In Physics, the Simple Harmonic Oscillator is represented by the equation

By using the characteristic polynomial, you get solutions of the form

Pam Stokes
2022-01-16
Answered

In Physics, the Simple Harmonic Oscillator is represented by the equation

By using the characteristic polynomial, you get solutions of the form

You can still ask an expert for help

kalupunangh

Answered 2022-01-17
Author has **29** answers

However, if you assume the function x(t) is real, then they are related as

This can be seen since

Rewriting the complex exponentials as sine and cosine we get

Now as proved before

And the solution becomes

where

Robert Pina

Answered 2022-01-18
Author has **42** answers

Remember that

Euler's fomula, along with the properties

Similarly, choosing

Finally, by linearity you can conclude that the linear combination

alenahelenash

Answered 2022-01-24
Author has **368** answers

the answers given are absolutely fine as they show the way from ${e}^{\pm i\omega t}$ to sin and cos.
It might be even shorter if you use that
$\mathrm{sin}(x)=\frac{1}{2i}({e}^{ix}-{e}^{ix})$
and $\mathrm{cos}(x)=\frac{1}{2}({e}^{ix}+{e}^{-ix})$
This can be interpreted as a definition for the functions or seen from its series definitions.

asked 2021-09-17

Find the inverse Laplace transform of

$G\left(s\right)=\frac{1}{{s}^{2}-4s+5}$

asked 2022-01-15

Find the orthogonal trajectory.

$y}^{2}=k{x}^{3$

asked 2022-01-18

Is there a closed form for any function f(x,y) satisfying:

$\frac{df}{dx}+\frac{df}{dy}=xy$

asked 2022-01-16

Suppose we have a mechanical system with 1 degree of freedom, i.e. an ODE

$\left(1\right)\ddot{q}+{V}^{\prime}\left(q\right)=0$ ,

where$V:R\to R$ is some smooth function (potential energy). We then easily see that any solution of this equation must satisfy

$\frac{{\dot{q}}^{2}}{2}+V\left(q\right)=\text{constant}$

In other words, if we put

$E(q,\dot{q})=\frac{{\dot{q}}^{2}}{2}+V\left(q\right)$

(energy), then the image of every solution of (1) must lie in a level set of E.

where

In other words, if we put

(energy), then the image of every solution of (1) must lie in a level set of E.

asked 2022-07-01

Consider the linear first order non-homogeneous partial differential equation

${U}_{x}+y{U}_{y}-y=y{e}^{-x}$

By using the method of characteristics show that its general solution is given by

$u(x,y)=-y{e}^{-x}+{e}^{x}g(y{e}^{-x})$

where g is any differentiable function of its argument

$A=1,B=y,d=-1,f=y{e}^{-x}\phantom{\rule{0ex}{0ex}}{\displaystyle \frac{dy}{dx}}={\displaystyle \frac{b}{a}}=y\phantom{\rule{0ex}{0ex}}dy=ydx\phantom{\rule{0ex}{0ex}}\int {\displaystyle \frac{1}{y}}dy=\int dx\phantom{\rule{0ex}{0ex}}\mathrm{ln}y=x+c\phantom{\rule{0ex}{0ex}}\eta (x,y)=c\phantom{\rule{0ex}{0ex}}\text{let}\xi =x$

Then

$\eta =\mathrm{ln}y-x$

$\xi =x$

those are my characteristics curves,

when I put it into $[a{\xi}_{x}+b{\xi}_{y}]{\omega}_{\xi}+d\omega =F$

I get the equation ${\omega}_{\xi}-\omega ={e}^{\eta}$

Is this all correct?

${U}_{x}+y{U}_{y}-y=y{e}^{-x}$

By using the method of characteristics show that its general solution is given by

$u(x,y)=-y{e}^{-x}+{e}^{x}g(y{e}^{-x})$

where g is any differentiable function of its argument

$A=1,B=y,d=-1,f=y{e}^{-x}\phantom{\rule{0ex}{0ex}}{\displaystyle \frac{dy}{dx}}={\displaystyle \frac{b}{a}}=y\phantom{\rule{0ex}{0ex}}dy=ydx\phantom{\rule{0ex}{0ex}}\int {\displaystyle \frac{1}{y}}dy=\int dx\phantom{\rule{0ex}{0ex}}\mathrm{ln}y=x+c\phantom{\rule{0ex}{0ex}}\eta (x,y)=c\phantom{\rule{0ex}{0ex}}\text{let}\xi =x$

Then

$\eta =\mathrm{ln}y-x$

$\xi =x$

those are my characteristics curves,

when I put it into $[a{\xi}_{x}+b{\xi}_{y}]{\omega}_{\xi}+d\omega =F$

I get the equation ${\omega}_{\xi}-\omega ={e}^{\eta}$

Is this all correct?

asked 2021-12-16

Hello,

i'm here because i didn't found anything general on this topic, i need to know what is the expression of the laplace transform a composed function like that :

my complete equation to transform is :

the goal is to isolate the

Thanks for your time,

Robin Luiz

asked 2021-03-08

Solve differential equation
$\frac{dy}{dx}=\frac{x}{y},\text{}y\left(0\right)=-8$