Step-by-step Solution

Solve the differential equation $\frac{dy}{dx}-\left(\frac{y}{x}\right)=\frac{x}{3y}$

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$y=\sqrt{x^{2}\left(\ln\left(\sqrt[3]{x^{2}}\right)+C_0\right)},\:y=-\sqrt{x^{2}\left(\ln\left(\sqrt[3]{x^{2}}\right)+C_0\right)}$

Step-by-step explanation

Problem to solve:

$\frac{dy}{dx}\:-\frac{y}{x}=\frac{x}{3y}$
1

Multiplying the fraction by $-1$

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

We identify that the differential equation $\frac{dy}{dx}+\frac{-y}{x}=\frac{x}{3y}$ is a Bernoulli differential equation since it's of the form $\frac{dy}{dx}+P(x)y=Q(x)y^n$, where $n$ is any real number different from $0$ and $1$. To solve this equation, we can apply the following substitution. Let's define a new variable $u$ and set it equal to

$u=y^{\left(1-n\right)}$
3

Plug in the value of $n$, which equals $-1$

$u=y^{\left(1-1\cdot -1\right)}$
4

Simplify

$u=y^{2}$

Rearrange the equation

$y^{2}=u$

Removing the variable's exponent

$y=\pm \sqrt{u}$

As in the equation we have the sign $\pm$, this produces two identical equations that differ in the sign of the term $\sqrt{u}$. We write and solve both equations, one taking the positive sign, and the other taking the negative sign

$y=\sqrt{u},\:y=-\sqrt{u}$

Let's just take the first equation

$y=\sqrt{u}$
5

Isolate the dependent variable $y$

$y=\sqrt{u}$

The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$

$\frac{dy}{dx}=\frac{1}{2}u^{-\frac{1}{2}}$
6

Differentiate both sides of the equation with respect to the independent variable $x$

$\frac{dy}{dx}=\frac{1}{2}u^{-\frac{1}{2}}\frac{du}{dx}$
7

Now, substitute $\frac{dy}{dx}=\frac{1}{2}u^{-\frac{1}{2}}\frac{du}{dx}$ and $y=\sqrt{u}$ on the original differential equation

$\frac{1}{2}u^{-\frac{1}{2}}\frac{du}{dx}+\frac{-\sqrt{u}}{x}=\frac{x}{3\sqrt{u}}$
8

Simplify

$\frac{1}{2}u^{-\frac{1}{2}}\frac{du}{dx}+\frac{-\sqrt{u}}{x}=\frac{x}{3\sqrt{u}}$
9

We need to cancel the term that is in front of $\frac{du}{dx}$. We can do that by multiplying the whole differential equation by $\frac{1}{2}u^{\frac{1}{2}}$

$\left(\frac{1}{2}u^{-\frac{1}{2}}\frac{du}{dx}+\frac{-\sqrt{u}}{x}=\frac{x}{3\sqrt{u}}\right)\left(\frac{1}{2}u^{\frac{1}{2}}\right)$
10

Multiply both sides by $\frac{1}{2}u^{\frac{1}{2}}$

$\frac{1}{2}\sqrt{u}\left(\frac{1}{2}u^{-\frac{1}{2}}\frac{du}{dx}+\frac{-\sqrt{u}}{x}\right)=\frac{x}{3\sqrt{u}}\frac{1}{2}\sqrt{u}$

Multiplying polynomials $\frac{1}{2}\sqrt{u}$ and $\frac{1}{2}u^{-\frac{1}{2}}\frac{du}{dx}+\frac{-\sqrt{u}}{x}$

$\frac{1}{4}u^{-\frac{1}{2}}\sqrt{u}\left(\frac{du}{dx}\right)+\frac{1}{2}\sqrt{u}\left(\frac{-\sqrt{u}}{x}\right)=\frac{1}{2}\left(\frac{x}{3\sqrt{u}}\right)\sqrt{u}$

When multiplying exponents with same base we can add the exponents

$\frac{1}{4}\left(\frac{du}{dx}\right)+\frac{1}{2}\sqrt{u}\left(\frac{-\sqrt{u}}{x}\right)=\frac{1}{2}\left(\frac{x}{3\sqrt{u}}\right)\sqrt{u}$
11

Expand and simplify. Now we see that the differential equation looks like a linear differential equation, because we removed the original $y^{-1}$ term

$\frac{1}{4}\left(\frac{du}{dx}\right)+\frac{1}{2}\sqrt{u}\left(\frac{-\sqrt{u}}{x}\right)=\frac{1}{2}\left(\frac{x}{3\sqrt{u}}\right)\sqrt{u}$
12

Multiplying the fraction by $\frac{1}{2}$

$\frac{1}{4}\left(\frac{du}{dx}\right)+\frac{-\frac{1}{2}\sqrt{u}}{x}\sqrt{u}=\frac{1}{2}\left(\frac{x}{3\sqrt{u}}\right)\sqrt{u}$
13

Multiplying the fraction by $\sqrt{u}$

$\frac{1}{4}\left(\frac{du}{dx}\right)+\frac{-\frac{1}{2}\sqrt{u}\sqrt{u}}{x}=\frac{1}{2}\left(\frac{x}{3\sqrt{u}}\right)\sqrt{u}$
14

When multiplying exponents with same base we can add the exponents

$\frac{1}{4}\left(\frac{du}{dx}\right)+\frac{-\frac{1}{2}u}{x}=\frac{1}{2}\left(\frac{x}{3\sqrt{u}}\right)\sqrt{u}$
15

Multiplying the fraction by $\frac{1}{2}$

$\frac{1}{4}\left(\frac{du}{dx}\right)+\frac{-\frac{1}{2}u}{x}=\frac{\frac{1}{6}x}{\sqrt{u}}\sqrt{u}$
16

Simplify the fraction $\frac{\frac{1}{6}x}{\sqrt{u}}$

$\frac{1}{4}\left(\frac{du}{dx}\right)+\frac{-\frac{1}{2}u}{x}=\frac{x}{6\sqrt{u}}\sqrt{u}$
17

Multiplying the fraction by $\sqrt{u}$

$\frac{1}{4}\left(\frac{du}{dx}\right)+\frac{-\frac{1}{2}u}{x}=\frac{x\sqrt{u}}{6\sqrt{u}}$
18

Simplify the fraction $\frac{x\sqrt{u}}{6\sqrt{u}}$ by $\sqrt{u}$

$\frac{1}{4}\left(\frac{du}{dx}\right)+\frac{-\frac{1}{2}u}{x}=\frac{x}{6}$
19

Divide all the terms of the differential equation by $\frac{1}{4}$

$\frac{\frac{1}{4}}{\frac{1}{4}}\frac{du}{dx}+\frac{\frac{-\frac{1}{2}u}{x}}{\frac{1}{4}}=\frac{\frac{x}{6}}{\frac{1}{4}}$

Divide $\frac{1}{4}$ by $\frac{1}{4}$

$1\left(\frac{du}{dx}\right)+\frac{\frac{-\frac{1}{2}u}{x}}{\frac{1}{4}}=\frac{\frac{x}{6}}{\frac{1}{4}}$

Any expression multiplied by $1$ is equal to itself

$\frac{du}{dx}+\frac{\frac{-\frac{1}{2}u}{x}}{\frac{1}{4}}=\frac{\frac{x}{6}}{\frac{1}{4}}$

Divide fractions $\frac{\frac{-\frac{1}{2}u}{x}}{\frac{1}{4}}$ with Keep, Change, Flip: $\frac{a}{b}\div c=\frac{a}{b}\div\frac{c}{1}=\frac{a}{b}\times\frac{1}{c}=\frac{a}{b\cdot c}$

$\frac{du}{dx}+\frac{-2u}{x}=\frac{\frac{x}{6}}{\frac{1}{4}}$

Divide fractions $\frac{\frac{x}{6}}{\frac{1}{4}}$ with Keep, Change, Flip: $\frac{a}{b}\div c=\frac{a}{b}\div\frac{c}{1}=\frac{a}{b}\times\frac{1}{c}=\frac{a}{b\cdot c}$

$\frac{du}{dx}+\frac{-2u}{x}=\frac{x}{\frac{3}{2}}$
20

Simplifying

$\frac{du}{dx}+\frac{-2u}{x}=\frac{x}{\frac{3}{2}}$
21

We can identify that the differential equation has the form: $\frac{dy}{dx} + P(x)\cdot y(x) = Q(x)$, so we can classify it as a linear first order differential equation, where $P(x)=\frac{-2}{x}$ and $Q(x)=\frac{x}{\frac{3}{2}}$. In order to solve the differential equation, the first step is to find the integrating factor $\mu(x)$

$\displaystyle\mu\left(x\right)=e^{\int P(x)dx}$

Compute the integral

$\int\frac{-2}{x}dx$

The integral of the inverse of the lineal function is given by the following formula, $\displaystyle\int\frac{1}{x}dx=\ln(x)$

$-2\ln\left(x\right)$
22

To find $\mu(x)$, we first need to calculate $\int P(x)dx$

$\int P(x)dx=\int\frac{-2}{x}dx=-2\ln\left(x\right)$

Simplify $e^{-2\ln\left(x\right)}$ by applying the properties of exponents and logarithms

$x^{-2}$
23

So the integrating factor $\mu(x)$ is

$\mu(x)=x^{-2}$

Multiplying the fraction by $x^{-2}$

$x^{-2}\frac{du}{dx}+\frac{-2ux^{-2}}{x}=x^{-2}\frac{x}{\frac{3}{2}}$

Multiplying the fraction by $x^{-2}$

$x^{-2}\frac{du}{dx}+\frac{-2ux^{-2}}{x}=\frac{xx^{-2}}{\frac{3}{2}}$

When multiplying exponents with same base you can add the exponents: $xx^{-2}$

$x^{-2}\frac{du}{dx}+\frac{-2ux^{-2}}{x}=\frac{x^{-1}}{\frac{3}{2}}$

Simplify the fraction by $x$

$x^{-2}\frac{du}{dx}-2ux^{-3}=\frac{x^{-1}}{\frac{3}{2}}$
24

Now, multiply all the terms in the differential equation by the integrating factor $\mu(x)$ and check if we can simplify

$x^{-2}\frac{du}{dx}-2ux^{-3}=\frac{x^{-1}}{\frac{3}{2}}$
25

We can recognize that the left side of the differential equation consists of the derivative of the product of $\mu(x)\cdot y(x)$

$\frac{d}{dx}\left(x^{-2}u\right)=\frac{x^{-1}}{\frac{3}{2}}$
26

Integrate both sides of the differential equation with respect to $dx$

$\int\frac{d}{dx}\left(x^{-2}u\right)dx=\int\frac{x^{-1}}{\frac{3}{2}}dx$
27

Simplify the left side of the differential equation

$x^{-2}u=\int\frac{x^{-1}}{\frac{3}{2}}dx$

Take the constant $\frac{1}{\frac{3}{2}}$ out of the integral

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

The integral of the inverse of the lineal function is given by the following formula, $\displaystyle\int\frac{1}{x}dx=\ln(x)$

$\frac{2}{3}\ln\left(x\right)$

Using the power rule of logarithms: $n\log_b(a)=\log_b(a^n)$

$\ln\left(\sqrt[3]{x^{2}}\right)$

Using the power rule of logarithms: $\log_a(x^n)=n\cdot\log_a(x)$

$\frac{2}{3}\ln\left(x\right)$
28

Solve the integral $\int\frac{x^{-1}}{\frac{3}{2}}dx$ and replace the result in the differential equation

$x^{-2}u=\ln\left(\sqrt[3]{x^{2}}\right)$
29

As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$

$x^{-2}u=\ln\left(\sqrt[3]{x^{2}}\right)+C_0$
30

Replace $u$ with the value $y^{2}$

$x^{-2}y^{2}=\ln\left(\sqrt[3]{x^{2}}\right)+C_0$
31

Multiply the equation by the reciprocal of $x^{-2}$

$y^{2}=x^{2}\left(\ln\left(\sqrt[3]{x^{2}}\right)+C_0\right)$
32

Removing the variable's exponent

$y=\pm \sqrt{x^{2}\left(\ln\left(\sqrt[3]{x^{2}}\right)+C_0\right)}$
33

As in the equation we have the sign $\pm$, this produces two identical equations that differ in the sign of the term $\sqrt{x^{2}\left(\ln\left(\sqrt[3]{x^{2}}\right)+C_0\right)}$. We write and solve both equations, one taking the positive sign, and the other taking the negative sign

$y=\sqrt{x^{2}\left(\ln\left(\sqrt[3]{x^{2}}\right)+C_0\right)},\:y=-\sqrt{x^{2}\left(\ln\left(\sqrt[3]{x^{2}}\right)+C_0\right)}$

$y=\sqrt{x^{2}\left(\ln\left(\sqrt[3]{x^{2}}\right)+C_0\right)},\:y=-\sqrt{x^{2}\left(\ln\left(\sqrt[3]{x^{2}}\right)+C_0\right)}$
$\frac{dy}{dx}\:-\frac{y}{x}=\frac{x}{3y}$