31.6 Solve the initial value problem

[karush]

$\tiny{31.6}$
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]

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 $$y'(x)= y''(x)= Ae^x$$ and the equation becomes $$Ae^x- 4Ae^x+ 5Ae^x= 2Ae^x= -e^x$$ so [tex]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'">$$y_2= y_1'- 2y_1- e^x$$.<br /> <br /> (It is odd that you posted this, which involves a differential equation, under "Linear and Abstract Algebra" while you posted <u>[FONT=Tahoma,Calibri,Verdana,Geneva,sans-serif]"https://mathhelpboards.com/differential-equations-17/14-1-find-vector-v-will-satisfy-system-26203.html",</u><br /> </span>[/tex]​

$$<span style="font-family: 'Tahoma'"><br /> </span>$$​

$$<span style="font-family: 'Tahoma'"><br /> which does not, under "differential equations"!)</span>$$

 

[karush]

(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