reproacht3

2021-12-26

Find the solution of the following Differential Equations
${y}^{\prime }=\left(y-x\right)$

puhnut1m

$\frac{dy}{dx}={\left(y-x\right)}^{2}$
Let $y-x=t$
$\frac{dy}{dx}-1=\frac{dt}{dx}$
$⇒\frac{dy}{dx}=\frac{dt}{dx}+1$
$\frac{dt}{dx}+1={t}^{2}$
$\frac{dt}{dx}={t}^{2}-1$
$\int \frac{dt}{{t}^{2}-1}=\int dx$
$\frac{1}{2}\mathrm{ln}\left(\frac{t-1}{t+1}\right)=x+c$
$\mathrm{ln}\left(\frac{y-x-1}{y-x+1}\right)=2x+2c$

Jimmy Macias

$y=x+\frac{{a}^{2}+{x}^{2}}{{a}^{2}-{x}^{2}}$
where a is a constant of integration.
Explanation:
Let us change variables to $u=y-x$
Then ${u}^{\prime }={y}^{\prime }-1$ and so the differential equation becomes
${u}^{\prime }={u}^{2}-1⇒\frac{du}{{u}^{2}-1}=dx⇒\frac{1}{2}\mathrm{log}|\frac{u-1}{u+1}|=\mathrm{log}x-\mathrm{log}a$
where we have written the constant of integration as $-\mathrm{log}a$. Thus $\frac{u-1}{u+1}=\frac{{x}^{2}}{{a}^{2}}⇒u=\frac{{a}^{2}+{x}^{2}}{{a}^{2}-{x}^{2}}$

karton

Explanation:
We have: ${y}^{\prime }=\left(y-x{\right)}^{2}\left[A\right]$
Perform the substitution:
u=y-x
Then differentiating wrt x, and applying the product rule:
$\frac{du}{dx}=\frac{dy}{dx}-1⇒\frac{dy}{dx}=1+\frac{du}{dx}$
Substituting into the DE:
$1+\frac{du}{dx}={u}^{2}$
$\therefore \frac{du}{dx}={u}^{2}-1$
Reducing the DE to a separable form:
$\frac{1}{{u}^{2}-1}\frac{du}{dx}=1$
So we can ''separate the variables'' to get:
$\int \frac{1}{{u}^{2}-1}du=\int dx$
We can perform partial fraction decomposition of the LHS integrand:
$\frac{1}{{u}^{2}-1}=\frac{1}{\left(u+1\right)\left(u-1\right)}$
$=\frac{A}{u+1}+\frac{B}{u-1}$
And using the ''cover up'' methods we get:
$A=-\frac{1}{2},B=\frac{1}{2}$
Allowing us to write the equation as:
$\frac{1}{2}\int \frac{1}{u-1}-\frac{1}{u+1}du=\int dx$
And integrating, we get:
$\begin{array}{}\\ \frac{1}{2}\mathrm{ln}\left(u-1\right)-\mathrm{ln}\left(u+1\right)=x+C\\ \mathrm{ln}\left(\frac{u-1}{u+1}\right)=2x+2C\\ \frac{u-1}{u+1}={e}^{2x+2C}\\ u-1=\left(u+1\right)A{e}^{2x}\\ u-1=uA{e}^{2x}+A{e}^{2x}\\ u-uA{e}^{2x}=1+A{e}^{2x}\\ u\left(1-A{e}^{2x}\right)=1+A{e}^{2x}\\ u=\frac{1+A{e}^{2x}}{1-A{e}^{2x}}\end{array}$

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