Final Answer
Step-by-step Solution
Problem to solve:
Multiplying the fraction by $-1$
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
Plug in the value of $n$, which equals $-1$
Simplify
Rearrange the equation
Removing the variable's exponent
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
Let's just take the first equation
Isolate the dependent variable $y$
The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$
Differentiate both sides of the equation with respect to the independent variable $x$
Now, substitute $\frac{dy}{dx}=\frac{1}{2}u^{-\frac{1}{2}}\frac{du}{dx}$ and $y=\sqrt{u}$ on the original differential equation
Simplify
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 $\frac{1}{2}\sqrt{u}$
Multiply both sides by $\frac{1}{2}\sqrt{u}$
Multiplying polynomials $\frac{1}{2}\sqrt{u}$ and $\frac{1}{2}u^{-\frac{1}{2}}\frac{du}{dx}+\frac{-\sqrt{u}}{x}$
When multiplying exponents with same base we can add the exponents
Simplify the fraction $\frac{\frac{1}{6}x\sqrt{u}}{\sqrt{u}}$ by $u$
Expand and simplify. Now we see that the differential equation looks like a linear differential equation, because we removed the original $y^{-1}$ term
Divide all the terms of the differential equation by $\frac{1}{4}$
Divide $\frac{1}{4}$ by $\frac{1}{4}$
Any expression multiplied by $1$ is equal to itself
Divide fractions $\frac{\frac{-\frac{1}{2}u}{x}}{\frac{1}{4}}$ with Keep, Change, Flip: $\frac{a}{b}\div c=\frac{a}{b}\div\frac{c}{1}=\frac{a}{b}\times\frac{1}{c}=\frac{a}{b\cdot c}$
Simplifying
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 $Q(x)=\frac{2}{3}x$. In order to solve the differential equation, the first step is to find the integrating factor $\mu(x)$
Compute the integral
The integral of the inverse of the lineal function is given by the following formula, $\displaystyle\int\frac{1}{x}dx=\ln(x)$
To find $\mu(x)$, we first need to calculate $\int P(x)dx$
Simplify $e^{-2\ln\left(x\right)}$ by applying the properties of exponents and logarithms
So the integrating factor $\mu(x)$ is
When multiplying exponents with same base you can add the exponents: $\frac{2}{3}xx^{-2}$
Multiplying the fraction by $x^{-2}$
Simplify the fraction by $x$
Now, multiply all the terms in the differential equation by the integrating factor $\mu(x)$ and check if we can simplify
We can recognize that the left side of the differential equation consists of the derivative of the product of $\mu(x)\cdot y(x)$
Integrate both sides of the differential equation with respect to $dx$
Simplify the left side of the differential equation
The integral of a constant by a function is equal to the constant multiplied by the integral of the function
The integral of the inverse of the lineal function is given by the following formula, $\displaystyle\int\frac{1}{x}dx=\ln(x)$
Using the power rule of logarithms: $n\log_b(a)=\log_b(a^n)$
Using the power rule of logarithms: $\log_a(x^n)=n\cdot\log_a(x)$
Solve the integral $\int\frac{2}{3}x^{-1}dx$ and replace the result in the differential equation
As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$
Replace $u$ with the value $y^{2}$
Using the power rule of logarithms: $n\log_b(a)=\log_b(a^n)$
Using the power rule of logarithms: $\log_a(x^n)=n\cdot\log_a(x)$
Divide both sides of the equation by $x^{-2}$
Removing the variable's exponent
The power of a quotient is equal to the quotient of the power of the numerator and denominator: $\displaystyle\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n}$
As in the equation we have the sign $\pm$, this produces two identical equations that differ in the sign of the term $\frac{\sqrt{\frac{2}{3}\ln\left(x\right)+C_0}}{x^{-1}}$. We write and solve both equations, one taking the positive sign, and the other taking the negative sign
Applying the property of exponents, $\displaystyle a^{-n}=\frac{1}{a^n}$, where $n$ is a number
Divide fractions $\frac{\sqrt{\frac{2}{3}\ln\left(x\right)+C_0}}{\frac{1}{x}}$ with Keep, Change, Flip: $a\div \frac{b}{c}=\frac{a}{1}\div\frac{b}{c}=\frac{a}{1}\times\frac{c}{b}=\frac{a\cdot c}{b}$
Applying the property of exponents, $\displaystyle a^{-n}=\frac{1}{a^n}$, where $n$ is a number
Divide fractions $\frac{-\sqrt{\frac{2}{3}\ln\left(x\right)+C_0}}{\frac{1}{x}}$ with Keep, Change, Flip: $a\div \frac{b}{c}=\frac{a}{1}\div\frac{b}{c}=\frac{a}{1}\times\frac{c}{b}=\frac{a\cdot c}{b}$
Find the explicit solution to the differential equation