$y^2dx=\left(xy-x^2\right)dy$

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Final answer to the problem

$\frac{y}{-x}=-\ln\left|y\right|+C_0$
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We can identify that the differential equation $y^2dx=\left(xy-x^2\right)dy$ is homogeneous, since it is written in the standard form $M(x,y)dx+N(x,y)dy=0$, 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

$y^2dx=\left(xy-x^2\right)dy$

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$y^2dx=\left(xy-x^2\right)dy$

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Learn how to solve equations différentielles problems step by step online. y^2dx=(xy-x^2)dy. We can identify that the differential equation y^2dx=\left(xy-x^2\right)dy is homogeneous, since it is written in the standard form M(x,y)dx+N(x,y)dy=0, 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: x=uy. Expand and simplify. Group the terms of the differential equation. Move the terms of the u variable to the left side, and the terms of the y variable to the right side of the equality.

Final answer to the problem

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

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Function Plot

Plotting: $y^2dx-\left(xy-x^2\right)dy$

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1
2
3
4
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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