Frank Guyton

2021-12-28

What is the general solution for the high-order differential equations ${y}^{5}+6{y}^{4}+12y+8y=0$

zurilomk4

Beginner2021-12-29Added 35 answers

The derivatives of this trail solution are: $y}^{\prime}=m{e}^{mx$

$y{m}^{2}{e}^{mx}$

Substitute the trail solution and its derivatives in the given D.E.

${y}^{5}+6{y}^{4}+12{y}^{3}+8y=0$

${m}^{5}{e}^{mx}+6{m}^{4}{e}^{mx}+12{m}^{3}{e}^{mx}+8m{e}^{mx}=0$

${e}^{mx}({m}^{5}+6{m}^{4}+12{m}^{3}+8m)=0$

${m}^{5}+6{m}^{4}+12{m}^{3}+8m=0$

$m({m}^{4}+6{m}^{3}+12{m}^{2}+8)=0$

$m=0,{m}^{4}+6{m}^{3}+12{m}^{2}+8=0$

$y=0$ , no solution for $4\in R$

Substitute the trail solution and its derivatives in the given D.E.

soanooooo40

Beginner2021-12-30Added 35 answers

karton

Expert2022-01-10Added 613 answers

where

The auxilary equation is given by

Here we see that the roots y the auxiliary equation are repeated.

Hence complementary function is given by

Hence general solution is

- 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)$\frac{1}{s}+2$

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(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:

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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

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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:

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

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

$K=0$

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