Solve the differential equation dy/dx+(-y)/x=x/(3y)

 Final answer to the problem

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

Got another answer? Verify it here!

 Step-by-step Solution  

 How should I solve this problem?

Choose an option
Exact Differential Equation
Linear Differential Equation
Separable Differential Equation
Homogeneous Differential Equation
Integrate by partial fractions
Product of Binomials with Common Term
FOIL Method
Integrate by substitution
Integrate by parts
Load more...

Can't find a method? Tell us so we can add it.

 

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

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

 

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

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

 

Simplify

$u=y^{2}$

Rearrange the equation

$y^{2}=u$

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

$y=\sqrt{u}$

 

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

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

 

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

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

 

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

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

 

Simplify

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

 

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

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

 

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 $1$

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

 

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: $a\div \frac{b}{c}=\frac{a}{1}\div\frac{b}{c}=\frac{a}{1}\times\frac{c}{b}=\frac{a\cdot c}{b}$

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

 

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

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

 

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

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

 

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

$x^{-2}$

 

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 $x$

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

 

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

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

 

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

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

 

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

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

 

Simplify the left side of the differential equation

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

 

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

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

 

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 $C$

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

 

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

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

 

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 $y^{2}$

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

 

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

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

 

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

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

 

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  

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

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

Step-by-step Solution

 Solution

 Videos

 Final answer to the problem

 Step-by-step Solution  

Final answer to the problem  

 Explore different ways to solve this problem

 Help us improve with your feedback!

 Function Plot

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

beta
Got a different answer? Verify it!

 How to improve your answer:

 Main Topic: Differential Equations

 Related Topics

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

Step-by-step Solution

 Solution

 Videos

 Final answer to the problem

 Step-by-step Solution  

 Final answer to the problem  

 Explore different ways to solve this problem

 Help us improve with your feedback!

 Share this solution with your friends!

 Function Plot

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

beta Got a different answer? Verify it!

 How to improve your answer:

Similar Problems Solved

 Main Topic: Differential Equations

 Related Topics

Supercharge your math learning

✨ Create your account today!

Supercharge your math learning

✨ New: AI whiteboard. Do your own calculations and find out if you made any errors. Try whiteboard.

 Your Personal Math Tutor. Powered by AI

Available 24/7, 365.

 Complete step-by-step math solutions. No ads.

 Choose between multiple solving methods.

 Download complete solutions and keep them forever.

 Unlimited practice with our AI whiteboard.

 Premium access on our iOS and Android apps.

 Join 500k+ students in problem solving.

Choose your plan. Cancel Anytime.

 Pay $39.97 USD securely with your payment method.

Please hold while your payment is being processed.

Final answer to the problem  

beta
Got a different answer? Verify it!