 # Determinants and Differential Equations Problem. "[Here] we explore a relationship between determinants and solutions to a differential equation. The 3 times 3 matrix consisting of solutions to a differential equation and their derivatives is called the Wronskian and, as we will see in later chapters, plays a pivotal role in the theory of differential equations." andg17o7 2022-09-11 Answered
Determinants and Differential Equations Problem
"[Here] we explore a relationship between determinants and solutions to a differential equation. The $3×3$ matrix consisting of solutions to a differential equation and their derivatives is called the Wronskian and, as we will see in later chapters, plays a pivotal role in the theory of differential equations."
This is the question following the description:
"Verify that ${y}_{1}\left(x\right)=\mathrm{cos}\left(2x\right),{y}_{2}\left(x\right)=\mathrm{sin}\left(2x\right),{y}_{3}\left(x\right)={e}^{x}$ are solutions to the differential equation: ${y}^{‴}-{y}^{″}+4{y}^{\prime }-4y=0$, and show that $\left\{\left\{{y}_{1},{y}_{2},{y}_{3}\right\},\left\{{y}_{1}^{\prime },{y}_{2}^{\prime },{y}_{3}^{\prime }\right\},\left\{{y}_{1}^{″},{y}_{2}^{″},{y}_{3}^{″}\right\}\right\}$ is nonzero on any interval."
Now I'm really just looking for an explanation on how I would show that it is nonzero on any interval. I completed the first part, I believe, by just taking up to the third derivative for each and plugging each set in to verify an identity of 0=0 at the end. I would then put the terms into the matrix which would give me:
$\left\{\left\{\mathrm{cos}\left(2x\right),\mathrm{sin}\left(2x\right),{e}^{x}\right\},\left\{-2\mathrm{sin}\left(2x\right),2\mathrm{cos}\left(2x\right),{e}^{x}\right\},\left\{-4\mathrm{cos}\left(2x\right),-4\mathrm{sin}\left(2x\right),{e}^{x}\right\}\right\}$
So I've gotten this far and if anyone could point me in the right direction on where to go from here it would be greatly appreciated.
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Step 1
Calculating the Wronskian involves taking the determinant of the matrix you've calculated. The typical formula given looks like
$|\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}|=a|\begin{array}{cc}e& f\\ h& i\end{array}|-b|\begin{array}{cc}d& f\\ g& i\end{array}|+c|\begin{array}{cc}d& e\\ g& h\end{array}|$
$=aei-afh-bdi+bfg+cdh-ceg.$
Step 2
When you calculate this for your matrix you'll get some function in x (since each entry is a function in x). This function is the Wronskian, and you simply examine it to make sure it is never zero.

We have step-by-step solutions for your answer! cubanwongux
Step 1
$\left\{\left\{\mathrm{cos}\left(2x\right),\mathrm{sin}\left(2x\right),{e}^{x}\right\},\left\{-2\mathrm{sin}\left(2x\right),2\mathrm{cos}\left(2x\right),{e}^{x}\right\},\left\{-4\mathrm{cos}\left(2x\right),-4\mathrm{sin}\left(2x\right),{e}^{x}\right\}\right\}$
Step 2
This is a matrix. You're asked to prove there is no x which makes the determinant of this matrix zero.
$det\left[\begin{array}{ccc}\mathrm{cos}\left(2x\right)& \mathrm{sin}\left(2x\right)& {e}^{x}\\ -2\mathrm{sin}\left(2x\right)& 2\mathrm{cos}\left(2x\right)& {e}^{x}\\ -4\mathrm{cos}\left(2x\right)& -4\mathrm{sin}\left(2x\right)& {e}^{x}\end{array}\right]=0$

We have step-by-step solutions for your answer!