deiteresfp

2022-01-17

Solve the following equation?

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

Piosellisf

Beginner2022-01-18Added 40 answers

For this one we look for an integrating factor which will make the equation exact.

Let$M=x-2xy-{y}^{2}\text{}\text{and}\text{}N={y}^{2}$ . Then ${M}_{x}=1-2y\text{}\text{and}\text{}{N}_{y}=2y$ .

Thus$\frac{{M}_{x}-{N}_{y}}{N}=\frac{1-2y}{{y}^{2}}=\frac{1}{{y}^{2}}-2\frac{1}{y}$

which is purely a function of y. This tells us that there is an integrating factor$\mu \left(y\right)$ which makes the equation exact which satisfies

$\frac{d\mu}{dy}=(\frac{1}{{y}^{2}}-2\frac{1}{y})\mu$

Clearly$\mu \left(y\right)={e}^{\int (\frac{1}{{y}^{2}}-2\frac{1}{y})dy}={e}^{\frac{-1}{y}-2\mathrm{log}y}={y}^{-2}{e}^{y-1}$ .

Try (first checking that my algebra is correct) and then multiplying the differential equation by this integrating factor, it should become an exact differential equation which I'm guessing you know how to solve if you are asking this question.

Let

Thus

which is purely a function of y. This tells us that there is an integrating factor

Clearly

Try (first checking that my algebra is correct) and then multiplying the differential equation by this integrating factor, it should become an exact differential equation which I'm guessing you know how to solve if you are asking this question.

Cleveland Walters

Beginner2022-01-19Added 40 answers

Hint: Your equation can be turned into a first order linear differential equation in $\frac{dx}{dy}$ (i.e., just multiply the equation by $\frac{dx}{dy}$ and it is a standard form....)

alenahelenash

Expert2022-01-24Added 556 answers

As it looks some people misunderstood (I apologize, the fault may be my side) my first answer, here is the complete solution after reducing the problem into standard form: As I wrote in the first answer, multiplying both sides by $\frac{dx}{dy}$ (the hint was that it will make the equation linear in x),
$\frac{dx}{dy}+(\frac{1-2y}{{y}^{2}})\cdot x=1$ . This is a linear equation (in x) of the standard form $\frac{dx}{dy}+P(y)\cdot x=Q(x)$ . So, ${e}^{\int P(y)dy}={e}^{\int (\frac{1-2y}{{y}^{2}})dy}=\frac{{e}^{\frac{-1}{y}}}{{y}^{2}}$ is an integrating factor (IF). So, multiplying by the IF and integrating the general solution is given by
$x\cdot \frac{{e}^{\frac{-1}{y}}}{{y}^{2}}=\int \frac{{e}^{\frac{-1}{y}}}{{y}^{2}}dy+c={e}^{\frac{-1}{y}}+c$
Or, $x={y}^{2}(1+c{e}^{\frac{1}{y}})$ .

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B) $\frac{5}{9}$ degree on fahrenheit scale

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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:

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

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

$K=0$

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