# -b.2.2.32 First order homogeneous ODE

• MHB
• karush
In summary: Since $u= y/x$, we get $y^2+ x^2= C_2x^3$.In summary, the conversation discusses rewriting a given ODE and using substitution to solve for the derivatives. The final solution is $y^2+ x^2= C_2x^3$.
karush
Gold Member
MHB
$\dfrac{dy}{dx} =\dfrac{x^2+3y^2}{2xy} =\dfrac{x^2}{2xy}+\dfrac{3y^2}{2xy} =\dfrac{x}{2y}+\dfrac{3y}{2x}$

ok not sure if this is the best first steip,,,, if so then do a $u=\dfrac{x}{y}$ ?

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I would write the ODE as:

$$\displaystyle \frac{dy}{dx}=\frac{1+3\left(\dfrac{y}{x}\right)^2}{2\dfrac{y}{x}}$$

And then use:

$$\displaystyle u=\frac{y}{x}\implies\frac{dy}{dx}=x\frac{du}{dx}+u$$

 $\displaystyle u=\frac{y}{x}\implies\frac{dy}{dx}=x\frac{du}{dx}+u$ so is that $$xu'+x'u=(xu)'$$

 $x\dfrac{du}{dx}=\dfrac{1+3u^2}{2u}$ corrections!

MarkFL said:
I would write the ODE as:

$$\displaystyle \frac{dy}{dx}=\frac{1+3\left(\dfrac{y}{x}\right)^2}{2\dfrac{y}{x}}$$

And then use:

$$\displaystyle u=\frac{y}{x}\implies\frac{dy}{dx}=x\frac{du}{dx}+u$$

We would then have:

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

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

$$\displaystyle x\frac{du}{dx}=\frac{1+u^2}{2u}$$

$x\dfrac{du}{dx}=\dfrac{1+u^2}{2u}$
$x\dfrac{1}{dx}=\dfrac{1+u^2}{2u}\dfrac{1}{du}$
tried to separate varibles but didn't see how it reach the answer of

Last edited:
$\dfrac{2u}{1+u^2} \, du = \dfrac{dx}{x}$

continue ...

I would next write:

$$\displaystyle \frac{2u}{u^2+1}\,du=\frac{1}{x}\,dx$$

Integrating, we get what?

$\ln |u^2+1|=\ln|x|$

$u=\dfrac{x}{y}$

You forgot the constant of integration:

$$\displaystyle \ln|u^2+1|=\ln|cx|$$

And:

$$\displaystyle u=\frac{y}{x}$$

$\displaystyle \ln|u^2+1|=\ln|cx|$
why did you put the c inside the ln?

so anyway
$\displaystyle \ln\left|\dfrac{u^2+1}{cx}\right|=0$

$\ln(u^2+1) = \ln|x| + C_1$

rewite $C_1=\ln|C_2|$ ...

$\ln\left(\dfrac{y^2}{x^2}+1\right)=\ln|C_2 x|$

note, $\ln{a}=\ln{b} \implies a=b$

$\dfrac{y^2}{x^2} + 1 = Cx$

$y^2+x^2=Cx^3$

wow... that was a lot of help..

skeeter said:
$\ln(u^2+1) = \ln|x| + C_1$

rewite $C_1=\ln|C_2|$ ...

$\ln\left(\dfrac{y^2}{x^2}+1\right)=\ln|C_2 x|$

note, $\ln{a}=\ln{b} \implies a=b$

$\dfrac{y^2}{x^2} + 1 = Cx$

$y^2+x^2=Cx^3$

Equivalently
$\ln(u^2+1) = \ln|x| + C_1$
Taking the exponetial of both sides
$e^{ln(u^2+ 1)}= u^2+ 1= e^{ln|x|+ C_1}= e^{ln|x|}e^{C_1}= e^{C_1}|x|$
$u^2+ 1= C_2|x|$ where $C_2= e^{C_1}$.

Now, technically e to any power is positive but we can get the general solution by allowing $C_2$ to be positive or negative and then we no longer need the absolute value: $u^2+ 1= C_2x$.

## 1. What is a first order homogeneous ODE?

A first order homogeneous ODE (ordinary differential equation) is an equation that involves an unknown function and its derivatives, where the function and its derivatives have the same degree of dependence. This means that the equation can be written in a form where all terms have the same power of the dependent variable.

## 2. What does it mean for an ODE to be homogeneous?

A homogeneous ODE is one where all terms in the equation have the same degree of dependence on the dependent variable. This means that the equation can be written in a form where all terms have the same power of the dependent variable.

## 3. How do you solve a first order homogeneous ODE?

The general method for solving a first order homogeneous ODE is to use the substitution method, where you substitute the dependent variable with a new variable and then solve for that variable. This will result in a separable equation, which can then be solved using integration.

## 4. What are some real-world applications of first order homogeneous ODEs?

First order homogeneous ODEs are commonly used in physics and engineering to model various natural phenomena, such as population growth, radioactive decay, and chemical reactions. They are also used in economics and finance to model interest rates and stock prices.

## 5. Can a first order homogeneous ODE have more than one solution?

Yes, a first order homogeneous ODE can have an infinite number of solutions. This is because the equation is not fully specified and requires an initial condition to determine a unique solution. Without an initial condition, the equation will have a general solution that includes all possible solutions.

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