Solve the differential equation $\frac{dy}{dx}=\frac{y^2}{2x^2+3xy}$

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Final answer to the problem

$\frac{1}{2}\ln\left(\frac{x}{y}\right)-\frac{1}{2}\ln\left(\frac{x}{y}+1\right)=\ln\left(y\right)+C_0$
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We can identify that the differential equation $\frac{dy}{dx}=\frac{y^2}{2x^2+3xy}$ 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

$\frac{dy}{dx}=\frac{y^2}{2x^2+3xy}$

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$\frac{dy}{dx}=\frac{y^2}{2x^2+3xy}$

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Learn how to solve problems step by step online. Solve the differential equation dy/dx=(y^2)/(2x^2+3xy). We can identify that the differential equation \frac{dy}{dx}=\frac{y^2}{2x^2+3xy} 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: x=uy. Expand and simplify. Integrate both sides of the differential equation, the left side with respect to u, and the right side with respect to y.

Final answer to the problem

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

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Function Plot

Plotting: $\frac{dy}{dx}+\frac{-y^2}{2x^2+3xy}$

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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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