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31.6 Solve the initial value problem

karush

Well-known member
Jan 31, 2012
2,886
$\tiny{31.6}$
Solve the initial value problem

$Y'=\left|\begin{array}{rr}2 & 1 \\-1 & 2 \end{array}\right|Y
+\left|\begin{array}{rr}e^x \\0 \end{array}\right|,
\quad Y(0)=\left|\begin{array}{rr} 1 \\1 \end{array}\right| $
ok so we have the form $y'=AY+G$
rewrite as
$$\displaystyle
\left|\begin{array}{rr}y_1^\prime \\y_2^\prime \end{array}\right|
=\left|\begin{array}{rr}2 & 1 \\-1 & 2 \end{array}\right|
\left|\begin{array}{rr}y_1 \\y_2\end{array}\right|
+\left|\begin{array}{rr}e^x \\0 \end{array}\right|$$
ok so the next thing to do is find eigenvalues of A so
$\left| \begin{array}{cc}
-\lambda+2&1\\-1&-\lambda+2\end{array}
\right|
=\left(-\lambda+2\right)^{2}+1$
so roots are
$\lambda_{1}=2 + i, \qquad \lambda_{2}=2 - i$


so far ???? hopefully
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
Yes, that is correct. Now, having found the eigenvalues, you need to find the corresponding eigenvectors.

Let's look at solving this with a slightly different method. The matrix equation is equivalent to the pair of equations [tex]y_1'= 2y_1+ y_2+ e^x[/tex] and [tex]y_2'= -y_1+ 2y_2[/tex]. Differentiate the first equation again to get [tex]y_1''= 2y_1'+ y_2'+ e^x[/tex]. Replace that [tex]y_2'[/tex] by [tex]-y_1+ 2y_2[/tex] from the second equation: [tex]y_1''= 2y_1'- y_1+ 2y_2+ e^x[/tex]. From the first equation, [tex]y_2= y_1'- 2y_1- e^x[/tex] so [tex]y_1''= 2y_1'- y_1+ 2(y_1'- 2y_1- e^x)+ e^x[/tex] or [tex]y_1''- 4y_1'+ 5y_1= - e^x[/tex]. The associated homogeneous equation is
y_1''- 4y_1'+ 5y_1= 0 which has characteristic equation [tex]r^2- 4r+ 5= 0[/tex]. By the quadratic formula, that has roots [tex]\frac{4\pm\sqrt{16- 20}}{2}= 2\pm i[/tex] just as you say.

That tells us that the general solution to the associated homogeneous equation is [tex]y_1(x)= e^{2x}(C_1cos(x)+ C_2sin(x))[/tex]. We look for a solution to the entire equation of the form [tex]y_1(x)= Ae^x[/tex]. Then [tex]y'(x)= y''(x)= Ae^x[/tex] and the equation becomes [tex]Ae^x- 4Ae^x+ 5Ae^x= 2Ae^x= -e^x[/tex] so [tex]A= -\frac{1}{2}. We have [tex]y_1(x)= e^{2x}(C_1cos(x)+ C_2sin(x))- \frac{1}{2}e^x[/tex].

Solve for [tex]y_2(x)[/tex] from [tex]y_2= y_1'- 2y_1- e^x[/tex].

(It is odd that you posted this, which involves a differential equation, under "Linear and Abstract Algebra" while you posted
[FONT=Tahoma,Calibri,Verdana,Geneva,sans-serif]"find a vector v that will satisfy the system",

[/FONT]
which does not, under "differential equations"!)
 
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karush

Well-known member
Jan 31, 2012
2,886
(It is odd that you posted this, which involves a differential equation, under "Linear and Abstract Algebra" while you posted "find a vector v that will satisfy the system",
which does not, under "differential equations"!)
View attachment 9086

will they combined the DE with LA so I often label it wrong
poor reason I know.

Just hope I will be ready for the next class, whatever commonly comes after this one

the last week they started the chapter "First Order Ordinary Differential Equations"

Which I posted some of earlier on MHB

ok well back in the sattle solve
$\displaystyle y_2= y_1'- 2y_1- e^x$
rewrite as
$\displaystyle y_2-e^x=y_1^\prime-\frac{1}{2}y_1$


ok assume the next step is the factor
 
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