6.1.1 Show that .... is a solution of.....

[bigwave]

$\textsf{6.1.1 Show that}\\$
$$y=\begin{bmatrix}
c_1e^{2x} + c_2e^{3x}\\
2c_1e^{2x} + c_2e^{3x}
\end{bmatrix}\\$$
$\textsf{is a solution of} \\$
$$Y'=\left[\begin{array}{rr}
4 & -1 \\2 & 1
\end{array}\right]Y\\$$
$\textsf{6.1.2 Show that}\\$
$$Y'=\left[\begin{array}{rrrr}
c_1e^{2x}& + c_2e^{3x}& -\frac{x}{6}&-\frac{11}{36} \\ \\
2c_1e^{2x}& + c_2e^{3x}& -\frac{2x}{6}&-\frac{1}{18}
\end{array}\right]\\$$
$\textsf{is a solution of} \\$
$$Y'=\left[\begin{array}{rr}
4 & -1 \\2 & 1
\end{array}\right]Y
+\left[\begin{array}{rr}
1 \\x
\end{array}\right]$$

ok I realize these are simple but still the examples given confused me.
I assume the first step is to span

 

[HallsofIvy]

To show that Y is a solution to a given equation, put that Y into the equation!

Here, $Y= \begin{bmatrix}c_1e^{2x}+ c_2e^{3x} \\ 2c_1e^{2x}+ c_2e^{3x}\end{bmatrix}$ and the equation is \(\displaystyle Y'= \begin{bmatrix}4 & -1 \\ 2 & 1 \end{bmatrix}Y\)

Do you understand that $Y'= \begin{bmatrix}2c_1e^{2x}+ 3c_2e^{3x} \\ 4c_1e^{2x}+ 3c_2e^{3x}\end{bmatrix}$?

What is $\begin{bmatrix}4 & -1 \\ 2 & 1 \end{bmatrix}\begin{bmatrix}c_1e^{2x}+ c_2e^{3x} \\ 2c_1e^{2x}+ c_2e^{3x}\end{bmatrix}$?

 

[bigwave]

$\begin{bmatrix}
4&-1\\ 2&1
\end{bmatrix}$
$\begin{bmatrix}
c_1e^{2x}&c_2e^{3x}\\
2c_1e^{2x}&c_2e^{3x}
\end{bmatrix}$
$=\begin{bmatrix}
2c_1e^{2x}&3c_2e^{3x}\\
4c_1e^{2x}&3c_2e^{3x}
\end{bmatrix}$
$\begin{bmatrix}
4c_1e^{2x}-2c_1e^{2x}&4c_2e^{3x}-c_2e^{3x}\\
2c_1e^{2x}+2c_1e^{2x}&2c_2e^{3x}+c_2e^{3x}
\end{bmatrix}$
$=\begin{bmatrix}
2c_1e^{2x}&3c_2e^{3x}\\
4c_1e^{2x}&3c_2e^{3x}
\end{bmatrix}$


hopefully
typos maybe

 

[HallsofIvy]

Yes, that is correct!

 

[bigwave]

These are too many tedious steps
Ok for 6.1.2 I'm not sure just what Y is
Since it's a different problem
Do we take the integral of Y' if so how?

 

[topsquark]

These are too many tedious steps
Ok for 6.1.2 I'm not sure just what Y is
Since it's a different problem
Do we take the integral of Y' if so how?

You don't need to do anything further with this. They simply gave you an example of a problem and just made it up. It can get tedious but the method does make things easier, especially if you are dealing with several simultanous differential equations.

Since you asked, the integral of a function \(\displaystyle W = \left ( \begin{matrix} Y \\ Z \end{matrix} \right )\) is simply
\(\displaystyle \int W ~ dx = \int \left ( \begin{matrix} Y \\ Z \end{matrix} \right ) ~ dx = \left ( \begin{matrix} \int Y ~ dx \\ \int Z ~ dx \end{matrix} \right )\)

You take the integral of the components just like to find the derivative of W you simply take the derivative of the components.

-Dan

 

[bigwave]

Really appreciate the help
I'll continue on this on Monday
When I use the university PC
I just have a cell phone on weekends

 

[bigwave]

so how do put in the "solved" prefix? I don't see that option in thread tools?

 

[topsquark]

so how do put in the "solved" prefix? I don't see that option in thread tools?

That feature was removed at some point. And for some silly reason spammers can do that, so it's still in the system somewhere.

-Dan