Solve the differential equation $\frac{dy}{dx}=\frac{y-x}{x-2y}$

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Final answer to the problem

$\frac{-1}{\sqrt{\left(2\right)^{3}}}\ln\left|\frac{\sqrt{2}y}{x}+1\right|+\frac{1}{\sqrt{\left(2\right)^{3}}}\ln\left|\frac{\sqrt{2}y}{x}-1\right|-\frac{1}{2}\ln\left|\frac{-x^2+2y^2}{2x^2}\right|=\ln\left|x\right|+C_0$
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Step-by-step Solution

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We can identify that the differential equation $\frac{dy}{dx}=\frac{y-x}{x-2y}$ 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

$\frac{dy}{dx}=\frac{y-x}{x-2y}$

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$\frac{dy}{dx}=\frac{y-x}{x-2y}$

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Learn how to solve problems step by step online. Solve the differential equation dy/dx=(y-x)/(x-2y). We can identify that the differential equation \frac{dy}{dx}=\frac{y-x}{x-2y} 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. Integrate both sides of the differential equation, the left side with respect to u, and the right side with respect to x.

Final answer to the problem

$\frac{-1}{\sqrt{\left(2\right)^{3}}}\ln\left|\frac{\sqrt{2}y}{x}+1\right|+\frac{1}{\sqrt{\left(2\right)^{3}}}\ln\left|\frac{\sqrt{2}y}{x}-1\right|-\frac{1}{2}\ln\left|\frac{-x^2+2y^2}{2x^2}\right|=\ln\left|x\right|+C_0$

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Function Plot

Plotting: $\frac{dy}{dx}+\frac{-y+x}{x-2y}$

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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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