MHB 31.6 Solve the initial value problem

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The discussion revolves around solving the initial value problem defined by the differential equation Y' = AY + G, where A is a matrix and G is a vector function. Participants detail the process of finding eigenvalues and eigenvectors of matrix A, leading to the characteristic equation with roots λ1 = 2 + i and λ2 = 2 - i. The general solution for the associated homogeneous equation is established, and a particular solution is sought, resulting in y1(x) = e^(2x)(C1cos(x) + C2sin(x)) - (1/2)e^x. The conversation also touches on the confusion regarding the categorization of the problem under linear algebra versus differential equations. The thread emphasizes the integration of concepts from both fields to solve the problem effectively.
karush
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$\tiny{31.6}$
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
 
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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 y_1'= 2y_1+ y_2+ e^x and y_2'= -y_1+ 2y_2. Differentiate the first equation again to get y_1''= 2y_1'+ y_2'+ e^x. Replace that y_2' by -y_1+ 2y_2 from the second equation: y_1''= 2y_1'- y_1+ 2y_2+ e^x. From the first equation, y_2= y_1'- 2y_1- e^x so y_1''= 2y_1'- y_1+ 2(y_1'- 2y_1- e^x)+ e^x or y_1''- 4y_1'+ 5y_1= - e^x. The associated homogeneous equation is
y_1''- 4y_1'+ 5y_1= 0 which has characteristic equation r^2- 4r+ 5= 0. By the quadratic formula, that has roots \frac{4\pm\sqrt{16- 20}}{2}= 2\pm i just as you say.

That tells us that the general solution to the associated homogeneous equation is y_1(x)= e^{2x}(C_1cos(x)+ C_2sin(x)). We look for a solution to the entire equation of the form y_1(x)= Ae^x. Then y&#039;(x)= y&#039;&#039;(x)= Ae^x and the equation becomes Ae^x- 4Ae^x+ 5Ae^x= 2Ae^x= -e^x so A= -\frac{1}{2}. We have y_1(x)= e^{2x}(C_1cos(x)+ C_2sin(x))- \frac{1}{2}e^x.<br /> <br /> Solve for y_2(x) from <span style="font-family: 'Tahoma'">y_2= y_1&amp;#039;- 2y_1- e^x.<br /> <br /> (It is odd that you posted this, which involves a differential equation, under &quot;Linear and Abstract Algebra&quot; while you posted <u>[FONT=Tahoma,Calibri,Verdana,Geneva,sans-serif]&quot;https://mathhelpboards.com/differential-equations-17/14-1-find-vector-v-will-satisfy-system-26203.html&quot;,</u><br /> </span>
<span style="font-family: 'Tahoma'"><br /> </span>
<span style="font-family: 'Tahoma'"><br /> which does not, under &quot;differential equations&quot;!)</span>
 
Last edited by a moderator:
HallsofIvy said:
(It is odd that you posted this, which involves a differential equation, under "Linear and Abstract Algebra" while you posted "https://mathhelpboards.com/differential-equations-17/14-1-find-vector-v-will-satisfy-system-26203.html",


which does not, under "differential equations"!)


https://www.physicsforums.com/attachments/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
 
Last edited:

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