^2-4a c}}{2a}\, t\right) \right]\exp\left(-\frac{b}{2a}\right)$$

 

[alan2]

You’re making this very complicated. Did your professor not derive the three cases for you?

 

[cookiemnstr510510]

You’re making this very complicated. Did your professor not derive the three cases for you?

no, unfortunately not

 

[cookiemnstr510510]

it is desired to solve
$$(a D^2+b D+c)y=f(t)$$
we complete the square
$$\frac{1}{4a}[ 4a^2 D^2+4a b D+b^2-b^2+4a c ]y= f(t)$$
$$\frac{1}{4a}[ (2a D+b)^2-(b^2-4a c)]y= f(t)$$
the homogeneous solution will have one of three forms
if b^2-4a c=0
$$y_H=C_1+C_2 t$$
if b^2-4a c<0
$$y_H=[ C_1 \cos(\sqrt{4a c-b^2}\, t)+C_2 \sin(\sqrt{4a c-b^2}\, t) ]\exp\left(-\frac{b}{2a}\right)$$
if b^2-4a c>0
$$y_H=[ C_1 \cosh(\sqrt{b^2-4a c}\, t)+C_2 \sinh(\sqrt{b^2-4a c}\, t) ]\exp\left(-\frac{b}{2a}\right)$$

Okay.
And what is your b^2-4ac in the equations representing?

 

[lurflurf]

The discriminant is formed from the coefficients of the differential equation
ie if we solve
a y''+b y'+c=0
b^2-4ac is found from those numbers
in your examples above
y''-y'+9y
a=1
b=-1
c=9
y''+2y'-y
a=1
b=2
c=-1

 

[alan2]

no, unfortunately not

Sad. I tried to upload my notes for 2nd order linear homogeneous equations but it says the file is too large. Maybe you can pm me and I can email it?

 

[vela]

after completing the square→(r+1)^2=2→r=-1±√2 [since when r=±a, we get cosh(ax) and sinh(ax), the solution becomes...]
Y_(H)=e^((-1±√2)t) →e^(-t)⋅e^((±√2)t)→e^-t[C1cosh((√2)t)+C2sinh((√2)t)] etc. etc.

It wouldn't hurt to work this out in detail. You get two solutions, $e^{(-1+\sqrt 2)t}$ and $e^{(-1-\sqrt 2)t}$, so the general solution is
\begin{align*}
y_h &= Ae^{(-1+\sqrt 2)t} + Be^{(-1-\sqrt 2)t} \\
&= A e^{-t} e^{\sqrt 2 t} + B e^{-t} e^{-\sqrt 2 t} \\
&= e^{-t} (A e^{\sqrt 2 t} + B e^{-\sqrt 2 t}) \\
&= e^{-t} [\underbrace{(A+B)}_C \cosh \sqrt 2 t +\underbrace{(A-B)}_D \sinh \sqrt 2 t]
\end{align*} Now work out the case where the roots are complex, $r = \alpha \pm \beta i$, and you'll see you get solutions of the form $y_h = e^{\alpha t}(A \cos \beta t + B \sin \beta t)$.