Okay so I have another Problem

[Lovely918]

IV. Solve

y''' - y'' - y' + y - x = 0
So then I moved the equation to r^3-r^2-r+y-x
So an obvious answer would be like r=1
So I need to somehow divide by r-1 in order to get a quadratic or something.
Then I need to find some specific solution maybe using y(x)= Ax+ B for some numbers A and B.
But I need help executing that.

 

[Deleted member 4993]

IV. Solve

y''' - y'' - y' + y - x = 0
So then I moved the equation to r^3-r^2-r+y-x
So an obvious answer would be like r=1 ← How did you get that from there?
So I need to somehow divide by r-1 in order to get a quadratic or something.
Then I need to find some specific solution maybe using y(x)= Ax+ B for some numbers A and B.
But I need help executing that.

.

 

[Lovely918]

Well I got

.

Well I got r=1 because if you look at the equation

[FONT=MathJax_Math]r[FONT=MathJax_Main]3[/FONT][/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math]r[FONT=MathJax_Main]2[/FONT][/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math]r[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]0
then you can see that one is in the problem so [/FONT][FONT=MathJax_Math]r[FONT=MathJax_Main]3[/FONT][/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math]r[FONT=MathJax_Main]2[/FONT][/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math]r[/FONT][FONT=MathJax_Main]=1 but I guess I miss said that its probably supposed to be [/FONT][FONT=MathJax_Math]r[/FONT][FONT=MathJax_Main]=-1
So what do I do from here? [/FONT]​

 

[Lovely918]

Okay to be honest HallsofIvy was the one who gave me r-1. I do have an example problem though-.
So then I have y''-4y+13y=0; y(0)=1,y(0)=0 so then they turned it into an equation like r^2-4r+13 then did a quadratic
of +4 (+or -) sqrt(16-4(13)/2 which turned into +4 (+or _) 6i/2
and then they proceeded into doing something about y and y'. I thought by trying to work out this equation.

Should I continue with the r-1? or no?

Well I got r=1 because if you look at the equation

[FONT=MathJax_Math]r[FONT=MathJax_Main]3[/FONT][/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math]r[FONT=MathJax_Main]2[/FONT][/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math]r[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]0
then you can see that one is in the problem so [/FONT][FONT=MathJax_Math]r[FONT=MathJax_Main]3[/FONT][/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math]r[FONT=MathJax_Main]2[/FONT][/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math]r[/FONT][FONT=MathJax_Main]=1 but I guess I miss said that its probably supposed to be [/FONT][FONT=MathJax_Math]r[/FONT][FONT=MathJax_Main]=-1
So what do I do from here? [/FONT]​

 

[Lovely918]

Yeah they were right.

Okay to be honest HallsofIvy was the one who gave me r-1. I do have an example problem though-.
So then I have y''-4y+13y=0; y(0)=1,y(0)=0 so then they turned it into an equation like r^2-4r+13 then did a quadratic
of +4 (+or -) sqrt(16-4(13)/2 which turned into +4 (+or _) 6i/2
and then they proceeded into doing something about y and y'. I thought by trying to work out this equation.

Should I continue with the r-1? or no?

 

[Deleted member 4993]

Well I got r=1 because if you look at the equation

[FONT=MathJax_Math]r[FONT=MathJax_Main]3[/FONT][/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math]r[FONT=MathJax_Main]2[/FONT][/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math]r[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]0
then you can see that one is in the problem so [/FONT][FONT=MathJax_Math]r[FONT=MathJax_Main]3[/FONT][/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math]r[FONT=MathJax_Main]2[/FONT][/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math]r[/FONT][FONT=MathJax_Main]=1 but I guess I miss said that its probably supposed to be [/FONT][FONT=MathJax_Math]r[/FONT][FONT=MathJax_Main]=-1
So what do I do from here? [/FONT]​

Your original ODE is:

y''' - y'' - y' + y - x = 0

or

y''' - y'' - y' + y = x

The homogeneous part of the equation is:

y''' - y'' - y' + y = 0

You need to find solution to homogeneous part first.

To do that, you write the characteristic equation:

r³ - r² - r + 1 = 0

You need find roots of this polynomial. This a polynomial of degree 3 - so it will have 3 roots.

You are correct in asserting that by observation, one of the roots is r = 1

That means that (r-1) is a factor of the polynomial r³ - r² - r + 1

That means

r³ - r² - r + 1 = (r-1) * f(r) .... where f(r) is some yet undefined function of r.

f(r) = (r³ - r² - r + 1)/(r-1)

Now do the polynomial division and find f(r). To review polynomial division - go to:

http://www.purplemath.com/modules/polydiv2.htm

f(r) will be quadratic and will give you the other two roots.

 

[Lovely918]

Well if you divide f(r) = (r³ - r² - r + 1)/(r-1) you can have a few alternate forms like

So would it be permissible to put that into a quadratic and you get

. Did you mean it in that context?

Your original ODE is:

y''' - y'' - y' + y - x = 0

or

y''' - y'' - y' + y = x

The homogeneous part of the equation is:

y''' - y'' - y' + y = 0

You need to find solution to homogeneous part first.

To do that, you write the characteristic equation:

r³ - r² - r + 1 = 0

You need find roots ofthis polynomial. This a polynomial of degree 3 - so it will have 3 roots.

You are correct in asserting that by observation, one of the roots is r = 1

That means that (r-1) is a factor of the polynomial r³ - r² - r + 1

That means

r³ - r² - r + 1 = (r-1) * f(r) .... where f(r) is some yet undefined function of r.

f(r) = (r³ - r² - r + 1)/(r-1)

Now do the polynomial division and find f(r). To review polynomial division - go to:

http://www.purplemath.com/modules/polydiv2.htm

f(r) will be quadratic and will give you the other two roots.

 

[Lovely918]

WOuld you then incorporate this equation : y(x)=c1 xe^-x +c2 e^-x

 

[mmm4444bot]

if you divide … (r³ - r² - r + 1)/(r-1) you can have a few alternate forms like

I do not understand the statement "you can have a few alternate forms". Can you explain what you're thinking about "alternate forms"?

 

[Deleted member 4993]

WOuld you then incorporate this equation : y(x)=c1 xe^-x +c2 e^-x

No - your given ODE is of third order - so you would have three part solution for the homogeneous part.

The roots of the characteristic equation are:

r = -1 and r = 1 and r = 1 (repeated roots)

Now what .....

 

[Lovely918]

Oh okay! now I know what I did wrong! Thank you so much!

No - your given ODE is of third order - so you would have three part solution for the homogeneous part.

The roots of the characteristic equation are:

r = -1 and r = 1 and r = 1 (repeated roots)

Now what .....