Carole Yarbrough

2021-12-20

Solve the following differential equations. Use the method of Bernoulli’s Equation. $x(2{x}^{3}+y)dy-6{y}^{2}dx=0$

Bob Huerta

Beginner2021-12-21Added 41 answers

Step 1

General Bernoullis

General Bernoullis

Mason Hall

Beginner2021-12-22Added 36 answers

Solve $-x\frac{dy\left(x\right)}{dx}(2{x}^{3}+y\left(x\right))+6y{\left(x\right)}^{2}=0$ :

Solve for$\frac{dy\left(x\right)}{dx}$ :

$\frac{dy\left(x\right)}{dx}=\frac{6y{\left(x\right)}^{2}}{x(2{x}^{3}+y\left(x\right))}$

Write the differential equation in terms of x. Since$\frac{dy}{dx}\frac{dx}{dy}=1,\frac{dy\left(x\right)}{dx}=\frac{1}{\frac{dx\left(y\right)}{dy}}$ :

$\frac{1}{\frac{dx\left(y\right)}{dy}}=\frac{6{y}^{2}}{(y+2x{\left(y\right)}^{3})x\left(y\right)}$

Raise both sides to the power -1 and expand:

$\frac{dx\left(y\right)}{dy}=x\frac{{\left(y\right)}^{4}}{3{y}^{2}}+x\frac{y}{6y}$

Subtract$x\frac{y}{6y}$ from both sides:

$\frac{dx\left(y\right)}{dy}-x\frac{y}{6y}=x\frac{{\left(y\right)}^{4}}{3{y}^{2}}$

Divide both sides by$-\frac{1}{3}x{\left(y\right)}^{4}$ :

$-\frac{3\frac{dx\left(y\right)}{dy}}{x}{\left(y\right)}^{4}+\frac{1}{2yx{\left(y\right)}^{3}}=-\frac{1}{{y}^{2}}$

Let$v\left(y\right)=\frac{1}{x}{\left(y\right)}^{3}$ , which gives $\frac{dv\left(y\right)}{dy}=-\frac{3\frac{dx\left(y\right)}{dy}}{x}{\left(y\right)}^{4}$

Solve for

Write the differential equation in terms of x. Since

Raise both sides to the power -1 and expand:

Subtract

Divide both sides by

Let

RizerMix

Expert2021-12-29Added 656 answers

The equation

Obtain the solution as

- 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:

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

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

$K=0$

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