Question

I need help with this homogeneous equation

Answer 1

Answer:

$ln [1 - (\frac{y}{x} )^{2} ] + ln x + c = 0$. This is the solution.

Step-by-step explanation:

The homogeneous differential equation is given by  

$\frac{dy}{dx} = \frac{x^{2} + y^{2} }{2xy}$

⇒ $\frac{dy}{dx} = \frac{1 + (\frac{y}{x} )^{2} }{2(\frac{y}{x} )}$ ........ (1)

Now to solve this differential equation we assume that y = vx where v is another variable.

So, differentiating with respect to x we get $\frac{dy}{dx} = v + x \frac{dv}{dx}$

Therefore, the above equation (1) becomes

$v + x \frac{dv}{dx} = \frac{1 + v^{2} }{2v}$ {Since $v = \frac{y}{x}$}

⇒ $x\frac{dv}{dx} = \frac{1 + v^{2} - 2v^{2} }{2v}$

⇒ $x\frac{dv}{dx} = \frac{1 - v^{2}}{2v}$

⇒ $\frac{2v}{1 - v^{2} } dv = \frac{dx}{x}$ {By separation of variables}

Now, integrating both sides we get,

$\int {\frac{2v}{ 1- v^{2}}} \, dv = \int {\frac{dx}{x} } \, dx$

⇒ $- \int {\frac{d(1 - v^{2} )}{1 - v^{2}}} = \int {\frac{dx}{x} }$

⇒ $- ln (1 - v^{2}) = ln x + c$ {Where c is the integration constant}

⇒ $ln [1 - (\frac{y}{x} )^{2} ] + ln x + c = 0$ (Answer)