Differential Equations-Method of Undetermined Coefficients

In summary, when solving second-order linear homogeneous equations, the discriminant can take on three possible values: zero, a positive real number, or a negative real number. Depending on the value of the discriminant, the homogeneous solution will have one of three forms: a linear function, a combination of cosine and sine functions, or a combination of hyperbolic cosine and hyperbolic sine functions. The discriminant is derived from the coefficients of the differential equation and helps determine the appropriate form for the homogeneous solution.
  • #1
cookiemnstr510510
162
14
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!
 
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  • #2
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
$$y_H=\left[ C_1 \cosh\left(\frac{\sqrt{b^2-4a c}}{2a}\, t\right)+C_2 \sinh\left(\frac{\sqrt{b^2-4a c}}{2a}\, t\right) \right]\exp\left(-\frac{b}{2a}\right)$$
 
Last edited:
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  • #3
You’re making this very complicated. Did your professor not derive the three cases for you?
 
  • #4
alan2 said:
You’re making this very complicated. Did your professor not derive the three cases for you?
no, unfortunately not
 
  • #5
lurflurf said:
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?
 
  • #6
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
 
  • #7
cookiemnstr510510 said:
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?
 
  • #8
cookiemnstr510510 said:
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)##.
 

1. What is the Method of Undetermined Coefficients?

The Method of Undetermined Coefficients is a technique used to solve non-homogeneous linear differential equations. It involves finding a particular solution by assuming a form for the solution and then determining the coefficients using the given initial conditions.

2. When should the Method of Undetermined Coefficients be used?

The Method of Undetermined Coefficients should be used when the non-homogeneous term of the differential equation can be expressed as a polynomial, exponential, sine, cosine, or a combination of these functions. It is not applicable for non-linear or non-constant coefficient equations.

3. How is the particular solution found using the Method of Undetermined Coefficients?

The particular solution is found by assuming a form for the solution based on the non-homogeneous term and then solving for the coefficients using the given initial conditions. The form of the solution is determined by the type of non-homogeneous term in the equation.

4. What are the steps involved in using the Method of Undetermined Coefficients?

The steps involved in using the Method of Undetermined Coefficients are:
1. Identify the type of non-homogeneous term in the equation.
2. Assume a form for the particular solution based on the type of non-homogeneous term.
3. Substitute the assumed solution into the original equation and solve for the coefficients.
4. Use the given initial conditions to determine the values of the coefficients.
5. Combine the homogeneous and particular solutions to get the general solution.

5. Can the Method of Undetermined Coefficients be used for all types of differential equations?

No, the Method of Undetermined Coefficients can only be used for non-homogeneous linear differential equations with constant coefficients. It is not applicable for non-linear or non-constant coefficient equations.

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