Pam Stokes

2022-01-20

How to find $x\frac{dy}{dx}=y+\frac{4}{x}$ ?

It's given that$x>1,y=1\text{}\text{when}\text{}x=1$ .

It's given that

macalpinee3

Beginner2022-01-20Added 29 answers

You want to solve $x{y}^{\prime}=y+\frac{4}{x}$ , which is equivalent to

$x{y}^{\prime}-y=\frac{4}{x}$

$\frac{x{y}^{\prime}-y}{{x}^{2}}=\frac{4}{{x}^{3}}$

Can you notice derivative of some familiar expression there? Are you able to continue from there?

Can you notice derivative of some familiar expression there? Are you able to continue from there?

godsrvnt0706

Beginner2022-01-21Added 31 answers

Integrating factors I can explain. You may be familiar with the product rule for derivatives.

$d\left(uv\right)=udv+udu$

The idea behind an integrating factor is to get one side of the equation to look like${\left(uy\right)}^{\prime}=u{y}^{\prime}+{u}^{\prime}y$ , at which point you'll be able to integrate both sides. First, let's get all y's on one side.

$x{y}^{\prime}-y=\frac{4}{x}$

$y}^{\prime}-\frac{y}{x}=\frac{4}{{x}^{2}$

Now we have

$\frac{{u}^{\prime}}{u}=-\frac{1}{x}$

$\mathrm{ln}u=-\mathrm{ln}x=\mathrm{ln}\frac{1}{x}$

$u=\frac{1}{x}$

If you multiply both sides by$\frac{1}{x}$ , you'll get what Martin has in his answer. You can verify that the left side is equal to $\left(\frac{y}{x}\right)}^{\prime$ .

The idea behind an integrating factor is to get one side of the equation to look like

Now we have

If you multiply both sides by

RizerMix

Expert2022-01-27Added 583 answers

As an alternative to the other answer, it is always worthwile to try and solve the homogenous problem $x{z}^{\prime}=\frac{z}{x}$ first which has $z=x$ as a solution and then doing variation of parameters on $y:=v(x)z(x)$ , which gives:$x({v}^{\prime}x+v)=vx+\frac{4}{x}$ which should simplify nicely. Note that this takes advantage of the fact that your equation consists of a homogeneous part, $x{y}^{\prime}-y$ . That's why variation of parameters will always work.

- A body falls from rest against resistance proportional to the square root of the speed at any instant. If the body attains speed V1 and V2 feet per second, after 1 and 2 seconds in motion, respectively, find an expression for the limiting velocity.

The Laplace transform of the product of two functions is the product of the Laplace transforms of each given function. True or False

The Laplace transform of

is$u(t-2)$

(a)$\frac{1}{s}+2$

(b)$\frac{1}{s}-2$

(c) ??$e}^{2}\frac{s}{s}\left(d\right)\frac{{e}^{-2s}}{s$ 1 degree on celsius scale is equal to

A) $\frac{9}{5}$ degree on fahrenheit scale

B) $\frac{5}{9}$ degree on fahrenheit scale

C) 1 degree on fahrenheit scale

D) 5 degree on fahrenheit scaleThe Laplace transform of $t{e}^{t}$ is A. $\frac{s}{(s+1{)}^{2}}$ B. $\frac{1}{(s-1{)}^{2}}$ C. $\frac{s}{(s+1{)}^{2}}$ D. $\frac{s}{(s-1)}$

What is the Laplace transform of

into the s domain?$t\mathrm{cos}t$ Find the general solution of the given differential equation:

${y}^{\u2033}-2{y}^{\prime}+y=0$The rate at which a body cools is proportional to the difference in

temperature between the body and its surroundings. If a body in air

at 0℃ will cool from 200℃ 𝑡𝑜 100℃ in 40 minutes, how many more

minutes will it take the body to cool from 100℃ 𝑡𝑜 50℃ ?A body falls from rest against a resistance proportional to the velocity at any instant. If the limiting velocity is 60fps and the body attains half that velocity in 1 second, find the initial velocity.

What's the correct way to go about computing the Inverse Laplace transform of this?

$\frac{-2s+1}{({s}^{2}+2s+5)}$

I Completed the square on the bottom but what do you do now?

$\frac{-2s+1}{(s+1{)}^{2}+4}$How to find inverse Laplace transform of the following function?

$X(s)=\frac{s}{{s}^{4}+1}$

I tried to use the definition: $f(t)={\mathcal{L}}^{-1}\{F(s)\}=\frac{1}{2\pi i}\underset{T\to \mathrm{\infty}}{lim}{\int}_{\gamma -iT}^{\gamma +iT}{e}^{st}F(s)\phantom{\rule{thinmathspace}{0ex}}ds$or the partial fraction expansion but I have not achieved results.How do i find the lapalace transorm of this intergral using the convolution theorem? ${\int}_{0}^{t}{e}^{-x}\mathrm{cos}x\phantom{\rule{thinmathspace}{0ex}}dx$

How can I solve this differential equation? : $xydx-({x}^{2}+1)dy=0$

Find the inverse Laplace transform of $\frac{{s}^{2}-4s-4}{{s}^{4}+8{s}^{2}+16}$

inverse laplace transform - with symbolic variables:

$F(s)=\frac{2{s}^{2}+(a-6b)s+{a}^{2}-4ab}{({s}^{2}-{a}^{2})(s-2b)}$

My steps:

$F(s)=\frac{2{s}^{2}+(a-6b)s+{a}^{2}-4ab}{(s+a)(s-a)(s-2b)}$

$=\frac{A}{s+a}+\frac{B}{s-a}+\frac{C}{s-2b}+K$

$K=0$

$A=F(s)\ast (s+a)$