Donald Johnson

2021-12-21

Stella Calderon

Beginner2021-12-22Added 35 answers

The given differential equation is

$(2xy-3{x}^{2})dx+({x}^{2}+y)dy=0$

Comparing it with$Mdx+Mdy=0$

$M=2xy-3{x}^{2}\text{}N={x}^{2}+y$

$\frac{dM}{dy}=2x\text{}\frac{dN}{dx}=2x$

Here,$\frac{dM}{dy}=\frac{dN}{dx}$

Then, it is exact differential equation, so solution is

$\int Mdx+\int N$ (term not contining x) $dy=c$

$\int (2x-3{x}^{2})dx+\int ydy=c$

$y\cdot \frac{2{x}^{2}}{2}-\frac{3{x}^{3}}{3}+\frac{{y}^{2}}{2}=c$

${x}^{2}y-{x}^{3}+\frac{{y}^{2}}{2}=c$

This is solution of given DE.

Comparing it with

Here,

Then, it is exact differential equation, so solution is

This is solution of given DE.

Charles Benedict

Beginner2021-12-23Added 32 answers

Assuming $(2xy-3{x}^{2})dx+({x}^{2}+y)dy=0$

Then$2xy+({x}^{2}+y)\frac{dy}{dx}=3{x}^{2}$ (1)

Notice that$d\frac{{x}^{2}y+\frac{{y}^{2}}{2}}{dx}$ is:

$x}^{2}\frac{dy}{dx}+2xy+2\frac{y}{2}\frac{dy}{dx$

$=2xy+({x}^{2}+y)\frac{dy}{dx}$ which is left hand side of (1)

So integrating (1) gives

${x}^{2}y+\frac{{y}^{2}}{2}={x}^{3}+k$

So${y}^{2}+2{x}^{2}y-2{x}^{3}+c=0$ is the solution

Then

Notice that

So integrating (1) gives

So

RizerMix

Expert2021-12-29Added 656 answers

Let

Let

since

The given Equation

Integrating

or,

Nick Camelot

Skilled2023-05-23Added 164 answers

Mr Solver

Skilled2023-05-23Added 147 answers

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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

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(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

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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?

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

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:

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

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

$K=0$

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