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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

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

Problem to solve:

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

Specify the solving method

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 $\frac{1}{2}$

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

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

When multiplying exponents with same base we can add the exponents

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

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

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

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

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

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

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

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

Any expression multiplied by $1$ is equal to itself

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

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

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

Simplifying

$\frac{du}{dx}+\frac{-u}{\frac{1}{2}x}=\frac{4}{1} \frac{1}{6}x$
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 $Q(x)=\frac{4}{1} \frac{1}{6}x$. 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 $x^{-2}$

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

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

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

When multiplying exponents with same base you can add the exponents: $\frac{1}{6}xx^{-2}$

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

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

$x^{-2}\frac{du}{dx}+\frac{2}{1} -ux^{-3}=\frac{4}{1} \frac{1}{6}x^{-1}$
16

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

$x^{-2}\frac{du}{dx}+\frac{2}{1} -ux^{-3}=\frac{4}{1} \frac{1}{6}x^{-1}$
17

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

$\frac{d}{dx}\left(x^{-2}u\right)=\frac{4}{1} \frac{1}{6}x^{-1}$
18

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

$\int\frac{d}{dx}\left(x^{-2}u\right)dx=\int\frac{4}{1} \frac{1}{6}x^{-1}dx$
19

Simplify the left side of the differential equation

$x^{-2}u=\int\frac{4}{1} \frac{1}{6}x^{-1}dx$
20

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

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

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, $\displaystyle\int\frac{1}{x}dx=\ln(x)$

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

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: $\log_a(x^n)=n\cdot\log_a(x)$

$\frac{2}{3}\ln\left(x\right)$
21

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

$x^{-2}u=\frac{2}{3}\ln\left(x\right)$
22

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

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

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

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

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

$y^{2}\frac{1}{x^{2}}=\frac{2}{3}\ln\left(x\right)+C_0$
25

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

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

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 $x$

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

Solve the equation ($2$)

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

Multiply both sides of the equation by $x$

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

The $2$ solutions of the equation are

$y=x\sqrt{\frac{2}{3}\ln\left(x\right)+C_0},\:y=-x\sqrt{\frac{2}{3}\ln\left(x\right)+C_0}$
26

Find the explicit solution to the differential equation

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

Final Answer

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

Tips on how to improve your answer:

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

Related Formulas:

2. See formulas

Time to solve it:

~ 0.27 s