Step-by-step Solution

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

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Final Answer

$y=\sqrt[3]{C_0+x^2}$
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Step-by-step Solution

Problem to solve:

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

Choose the solving method

1

Take $\frac{2}{3}$ out of the fraction

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

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

$y^2dy-\frac{2}{3}xdx=0$
3

The differential equation $y^2dy-\frac{2}{3}xdx=0$ 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 $f(x,y)=C$

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

Find the derivative of $M(x,y)$ with respect to $y$

$\frac{d}{dy}\left(-\frac{2}{3}x\right)$

The derivative of the constant function ($-\frac{2}{3}x$) is equal to zero

0

Find the derivative of $N(x,y)$ with respect to $x$

$\frac{d}{dx}\left(y^2\right)$

The derivative of the constant function ($y^2$) is equal to zero

0
4

Using the test for exactness, we check that the differential equation is exact

$0=0$

The integral of a constant by a function is equal to the constant multiplied by the integral of the function

$-\frac{2}{3}\int xdx$

Applying the power rule for integration, $\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}$, where $n$ represents a number or constant function, in this case $n=1$

$-\frac{1}{3}x^2$

Since $y$ is treated as a constant, we add a function of $y$ as constant of integration

$-\frac{1}{3}x^2+g(y)$
5

Integrate $M(x,y)$ with respect to $x$ to get

$-\frac{1}{3}x^2+g(y)$

The derivative of the constant function ($-\frac{1}{3}x^2$) is equal to zero

0

The derivative of $g(y)$ is $g'(y)$

$0+g'(y)$
6

Now take the partial derivative of $-\frac{1}{3}x^2$ with respect to $y$ to get

$0+g'(y)$

Simplify and isolate $g'(y)$

$y^2=0+g$

$x+0=x$, where $x$ is any expression

$y^2=g$

Rearrange the equation

$g=y^2$
7

Set $y^2$ and $0+g'(y)$ equal to each other and isolate $g'(y)$

$g'(y)=y^2$

Integrate both sides with respect to $y$

$g=\int y^2dy$

Apply the power rule for integration, $\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}$, where $n$ represents a number or constant function, such as $2$

$g=\frac{y^{3}}{3}$
8

Find $g(y)$ integrating both sides

$g(y)=\frac{y^{3}}{3}$
9

We have found our $f(x,y)$ and it equals

$f(x,y)=-\frac{1}{3}x^2+\frac{y^{3}}{3}$
10

Then, the solution to the differential equation is

$-\frac{1}{3}x^2+\frac{y^{3}}{3}=C_0$

We need to isolate the dependent variable $y$, we can do that by subtracting $-\frac{1}{3}x^2$ from both sides of the equation

$\frac{y^{3}}{3}=C_0+\frac{1}{3}x^2$

Simplify the fraction

$\frac{1}{3}y^{3}=C_0+\frac{1}{3}x^2$

Eliminate the $\frac{1}{3}$ from the left, multiplying both sides of the equation by the inverse of $\frac{1}{3}$

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

Solve the product $3\left(C_0+\frac{1}{3}x^2\right)$

$y^{3}=3C_0+x^2$

We can rename $3C_0$ as other constant

$y^{3}=C_0+x^2$

Removing the variable's exponent

$y=\sqrt[3]{C_0+x^2}$
11

Find the explicit solution to the differential equation

$y=\sqrt[3]{C_0+x^2}$

Final Answer

$y=\sqrt[3]{C_0+x^2}$
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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

Tips on how to improve your answer:

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

Time to solve it:

~ 0.14 s