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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 to the problem

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

How should I solve this problem?

  • Homogeneous Differential Equation
  • Exact Differential Equation
  • Linear Differential Equation
  • Separable Differential Equation
  • Integrate by partial fractions
  • Product of Binomials with Common Term
  • FOIL Method
  • Integrate by substitution
  • Integrate by parts
  • Integrate using tabular integration
  • Load more...
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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

$u=y^{2}$
4

Isolate the dependent variable $y$

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

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

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

Simplifying

$\frac{du}{dx}+\frac{-u}{\frac{1}{2}x}=\frac{2}{3}x$
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 $Q(x)=\frac{2}{3}x$. In order to solve the differential equation, the first step is to find the integrating factor $\mu(x)$

$\displaystyle\mu\left(x\right)=e^{\int P(x)dx}$
14

To find $\mu(x)$, we first need to calculate $\int P(x)dx$

$\int P(x)dx=\int\frac{-1}{\frac{1}{2}x}dx=-2\ln\left|x\right|$
15

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

$\mu(x)=x^{-2}$
16

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

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

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

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

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

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

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Function Plot

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

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a
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d
f
g
m
n
u
v
w
x
y
z
.
(◻)
+
-
×
◻/◻
/
÷
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

How to improve your answer:

Main Topic: Differential Equations

A differential equation is a mathematical equation that relates some function with its derivatives.

Used Formulas

See formulas (2)

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