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

[chwala]

Homework Statement

kindly see attached below:

Relevant Equations

$y=vx$

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.

 

[Delta2]

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).

 

[chwala]

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

 

[kuruman]

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.