Solve the homogeneous ODE: dy/dx = (x^2 + y^2)/xy

In summary, the conversation discusses the placement of the constant in a formula involving natural logarithms and integrals. It is concluded that the placement of the constant does not affect the final result, but the exact value of the constant may differ depending on the initial conditions. One person prefers to incorporate the initial conditions explicitly through definite integrals.
  • #1
chwala
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Homework Statement
kindly see attached below:
Relevant Equations
##y=vx##
1625014099464.png

1625014177894.png


this is pretty easy for me to solve, no doubt on that. My question is on the constant. Alternatively, is it correct to have,
##ln x= \frac {v^2}{2}##+ C, then work it from there...
secondly, we are 'making" ##c= ln k##, is it for convenience purposes?, supposing i left the constant as it is, would that be wrong?
##ln x= \frac {v^2}{2}##+ C
##ln x-c= \frac {v^2}{2}##
##2[ln x-c]=v^2##
##2x^2[ln x-c]=y^2## would this be correct?...the question that i am trying to ask is, does it matter where one places the constant after integration.
 
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  • #2
No it doesn't matter where you put the constant neither if you set it equal to ##\ln k## or ##\ln \frac{1}{k}##, because given an initial condition you will get the same formula for v in both cases, but the exact value of the constants will be different (for example if you find k=2 for the one case, you ll find k=1/2 for the other case).
 
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  • #3
Delta2 said:
No it doesn't matter where you put the constant neither if you set it equal to ##\ln k## or ##\ln \frac{1}{k}##, because given an initial condition you will get the same formula for v in both cases, but the exact value of the constants will be different (for example if you find k=2 for the one case, you ll find k=1/2 for the other case).
nice delta :cool:
 
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  • #4
My personal preference is to insert the initial conditions explicitly by doing definite integrals
$$\int_{v_0}^v v~dv=\int_{x_0}^x \frac{1}{x}~dx$$ $$\frac{1}{2}\left(v^2-v_0^2\right)=\ln\left(\frac{x}{x_0}\right) \implies v=\pm \sqrt{ 2\left[v_0^2+ \ln\left(\frac{x}{x_0}\right)\right]}.$$If the equation is related to a physics problem, this method takes care of the dimensions automatically.
 
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FAQ: Solve the homogeneous ODE: dy/dx = (x^2 + y^2)/xy

1. What is a homogeneous ODE?

A homogeneous ODE (ordinary differential equation) is an equation that can be written in the form dy/dx = f(x,y), where f(x,y) is a function of x and y only. In other words, the dependent variable and its derivative appear in the equation in the same form.

2. How do you solve a homogeneous ODE?

To solve a homogeneous ODE, you can use the substitution y = vx. This will transform the equation into a separable ODE, which can be solved by integrating both sides and then solving for y.

3. What is the general solution to dy/dx = (x^2 + y^2)/xy?

The general solution to this homogeneous ODE is y = Cx, where C is a constant.

4. Can a homogeneous ODE have non-constant solutions?

Yes, a homogeneous ODE can have non-constant solutions. In fact, the general solution to a homogeneous ODE will always include a constant term.

5. Are there any special cases when solving a homogeneous ODE?

Yes, there are a few special cases when solving a homogeneous ODE. One example is when the equation is separable, meaning that it can be written as dy/dx = g(x)/h(y). Another special case is when the equation can be solved using the substitution y = vx, as mentioned in question 2.

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