avissidep
2021-01-05
Answered

Give the correct answer and solve the given equation

$(x+y)dx+(x-y)dy=0$

You can still ask an expert for help

SchulzD

Answered 2021-01-06
Author has **83** answers

First rewrite it as $(x-y)dy=-(x+y)dx$ (1)

Suppose that$x-y=0,\text{}\text{so}\text{}y=x$ . Then

$0=-2x$ ,

so$y=x$ is not a solution of the starding equation. Therefore, $y\ne x$ , and we can divide (1) by $x-y$ :

$dy=\frac{-(x+y)}{x-y}dx=\frac{x+y}{y-x}dx$

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,

$u}^{\prime}x=1-u+\frac{2}{u-1}=\frac{(1-u)(u-1)+2}{u-1}=\frac{-{(u-1)}^{2}+2}{u-1}=-\frac{{(u-1)}^{2}+2}{u-1$

Now write${u}^{\prime}=d\frac{u}{dx}:$

$d\frac{u}{dx}\cdot x=-\frac{{(u-1)}^{2}+2}{u-1}$

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$\frac{{(u-1)}^{2}+2}{u-1}$ (notice that this is always nonzero!):

$\frac{u-1}{{(u-1)}^{2}+2}\cdot d\frac{u}{dx}\cdot x=-1$

Now divide the equation by x:

$\frac{u-1}{{(u-1)}^{2}+2}\cdot \frac{du}{dx}=-\frac{1}{x}$

Finally, "multiply" thq equation by dx:

Suppose that

so

We can also write this as

Now use the substitution

which we can also write as

to get that

From(2),

Combining this and (3) we get

Furthermore,

Now write

Whis is a separable equation, because we can write it in the form h

To get this, first divide the equation by

Now divide the equation by x:

Finally, "multiply" thq equation by dx:

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