Coupled differential equations

[aaaa202]

A bit related to my other topic but here goes:

I have the set of differential equations:

x1' = a-ax1-ax2
x2' = b-bx1-bx2

What is the solution to such a set? I could google systems of differential equations, but that turns up large texts on the theory. I just wonna know in a few lines what the solution to such a set is. I know how to diagonalize the coefficient matrix etc. if I need that.

 

[Mark44]

A bit related to my other topic but here goes:

I have the set of differential equations:

x1' = a-ax1-ax2
x2' = b-bx1-bx2

What is the solution to such a set? I could google systems of differential equations, but that turns up large texts on the theory. I just wonna know in a few lines what the solution to such a set is. I know how to diagonalize the coefficient matrix etc. if I need that.

I did a search for "nonhomogeneous system of linear equations". Here is one link.
http://tutorial.math.lamar.edu/classes/de/NonhomogeneousSystems.aspx

 

[Mark44]

It takes more that "a few lines" to describe how to solve systems such as yours...

 

[Ray Vickson]

A bit related to my other topic but here goes:

I have the set of differential equations:

x1' = a-ax1-ax2
x2' = b-bx1-bx2

What is the solution to such a set? I could google systems of differential equations, but that turns up large texts on the theory. I just wonna know in a few lines what the solution to such a set is. I know how to diagonalize the coefficient matrix etc. if I need that.

Are the three 'a's in the first equation really supposed to be all the same (and ditto for the 'b's in the second equation)? If so, try looking at y = x1+x2.

BTW: what does the word "wonna" mean?

 

[SteamKing]

'wonna' is a variant of 'wanna' = 'want to'

 

[Mark44]

'wonna' is a variant of 'wanna' = 'want to'

Not like in "wonna these days"?

 

[Ray Vickson]

'wonna' is a variant of 'wanna' = 'want to'

I know that of course; but it does no harm to try to encourage proper language.

 

[HallsofIvy]

There several different ways to handle
$x_1'= a- ax_1- ax_2$
$x_2'= b- bx_1- bx_2$

The more "elementary" method would be to differentiate the first equation again:
[itex]x_1''= -ax_1'- ax_2'= ax_1'- a(b- bx_1- bx_2)= (a+ ab)x_1+ abx_2- ab[itex] but, from the first equation, multiplied by b, [itex]abx_2= ab- bx_1'- abx_1[itex]so <br /> $x_1''= (a+ ab)x_1+ (ab- bx_1'- abx_1)- ab$<br /> $x_1''+ bx_1'- ax_1= 0$.<br /> That has characteristic equation $r^2+ br- a= 0$ with roots $r= \frac{-b\pm\sqrt{b^2+ 4a}}{2}$ so that $x_1(t)= e^{-bt}(C_1e^{\sqrt{b^2+ 4a}t}+ C_2e^{-\sqrt{b^2+ 4a}t}$ if $b^2+ 4a>0$, $x_1(t)= e^{-bt}(C_1 cos(\sqrt{-b^2- 4a}t)+ C_2 sin(\sqrt{-b^2- 4a}t)$ if $b^2+ 4a< 0$, or $x_1(t)= e^{-bt}(C_1+ C_2t)$ if $b^2+ 4a= 0$.<br /> <br /> In any case, $x_2= (ab- bx_1'- abx_1)/ab$.<br /> <br /> A more "sophisticated" method would be to write the differential equations as a matrix equation:<br /> $$\begin{pmatrix}x_1 \\ x_2\end{pmatrix}'= \begin{pmatrix}-a & -a \\ -b & -b\end{pmatrix}\begin{pmatrix}x_1 \\ x_2 \end{pmatrix}+ \begin{pmatrix}a \\ b\end{pmatrix}$$<br /> <br /> The characteristic equation for that matrix is <br /> $$\left|\begin{array}{cc}-a- \lambda & -a \\ -b & -b- \lambda\end{array}\right|= (-a- \lambda)(-b- \lambda)- ab= \lambda^2+ (a+ b)\lambda= 0$$<br /> which has roots $\lambda= 0$ and [itex]\lambda= -a- b$.<br /> We can find the eigenvectors corresponding to those eigenvectors and so diagonalize the matrix,$[/itex][/itex][/itex][/itex][/itex]

 

[Ray Vickson]

There several different ways to handle
$x_1'= a- ax_1- ax_2$
$x_2'= b- bx_1- bx_2$

The more "elementary" method would be to differentiate the first equation again:
[itex]x_1''= -ax_1'- ax_2'= ax_1'- a(b- bx_1- bx_2)= (a+ ab)x_1+ abx_2- ab[itex] but, from the first equation, multiplied by b, [itex]abx_2= ab- bx_1'- abx_1[itex]so <br /> $x_1''= (a+ ab)x_1+ (ab- bx_1'- abx_1)- ab$<br /> $x_1''+ bx_1'- ax_1= 0$.<br /> That has characteristic equation $r^2+ br- a= 0$ with roots $r= \frac{-b\pm\sqrt{b^2+ 4a}}{2}$ so that $x_1(t)= e^{-bt}(C_1e^{\sqrt{b^2+ 4a}t}+ C_2e^{-\sqrt{b^2+ 4a}t}$ if $b^2+ 4a>0$, $x_1(t)= e^{-bt}(C_1 cos(\sqrt{-b^2- 4a}t)+ C_2 sin(\sqrt{-b^2- 4a}t)$ if $b^2+ 4a< 0$, or $x_1(t)= e^{-bt}(C_1+ C_2t)$ if $b^2+ 4a= 0$.<br /> <br /> In any case, $x_2= (ab- bx_1'- abx_1)/ab$.<br /> <br /> A more "sophisticated" method would be to write the differential equations as a matrix equation:<br /> $$\begin{pmatrix}x_1 \\ x_2\end{pmatrix}'= \begin{pmatrix}-a & -a \\ -b & -b\end{pmatrix}\begin{pmatrix}x_1 \\ x_2 \end{pmatrix}+ \begin{pmatrix}a \\ b\end{pmatrix}$$<br /> <br /> The characteristic equation for that matrix is <br /> $$\left|\begin{array}{cc}-a- \lambda & -a \\ -b & -b- \lambda\end{array}\right|= (-a- \lambda)(-b- \lambda)- ab= \lambda^2+ (a+ b)\lambda= 0$$<br /> which has roots $\lambda= 0$ and [itex]\lambda= -a- b$.<br /> We can find the eigenvectors corresponding to those eigenvectors and so diagonalize the matrix,$[/itex][/itex][/itex][/itex][/itex]

[itex][itex][itex][itex][itex][itex] <br /> The two DEs added together give $y' = c - cy$ where $c = a+b$ and $y = x_1 + x_2$. Once we know $y$ we can get $x_1, x_2$ from $x_1' = a - ay$, etc.[/itex][/itex][/itex][/itex][/itex][/itex]