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Separable differential equations Calculator

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1

Solved example of separable differential equations

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

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

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

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

$y^2dy-\frac{2}{3}xdx=0$
4

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
5

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

$-\frac{1}{3}x^2+g(y)$
6

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 $g'(y)$

$0+g'(y)$
7

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

$g=y^2$
8

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 $2$

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

Find $g(y)$ integrating both sides

$g(y)=\frac{y^{3}}{3}$
10

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

$f(x,y)=-\frac{1}{3}x^2+\frac{y^{3}}{3}$
11

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

$y=\sqrt[3]{C_0+x^2}$
12

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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