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

## Step-by-step Solution

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###  Videos

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

Specify the solving method

1

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

$3y^2dy-2xdx=0$
2

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

$3y^2dy-2xdx=0$

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

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

The derivative of the constant function ($-2x$) is equal to zero

0

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

$\frac{d}{dx}\left(3y^2\right)$

The derivative of the constant function ($3y^2$) is equal to zero

0
3

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

$0=0$

The integral of a function times a constant ($-2$) is equal to the constant times the integral of the function

$-2\int xdx$

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 $n=1$

$-x^2$

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

$-x^2+g(y)$
4

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

$-x^2+g(y)$

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

0

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

$0+g'(y)$
5

Now take the partial derivative of $-x^2$ with respect to $y$ to get

$0+g'(y)$

Simplify and isolate $g'(y)$

$3y^2=0+g$

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

$3y^2=g$

Rearrange the equation

$g=3y^2$
6

Set $3y^2$ and $0+g'(y)$ equal to each other and isolate $g'(y)$

$g'(y)=3y^2$

Integrate both sides with respect to $y$

$g=\int3y^2dy$

The integral of a function times a constant ($3$) is equal to the constant times the integral of the function

$g=3\int y^2dy$

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 $2$

$g=1y^{3}$

Any expression multiplied by $1$ is equal to itself

$g=y^{3}$
7

Find $g(y)$ integrating both sides

$g(y)=y^{3}$
8

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

$f(x,y)=-x^2+y^{3}$
9

Then, the solution to the differential equation is

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

We need to isolate the dependent variable $y$, we can do that by simultaneously subtracting $-x^2$ from both sides of the equation

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

Multiply $-1$ times $-1$

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

Removing the variable's exponent raising both sides of the equation to the power of $\frac{1}{3}$

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

Divide $1$ by $3$

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

Simplify $\sqrt[3]{y^{3}}$ using the power of a power property: $\left(a^m\right)^n=a^{m\cdot n}$. In the expression, $m$ equals $3$ and $n$ equals $\frac{1}{3}$

$y^{3\frac{1}{3}}$

Multiply $3$ times $\frac{1}{3}$

$y$

Multiply $3$ times $\frac{1}{3}$

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

Divide $1$ by $3$

$y=\sqrt[3]{x^2+C_0}$
10

Find the explicit solution to the differential equation. We need to isolate the variable $y$

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

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

##  Explore different ways to solve this problem

Solving a math problem using different methods is important because it enhances understanding, encourages critical thinking, allows for multiple solutions, and develops problem-solving strategies. Read more

Linear Differential EquationExact Differential EquationSeparable Differential EquationHomogeneous Differential Equation

SnapXam A2

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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

### Main Topic: Integral Calculus

Integration assigns numbers to functions in a way that can describe displacement, area, volume, and other concepts that arise by combining infinitesimal data.