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2022-09-20
Answered

Find the equation of the line perpendicular to the line y=5 and passing through (5,4)

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pedradauy

Answered 2022-09-21
Author has **8** answers

Since the graph of y=5 is a horizontal line, any line perpendicular to a horizontal line is a vertical line, whose equation can be written in the form x=a. Since it passes through the point (5,4), so the equation is:

x=5

x=5

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Consider a 3rd order linear homogeneous DE of the form

$Lu={u}^{\u2034}+{a}_{2}(x){u}^{\u2033}+{a}_{1}(x){u}^{\prime}+{a}_{0}(x)u=f(x)\text{}\text{}\text{}\text{}(1)$

and for which ${u}_{1}={e}^{-x}$ and ${u}_{2}={e}^{-2x}$ are solutions to the homogeneous form of (1).

Let $f(x)=10{e}^{-2x}$. Give an example of a form of ${a}_{2}$,${a}_{1}$ and ${a}_{0}$ such that (1) has a stable equilibrium point and an example such that (1) has no stable equilibrium point.

My attempt:

When I think of stability, I immediately think of eigenvalues (nodes etc). Hence I reduced (1) into a system of linear equations:

$\begin{array}{rl}\frac{du}{dt}& =y,\\ \frac{dy}{dt}& =z,\\ \frac{dz}{dt}& =-{a}_{2}z-{a}_{1}y-{a}_{0}u.\end{array}$

This gives a corresponding matrix

$A=\left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ -{a}_{0}& -{a}_{1}& -{a}_{2}\end{array}\right).$

But after working with this, I feel as if I'm not on the right track. Any advice would be greatly appreciated.

$Lu={u}^{\u2034}+{a}_{2}(x){u}^{\u2033}+{a}_{1}(x){u}^{\prime}+{a}_{0}(x)u=f(x)\text{}\text{}\text{}\text{}(1)$

and for which ${u}_{1}={e}^{-x}$ and ${u}_{2}={e}^{-2x}$ are solutions to the homogeneous form of (1).

Let $f(x)=10{e}^{-2x}$. Give an example of a form of ${a}_{2}$,${a}_{1}$ and ${a}_{0}$ such that (1) has a stable equilibrium point and an example such that (1) has no stable equilibrium point.

My attempt:

When I think of stability, I immediately think of eigenvalues (nodes etc). Hence I reduced (1) into a system of linear equations:

$\begin{array}{rl}\frac{du}{dt}& =y,\\ \frac{dy}{dt}& =z,\\ \frac{dz}{dt}& =-{a}_{2}z-{a}_{1}y-{a}_{0}u.\end{array}$

This gives a corresponding matrix

$A=\left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ -{a}_{0}& -{a}_{1}& -{a}_{2}\end{array}\right).$

But after working with this, I feel as if I'm not on the right track. Any advice would be greatly appreciated.

asked 2022-05-21

How to solve this linear pde for $y(x)$ (other functions are known and $\lambda $ is a constant):

$\frac{d}{dx}(g(x){y}^{\prime}(x))={\lambda}^{2}g(x)y(x)$

Everything I know about this equation is that it is called the Sturmian equation. I did some research, but the theory, which is for a more general form of the equation, is too hard for me to understand.

$\frac{d}{dx}(g(x){y}^{\prime}(x))={\lambda}^{2}g(x)y(x)$

Everything I know about this equation is that it is called the Sturmian equation. I did some research, but the theory, which is for a more general form of the equation, is too hard for me to understand.