avissidep

2021-01-05

Give the correct answer and solve the given equation
$\left(x+y\right)dx+\left(x-y\right)dy=0$

SchulzD

First rewrite it as $\left(x-y\right)dy=-\left(x+y\right)dx$ (1)
Suppose that . Then
$0=-2x$,
so $y=x$ is not a solution of the starding equation. Therefore, $y\ne x$, and we can divide (1) by $x-y$:
$dy=\frac{-\left(x+y\right)}{x-y}dx=\frac{x+y}{y-x}dx$
We can also write this as
${y}^{\prime }=\frac{dy}{dx}=\frac{x+y}{y-x}=\frac{y-x+2x}{y-x}=1+2\frac{x}{y-x}=1+\frac{2}{\frac{y}{x}-1}$ (2)
Now use the substitution
$u=\frac{y}{x}$,
which we can also write as
$y-ux,$
to get that
${y}^{\prime }={u}^{\prime }x+u$ (3)
From(2),
${y}^{\prime }=1+\frac{2}{\frac{y}{x}-1}=1+\frac{2}{u-1}$
Combining this and (3) we get
${u}^{\prime }x+u=1+\frac{2}{u-1}$
Furthermore,
${u}^{\prime }x=1-u+\frac{2}{u-1}=\frac{\left(1-u\right)\left(u-1\right)+2}{u-1}=\frac{-{\left(u-1\right)}^{2}+2}{u-1}=-\frac{{\left(u-1\right)}^{2}+2}{u-1}$
Now write ${u}^{\prime }=d\frac{u}{dx}:$
$d\frac{u}{dx}\cdot x=-\frac{{\left(u-1\right)}^{2}+2}{u-1}$
Whis is a separable equation, because we can write it in the form h
$h\left(u\right)du=g\left(x\right)dx$
To get this, first divide the equation by $\frac{{\left(u-1\right)}^{2}+2}{u-1}$ (notice that this is always nonzero!):
$\frac{u-1}{{\left(u-1\right)}^{2}+2}\cdot d\frac{u}{dx}\cdot x=-1$
Now divide the equation by x:
$\frac{u-1}{{\left(u-1\right)}^{2}+2}\cdot \frac{du}{dx}=-\frac{1}{x}$
Finally, "multiply" thq equation by dx:

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