Topics

More AI tools from SnapXam

Exercise

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

Step-by-step Solution

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 $-1$

$u=y^{\left(1+1\right)}$
3

Simplify

$u=y^{2}$
4

Isolate the dependent variable $y$

$y=\sqrt{u}$
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

$\frac{1}{4}\frac{du}{dx}+\frac{-u}{2x}=\frac{x}{6}$
11

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

$\frac{du}{dx}+\frac{-u}{\frac{1}{2}x}=\frac{2x}{3}$
12

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 $\mu(x)$

$\displaystyle\mu\left(x\right)=e^{\int P(x)dx}$
13

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

$\int P(x)dx=\int\frac{-1}{\frac{1}{2}x}dx=-2\ln\left(x\right)$
14

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

$\mu(x)=x^{-2}$
15

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

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

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

$\frac{d}{dx}\left(x^{-2}u\right)=\frac{2x^{-1}}{3}$
17

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

$\int\frac{d}{dx}\left(x^{-2}u\right)dx=\int\frac{2x^{-1}}{3}dx$
18

Simplify the left side of the differential equation

$x^{-2}u=\int\frac{2x^{-1}}{3}dx$
19

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

$x^{-2}u=\int\frac{2}{3x^{1}}dx$
20

Any expression to the power of $1$ is equal to that same expression

$x^{-2}u=\int\frac{2}{3x}dx$
21

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

$x^{-2}u=\frac{2}{3}\ln\left|x\right|+C_0$
22

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

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

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

$\frac{1}{x^{2}}y^{2}=\frac{2}{3}\ln\left|x\right|+C_0$
24

Multiply the fraction by the term

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

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 exercise

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

Try other ways to solve this exercise

  • Choose an option
  • Exact Differential Equation
  • Linear Differential Equation
  • Separable Differential Equations
  • 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.

Other solved exercises

$\tan^2a-\sin^2a=\tan^2a\cdot\sin^2a$

$-2h\ge28$

$\int xe^{3x^2+7}dx$

$6^8+7+8$

$\sqrt{2^{3x-1}}\cdot4^{\frac{-3}{5}\left(x-2\right)}$

$\left(523\right)\cdot\left(-2\right)\cdot\left(-9\right)$

$4x^2+12x+28$
Symbolic mode
Text mode
Go!
1
2
3
4
5
6
7
8
9
0
a
b
c
d
f
g
m
n
u
v
w
x
y
z
.
(◻)
+
-
×
◻/◻
/
÷
◻2
◻◻
√◻
√

◻√◻
◻√
∞
e
π
ln
log
log◻
lim
d/dx
D□x
∫
∫◻
|◻|
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

Empowering students and teachers through AI

By signing up, you agree to SnapXam's Terms and Privacy Policy.

Already have an account? Login here
Forgot your password?

Empowering students and teachers through AI



Register here if you don't have an account.
Forgot your password?

Your Personal Math Tutor. Powered by AI

Available 24/7, 365 days a year.

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

Access in depth explanations with descriptive diagrams.

Choose between multiple solving methods.

Download unlimited solutions in PDF format.

Premium access on our iOS and Android apps.

Join 1M+ students worldwide in problem solving.

Choose the plan that suits you best:
Pay $39.97 USD securely with your payment method.
Please hold while your payment is being processed.
Create an Account