Solve the differential equation $\frac{dy}{dx}=\frac{-2xy}{x^2-8y^2}$

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Final answer to the problem

$yx^2+\frac{-8y^{3}}{3}=C_0$
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Step-by-step Solution

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Rewrite the differential equation in the standard form $M(x,y)dx+N(x,y)dy=0$

$x^2-8y^2dy2xydx=0$

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$x^2-8y^2dy2xydx=0$

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Learn how to solve problems step by step online. Solve the differential equation dy/dx=(-2xy)/(x^2-8y^2). Rewrite the differential equation in the standard form M(x,y)dx+N(x,y)dy=0. The differential equation x^2-8y^2dy2xydx=0 is exact, 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 they satisfy the test for exactness: \displaystyle\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}. In other words, their second partial derivatives are equal. The general solution of the differential equation is of the form f(x,y)=C. Using the test for exactness, we check that the differential equation is exact. Integrate M(x,y) with respect to x to get.

Final answer to the problem

$yx^2+\frac{-8y^{3}}{3}=C_0$

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Function Plot

Plotting: $\frac{dy}{dx}+\frac{2xy}{x^2-8y^2}$

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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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