# 31.6 Solve the initial value problem

#### karush

##### Well-known member
$\tiny{31.6}$
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
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 $$A= -\frac{1}{2}. We have [tex]y_1(x)= e^{2x}(C_1cos(x)+ C_2sin(x))- \frac{1}{2}e^x$$.

Solve for $$y_2(x)$$ from $$y_2= y_1'- 2y_1- e^x$$.

(It is odd that you posted this, which involves a differential equation, under "Linear and Abstract Algebra" while you posted
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which does not, under "differential equations"!)

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#### karush

##### Well-known member
(It is odd that you posted this, which involves a differential equation, under "Linear and Abstract Algebra" while you posted
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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