killjoy1990xb9

2022-01-20

Solving simple differential equation

$(t+4)dx=4(1+{x}^{2})dt$

Bob Huerta

Beginner2022-01-20Added 41 answers

As you rearranged it in your post, we have

$\int \frac{dx}{1+{x}^{2}}=4\int \frac{dt}{t+4}$

As you discovered, the LHS is arctan. To find the value of the RHS, we may say$t+4=u\text{}\text{so}\text{}du=dt$ . Thus the RHS can be rewritten as

$4\int \frac{du}{u}=4\mathrm{log}\left|u\right|+{C}_{2}$

Because we want this integral to be in respect to t, we have$u=t+4$ , so

$\mathrm{arctan}x+{C}_{1}=4\mathrm{log}|t+4|+{C}_{2}$

To confirm that the integral we solved is indeed correct, you can simply differentiate$4\mathrm{log}|t+4|+{C}_{2}$

Thus$x=\mathrm{tan}(4\mathrm{log}|t+4|+{C}_{2}-{C}_{1})$

and$t=\mathrm{exp}\left(\frac{1}{4}(\mathrm{arctan}x+{C}_{1}-{C}_{2})\right)-4$

As you discovered, the LHS is arctan. To find the value of the RHS, we may say

Because we want this integral to be in respect to t, we have

To confirm that the integral we solved is indeed correct, you can simply differentiate

Thus

and

Dawn Neal

Beginner2022-01-21Added 35 answers

Here are some proceeding steps:

- Seperating the variables you get

$\frac{dx}{1+{x}^{2}}=4\cdot \frac{dt}{t+4}$

- Integrating both sides you have:

$\mathrm{tan}}^{-1}\left(x\right)+{C}_{1}=4\cdot \mathrm{log}|t+4|+{C}_{2$ .

- Seperating the variables you get

- Integrating both sides you have:

- 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

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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:

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

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

$K=0$

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