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

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Final answer to the problem

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

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

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

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

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Learn how to solve problems step by step online. dy/dx=(xy)/(x^2+y^2). We can identify that the differential equation \frac{dy}{dx}=\frac{xy}{x^2+y^2} 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. Apply the formula: b\cdot dy=a\cdot dx\to \int bdy=\int adx, where a=\frac{1}{y}, b=u, dx=dy, dy=du, dyb=dxa=u\cdot du=\frac{1}{y}dy, dyb=u\cdot du and dxa=\frac{1}{y}dy.

Final answer to the problem

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

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Function Plot

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

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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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