System of Differential equation to solve.

[DaxInvader]

Homework Statement

Solve the following differential equation.

y'= {{2,-1},{3,-2}}y + {{1},{-1}}(e)$^{x}$

If it's not clear, I made an image for it.

http://i.imgur.com/lypTxqf.jpg

Homework Equations

$y{g}$ = $y{h}$ + $y{p}$

The Attempt at a Solution

So basically, I am looking for a homogeneous solution and a particular solution.

I started by looking at the eigen values and eigen vectors.

I found $\lambda$1 = 1 and $\lambda$2 = -1

And the vectors that go with them : $\stackrel{\rightarrow}{V1}$ = (1, 1) and $\stackrel{\rightarrow}{V2}$ = (1,3). with give me the homogeneous solution

$y{h}$ = C1(e)$^{x}$(1, 1) + C2(e)$^{-x}$(1, 3)

But I have a hard time looking for a particular solution..

I supposed that Yp was something like $\stackrel{\rightarrow}{a}$ * x * (e)$^{x}$ + $\stackrel{\rightarrow}{b}$ * (e)$^{x}$ and... I'm having a lot of trouble from there.. I found Y'p and. it start to get pretty complicated.. is there an easier way? I'm I doing this right?

thanks!

Dario

 

[Simon Bridge]

Homework Statement

Solve the following differential equation.

y'= {{2,-1},{3,-2}}y + {{1},{-1}}(e)$^{x}$

If it's not clear, I made an image for it.

http://i.imgur.com/lypTxqf.jpg

The built-in equation editor is so clumsy isn't it?
Here's how to write that out in latex: $\renewcommand{\dyx}[1]{\frac{d #1}{dx}}$ $$\dyx{\vec{y}} = \begin{pmatrix} 2 & -1 \\ 3 & -2 \end{pmatrix} \vec{y} + \begin{pmatrix} 1 \\ -1 \end{pmatrix} e^x = \text{A}\vec{y}+\vec{w}e^x$$ (Ill do this all the way through just to make sure I have properly understood you.)

Homework Equations

$y{g}$ = $y{h}$ + $y{p}$

You mean $$y_g=y_h+y_p$$ ??
$y_h$ would be commonly referred to as the "complimentary solution".

The Attempt at a Solution

So basically, I am looking for a homogeneous solution and a particular solution.

I started by looking at the eigenvalues and eigenvectors.

I found $\lambda$1 = 1 and $\lambda$2 = -1

And the vectors that go with them : $\stackrel{\rightarrow}{V1}$ = (1, 1) and $\stackrel{\rightarrow}{V2}$ = (1,3). with give me the homogeneous solution

$y{h}$ = C1(e)$^{x}$(1, 1) + C2(e)$^{-x}$(1, 3)

You got eigenvalues $\lambda_1=1$ and $\lambda_2=-1$ with corresponding eigenvectors $$\vec{v}_1= \begin{pmatrix} 1 \\ 1 \end{pmatrix} \; ;\; \vec{v}_2= \begin{pmatrix} 1 \\ 3 \end{pmatrix} \; \Rightarrow \vec{y}_h=C_1 \vec{v}_1 e^x + C_2 \vec{v}_2 e^{-x}$$

But I have a hard time looking for a particular solution..

I supposed that Yp was something like $\stackrel{\rightarrow}{a}$ * x * (e)$^{x}$ + $\stackrel{\rightarrow}{b}$ * (e)$^{x}$ and... I'm having a lot of trouble from there.. I found Y'p and. it start to get pretty complicated.. is there an easier way? I'm I doing this right?

method of undertermined coefficients?
maybe: $\vec{y}_p = \vec{a}e^x + \vec{b}xe^{x}$ ?
then $$\vec{y}^\prime_p = (\vec{a}+\vec{b})e^x + \vec{b}xe^{x} = \text{A}(\vec{a}e^x + \vec{b}xe^{x}) + \vec{w}e^x$$ ... so that $$\vec{a}+\vec{b}=\text{A}\vec{a}+\vec{w}\\ \vec{b} = \text{A}\vec{b}$$... something like that?
Where do you get stuck?

ref.
http://tutorial.math.lamar.edu/Classes/DE/NonhomogeneousSystems.aspx

 

[DaxInvader]

Haha sorry for my bad English, I'm still learning all the terms since I'm a french student.

The method I'm doing is pretty much the same thing, But I'm just maybe confused on how it actually works..

I don't get how a + b = Aa + w and b = Ab.. But I'm going to do the problem again and maybe i'll wake up haha

 

[LCKurtz]

$y_h$ would be commonly referred to as the "complimentary solution".

That would be complementary.

 

[SithsNGiggles]

maybe: $\vec{y}_p = \vec{a}e^x + \vec{b}xe^{x}$ ?
then $$\vec{y}^\prime_p = (\vec{a}+\vec{b})\color{red}{e^x} + \vec{b}\color{blue}{xe^{x}} = \text{A}(\vec{a}\color{red}{e^x} + \vec{b}\color{blue}{xe^{x}}) + \vec{w}\color{red}{e^x}$$

$$\vec{a}+\vec{b}=\text{A}\vec{a}+\vec{w}\\ \vec{b} = \text{A}\vec{b}$$

@DaxInvader, you get these equations by matching up the vector terms containing $e^x$ and $xe^x$ on both sides.

 

[Mark44]

That would be complementary.

Yeah - they don't give them away for free.

 

[Simon Bridge]

That would be complementary.

Why thank you, that's nice of you to say so...
(There is no i in complementary huh... it's always about the other person...)

Haha sorry for my bad English, I'm still learning all the terms since I'm a french student.

Your English is better than a lot of native speakers, I don't want you to think I was correcting you when I was checking my understanding ;)

I don't get how a + b = Aa + w and b = Ab.

From grouping like terms as SithsNGiggles points out. It's just the same as you are used to only the coefficients are vectors. I was wondering if that was the trouble.