Final Answer
Step-by-step Solution
Problem to solve:
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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
Raise both sides of the equation to the exponent $\frac{1}{2}$
Divide $1$ by $2$
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 $\frac{1}{2}u^{-\frac{1}{2}}\frac{du}{dx}+\frac{-\sqrt{u}}{x}$
Multiplying the fraction by $\frac{1}{2}\sqrt{u}$
When multiplying exponents with same base we can add the exponents
Any expression (except $0$ and $\infty$) to the power of $0$ is equal to $1$
Rewrite the fraction $\frac{-\frac{1}{2}u}{x}$
Multiplying the fraction 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 $\frac{1}{4}$
Divide fractions $\frac{\frac{-u}{2x}}{\frac{1}{4}}$ with Keep, Change, Flip: $\frac{a}{b}\div c=\frac{a}{b}\div\frac{c}{1}=\frac{a}{b}\times\frac{1}{c}=\frac{a}{b\cdot c}$
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 $P(x)=\frac{-1}{\frac{1}{2}x}$ 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 $\mu(x)$
Compute the integral
Take the constant $\frac{1}{\frac{1}{2}}$ out of 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
Multiplying the fraction by $x^{-2}$
Multiplying the fraction by $x^{-2}$
When multiplying exponents with same base you can add the exponents: $\frac{2}{3}x\cdot x^{-2}$
Simplify the fraction $\frac{-ux^{-2}}{\frac{1}{2}x}$ 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 function times a constant ($\frac{2}{3}$) is equal to the constant times the integral of the function
The integral of the inverse of the lineal function is given by the following formula, $\displaystyle\int\frac{1}{x}dx=\ln(x)$
As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$
Solve the integral $\int\frac{2}{3}x^{-1}dx$ and replace the result in the differential equation
Replace $u$ with the value $y^{2}$
Applying the property of exponents, $\displaystyle a^{-n}=\frac{1}{a^n}$, where $n$ is a number
Multiply the fraction and term
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 $2$
Removing the variable's exponent
Cancel exponents $2$ and $\frac{1}{2}$
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
Solve the equation ($1$)
Multiply both sides of the equation by $x$
Solve the equation ($2$)
Multiply both sides of the equation by $x$
Combining all solutions, the $2$ solutions of the equation are
Find the explicit solution to the differential equation. We need to isolate the variable $y$