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Solve the differential equation $\frac{dy}{dx}+\frac{-y}{x}=\frac{x}{3y}$

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Solution

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Final answer to the problem

$y=\sqrt{\frac{2}{3}\ln\left(x\right)+c_0}x,\:y=-\sqrt{\frac{2}{3}\ln\left(x\right)+c_0}x$
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Step-by-step Solution

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1

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

$u=y^{\left(1-n\right)}$
2

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

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

Simplify

$u=y^{2}$

Rearrange the equation

$y^{2}=u$

Raise both sides of the equation to the exponent $\frac{1}{2}$

$y=\sqrt{u}$
4

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 $f'(x) = nx^{n-1}$

$\frac{dy}{dx}=\frac{1}{2}u^{-\frac{1}{2}}$
5

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

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

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

$\frac{1}{2}u^{-\frac{1}{2}}\frac{du}{dx}+\frac{-\sqrt{u}}{x}=\frac{x}{3\sqrt{u}}$
7

Simplify

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

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 $\frac{1}{2}\sqrt{u}$

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

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

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

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

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

Applying the property of exponents, $\displaystyle a^{-n}=\frac{1}{a^n}$, where $n$ is a number

$\frac{1}{4}\sqrt{u}\frac{1}{\sqrt{u}}\frac{du}{dx}+\frac{1}{2}\sqrt{u}\frac{-\sqrt{u}}{x}=\frac{1}{2}\frac{x}{3\sqrt{u}}\sqrt{u}$

Multiplying fractions $\frac{1}{4} \times \frac{1}{\sqrt{u}}$

$\frac{1}{4\sqrt{u}}\sqrt{u}\frac{du}{dx}+\frac{1}{2}\sqrt{u}\frac{-\sqrt{u}}{x}=\frac{1}{2}\frac{x}{3\sqrt{u}}\sqrt{u}$

Multiplying fractions $\frac{1}{2} \times \frac{-\sqrt{u}}{x}$

$\frac{1}{4\sqrt{u}}\sqrt{u}\frac{du}{dx}+\frac{-\sqrt{u}}{2x}\sqrt{u}=\frac{1}{2}\frac{x}{3\sqrt{u}}\sqrt{u}$

Multiplying fractions $\frac{x}{3\sqrt{u}} \times \frac{1}{2}$

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

Multiplying the fraction by $\sqrt{u}$

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

Multiply the fraction by the term

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

Multiplying the fraction by $\sqrt{u}$

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

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

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

Combine fractions with common denominator $2$

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

Add the values $1$ and $-1$

$\frac{1}{4}\frac{du}{dx}+\frac{-u}{2x}=\frac{xu^{\frac{0}{2}}}{6}$

Divide $0$ by $2$

$\frac{1}{4}\frac{du}{dx}+\frac{-u}{2x}=\frac{xu^{0}}{6}$

Any expression (except $0$ and $\infty$) to the power of $0$ is equal to $1$

$\frac{1}{4}\frac{du}{dx}+\frac{-u}{2x}=\frac{x}{6}$
10

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

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

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

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

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

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

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

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

Divide fractions $\frac{x}{\frac{3}{2}}$ with Keep, Change, Flip: $a\div \frac{b}{c}=\frac{a}{1}\div\frac{b}{c}=\frac{a}{1}\times\frac{c}{b}=\frac{a\cdot c}{b}$

$\frac{du}{dx}+\frac{-u}{\frac{1}{2}x}=\frac{2x}{3}$
11

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

$\frac{du}{dx}+\frac{-u}{\frac{1}{2}x}=\frac{2x}{3}$
12

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 $P(x)=\frac{-1}{\frac{1}{2}x}$ and $Q(x)=\frac{2x}{3}$. 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

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

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

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

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

$-2\ln\left|x\right|$
13

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

$\int P(x)dx=\int\frac{-1}{\frac{1}{2}x}dx=-2\ln\left(x\right)$

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

$x^{-2}$
14

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

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

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

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

Multiplying the fraction by $4$

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

Multiplying the fraction by $4$

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

Multiply $-1$ times $4$

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

Take $\frac{-4}{2}$ out of the fraction

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

Take $\frac{4}{6}$ out of the fraction

$16\left(\frac{du}{dx}\right)+\frac{-2u}{x}=\frac{2}{3}x$

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

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

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

$2x^{-2+1}$

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

$2x^{-2+1}$

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

$2x^{-2+1}$

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

$2x^{-2+1}$

Add the values $-2$ and $1$

$2x^{-1}$

Add the values $-2$ and $1$

$2x^{-1}$

Add the values $-2$ and $1$

$2x^{-1}$

Add the values $-2$ and $1$

$2x^{-1}$

Add the values $-2$ and $1$

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

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

$\frac{- 2u}{x}x^{-2}$

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

$\frac{x^{-2}-2u}{x}$

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

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

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

$\frac{x^{-2}- 2u}{x}$

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

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

Multiplying the fraction by $4$

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

Multiplying the fraction by $4$

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

Multiply $-1$ times $4$

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

Take $\frac{-4}{2}$ out of the fraction

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

Take $\frac{4}{6}$ out of the fraction

$16\left(\frac{du}{dx}\right)+\frac{-2u}{x}=\frac{2}{3}x$

Multiply $-1$ times $2$

$\frac{x^{-2}-2u}{x}$

Simplify the fraction $\frac{-2ux^{-2}}{x}$ by $x$

$\frac{du}{dx}x^{-2}-2ux^{-3}=\frac{2x^{-1}}{3}$
15

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

$\frac{du}{dx}x^{-2}-2ux^{-3}=\frac{2x^{-1}}{3}$
16

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

$\frac{d}{dx}\left(x^{-2}u\right)=\frac{2x^{-1}}{3}$
17

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

$\int\frac{d}{dx}\left(x^{-2}u\right)dx=\int\frac{2x^{-1}}{3}dx$
18

Simplify the left side of the differential equation

$x^{-2}u=\int\frac{2x^{-1}}{3}dx$
19

Applying the property of exponents, $\displaystyle a^{-n}=\frac{1}{a^n}$, where $n$ is a number

$x^{-2}u=\int\frac{2}{3x^{1}}dx$
20

Any expression to the power of $1$ is equal to that same expression

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

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

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

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

$2\left(\frac{1}{3}\right)\ln\left|x\right|$

Multiply the fraction and term in $2\left(\frac{1}{3}\right)\ln\left|x\right|$

$\frac{2}{3}\ln\left|x\right|$

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

$\frac{2}{3}\ln\left|x\right|+C_0$
21

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

$x^{-2}u=\frac{2}{3}\ln\left|x\right|+C_0$
22

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

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

Applying the property of exponents, $\displaystyle a^{-n}=\frac{1}{a^n}$, where $n$ is a number

$\frac{1}{x^{\left|-2\right|}}y^{2}$

Multiplying the fraction by $y^{2}$

$\frac{y^{2}}{x^{\left|-2\right|}}$
23

Applying the property of exponents, $\displaystyle a^{-n}=\frac{1}{a^n}$, where $n$ is a number

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

Multiply the fraction by the term

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

Any expression multiplied by $1$ is equal to itself

$\frac{y^{2}}{x^{2}}=\frac{2}{3}\ln\left|x\right|+C_0$
24

Multiply the fraction by the term

$\frac{y^{2}}{x^{2}}=\frac{2}{3}\ln\left|x\right|+C_0$

Apply the property of the quotient of two powers with the same exponent, inversely: $\frac{a^m}{b^m}=\left(\frac{a}{b}\right)^m$, where $m$ equals $2$

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

Removing the variable's exponent

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

Cancel exponents $2$ and $1$

$\frac{y}{x}=\pm \sqrt{\frac{2}{3}\ln\left(x\right)+C_0}$

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

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

Solve the equation ($1$)

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

Multiply both sides of the equation by $x$

$y=\sqrt{\frac{2}{3}\ln\left(x\right)+c_0}x$

Solve the equation ($2$)

$\frac{y}{x}=-\sqrt{\frac{2}{3}\ln\left(x\right)+C_0}$

Multiply both sides of the equation by $x$

$y=-\sqrt{\frac{2}{3}\ln\left(x\right)+c_0}x$

Combining all solutions, the $2$ solutions of the equation are

$y=\sqrt{\frac{2}{3}\ln\left(x\right)+c_0}x,\:y=-\sqrt{\frac{2}{3}\ln\left(x\right)+c_0}x$
25

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

$y=\sqrt{\frac{2}{3}\ln\left(x\right)+c_0}x,\:y=-\sqrt{\frac{2}{3}\ln\left(x\right)+c_0}x$

Final answer to the problem

$y=\sqrt{\frac{2}{3}\ln\left(x\right)+c_0}x,\:y=-\sqrt{\frac{2}{3}\ln\left(x\right)+c_0}x$

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Function Plot

Plotting: $\frac{dy}{dx}+\frac{-y}{x}+\frac{-x}{3y}$

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Main Topic: Factorization

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