$\frac{dy}{dx}=\frac{x+3y}{x-y}$

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Final answer to the problem

$\frac{-2x}{y+x}-\ln\left|\frac{y}{x}+1\right|=\ln\left|x\right|+C_0$
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We can identify that the differential equation $\frac{dy}{dx}=\frac{x+3y}{x-y}$ is homogeneous, since it is written in the standard form $\frac{dy}{dx}=\frac{M(x,y)}{N(x,y)}$, where $M(x,y)$ and $N(x,y)$ are the partial derivatives of a two-variable function $f(x,y)$ and both are homogeneous functions of the same degree

$\frac{dy}{dx}=\frac{x+3y}{x-y}$

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$\frac{dy}{dx}=\frac{x+3y}{x-y}$

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Learn how to solve problems step by step online. dy/dx=(x+3y)/(x-y). We can identify that the differential equation \frac{dy}{dx}=\frac{x+3y}{x-y} is homogeneous, since it is written in the standard form \frac{dy}{dx}=\frac{M(x,y)}{N(x,y)}, where M(x,y) and N(x,y) are the partial derivatives of a two-variable function f(x,y) and both are homogeneous functions of the same degree. Use the substitution: y=ux. Expand and simplify. Apply the formula: b\cdot dy=a\cdot dx\to \int bdy=\int adx, where a=\frac{1}{x}, b=\frac{1-u}{\left(u+1\right)^{2}}, dy=du, dyb=dxa=\frac{1-u}{\left(u+1\right)^{2}}du=\frac{1}{x}dx, dyb=\frac{1-u}{\left(u+1\right)^{2}}du and dxa=\frac{1}{x}dx.

Final answer to the problem

$\frac{-2x}{y+x}-\ln\left|\frac{y}{x}+1\right|=\ln\left|x\right|+C_0$

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Function Plot

Plotting: $\frac{dy}{dx}+\frac{-x-3y}{x-y}$

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5
6
7
8
9
0
a
b
c
d
f
g
m
n
u
v
w
x
y
z
.
(◻)
+
-
×
◻/◻
/
÷
2

e
π
ln
log
log
lim
d/dx
Dx
|◻|
θ
=
>
<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

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