David Troyer

2022-01-17

Find the general solution to the differential equation

$\frac{dy}{dt}={t}^{3}+2{t}^{2}-8t$

Also, part 8B. asks: Show that the constant function$y\left(t\right)=0$ is a solution.

Ive

Also, part 8B. asks: Show that the constant function

Ive

Maria Lopez

Beginner2022-01-18Added 32 answers

Parts of problems that say ''show that [explicitly given function] is a solution'' can be solved by simply ''plugging in''. E.g., if you're supposed to show that $y\left(t\right)=\frac{1}{t-t}$ is a solution to the differential equation $\frac{dy}{dt}={y}^{2}$ , then you should compute $\frac{dy}{dt}=\frac{1}{{(1-t)}^{2}}$ , compute $y}^{2}=\frac{1}{{(1-t)}^{2}$ , and ''plug them in'' to check that the differential equation is satisfied for this choice of y.

You're correct that the difference in the second problem is that you no longer have the derivative with respect to t expressed as a function of t. When the equation involves both a function y and its derivative (or derivatives), it can generally be more difficult to solve. In this case, a hint I'll give you is that$\frac{dy}{dt}=\frac{1}{\frac{dt}{dy}}$

You can first turn things around and think of t as a function of y. You can find t(y) using similar methods to the first problem (but with a more complicated function), and then find the inverse function to get y(t).

You're correct that the difference in the second problem is that you no longer have the derivative with respect to t expressed as a function of t. When the equation involves both a function y and its derivative (or derivatives), it can generally be more difficult to solve. In this case, a hint I'll give you is that

You can first turn things around and think of t as a function of y. You can find t(y) using similar methods to the first problem (but with a more complicated function), and then find the inverse function to get y(t).

Janet Young

Beginner2022-01-19Added 32 answers

The difference is that the main variable is y and you separate the variables to integrate and obtain the solution

Hint 1:

The first equation is very easy, so:

$\frac{dy}{dt}={t}^{3}+2{t}^{2}-8t\Rightarrow dy=({t}^{3}+2{t}^{2}-8t)dt\Rightarrow \int dy=\int ({t}^{3}+2{t}^{2}-8t)dt$

Hint 2:

For the other.

$\frac{dy}{dt}={y}^{3}+2{y}^{2}-8y\Rightarrow \frac{1}{{y}^{3}+2{y}^{2}-8y}dy=dt$

Then proceed integrating.

Hint 1:

The first equation is very easy, so:

Hint 2:

For the other.

Then proceed integrating.

alenahelenash

Skilled2022-01-24Added 474 answers

In the first question, you are given the derivative in terms of the variable. But in the second question, you are given an expression for the derivative that involves the function. For instance, it would be one thing if you were told $\frac{dy}{dx}=x$
(which would mean that $y=\frac{1}{2}{x}^{2}+C$ ), and a completely different thing if you are told $\frac{dy}{dx}=y$ (this tells you that the function y is equal to its derivative; which means that $y=A{e}^{x}$ for some constant A).
Actually, we can solve the second differential equation; but we dont

- 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:

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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:

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

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

$K=0$

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