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

Step-by-step Solution

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Final Answer

$y=\sqrt{\frac{2}{3}\ln\left(x\right)+C_0}x,\:y=-\sqrt{\frac{2}{3}\ln\left(x\right)+C_0}x$
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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 $-1$

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

Simplify

$u=y^{2}$

Rearrange the equation

$y^{2}=u$

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

$y=u^{\frac{1}{2}}$

Divide $1$ by $2$

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

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

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

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

When multiplying exponents with same base we can add the exponents

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

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

$1\cdot \frac{1}{4}\left(\frac{du}{dx}\right)$

Any expression multiplied by $1$ is equal to itself

$\frac{1}{4}\left(\frac{du}{dx}\right)$

Rewrite the fraction $\frac{-\frac{1}{2}u}{x}$

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

Multiplying the fraction by $u$

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

$\frac{1}{4}\left(\frac{du}{dx}\right)+\frac{-u}{2x}=\frac{1}{6}x$
11

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

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

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

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

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

$\frac{du}{dx}+\frac{-u}{\frac{1}{2}x}=\frac{\frac{1}{6}x}{\frac{1}{4}}$
12

Simplifying

$\frac{du}{dx}+\frac{-u}{\frac{1}{2}x}=\frac{\frac{1}{6}x}{\frac{1}{4}}$
13

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 $Q(x)=\frac{\frac{1}{6}x}{\frac{1}{4}}$. 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

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

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

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

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

$-2\ln\left(x\right)$
14

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

$x^{-2}$
15

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

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

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

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

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

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

Take $\frac{\frac{1}{6}}{\frac{1}{4}}$ out of the fraction

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

When multiplying exponents with same base you can add the exponents: $\frac{2}{3}x\cdot x^{-2}$

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

Simplify the fraction $\frac{-ux^{-2}}{\frac{1}{2}x}$ by $x$

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

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

$\frac{du}{dx}x^{-2}+\frac{-ux^{-3}}{\frac{1}{2}}=\frac{2}{3}x^{-1}$
17

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

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

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

$\int\frac{d}{dx}\left(x^{-2}u\right)dx=\int\frac{2}{3}x^{-1}dx$
19

Simplify the left side of the differential equation

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

The integral of a function times a constant ($\frac{2}{3}$) is equal to the constant times the integral of the function

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

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

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

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

$\frac{2}{3}\ln\left(x\right)+C_0$
20

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

$x^{-2}u=\frac{2}{3}\ln\left(x\right)+C_0$
21

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 $y^{2}$

$\frac{y^{2}}{x^{\left|-2\right|}}$
22

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

$\frac{1}{x^{2}}y^{2}=\frac{2}{3}\ln\left(x\right)+C_0$
23

Multiply the fraction and 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 $\frac{1}{2}$

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

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

$y=\sqrt{\frac{2}{3}\ln\left(x\right)+C_0}x,\:y=-\sqrt{\frac{2}{3}\ln\left(x\right)+C_0}x$
24

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

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

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

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