Respuesta :
Answer:
[tex]ln [1 - (\frac{y}{x} )^{2} ] + ln x + c = 0[/tex]. This is the solution.
Step-by-step explanation:
The homogeneous differential equation is given by
[tex]\frac{dy}{dx} = \frac{x^{2} + y^{2} }{2xy}[/tex]
⇒ [tex]\frac{dy}{dx} = \frac{1 + (\frac{y}{x} )^{2} }{2(\frac{y}{x} )}[/tex] ........ (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 [tex]\frac{dy}{dx} = v + x \frac{dv}{dx}[/tex]
Therefore, the above equation (1) becomes
[tex]v + x \frac{dv}{dx} = \frac{1 + v^{2} }{2v}[/tex] {Since [tex]v = \frac{y}{x}[/tex]}
⇒ [tex]x\frac{dv}{dx} = \frac{1 + v^{2} - 2v^{2} }{2v}[/tex]
⇒ [tex]x\frac{dv}{dx} = \frac{1 - v^{2}}{2v}[/tex]
⇒ [tex]\frac{2v}{1 - v^{2} } dv = \frac{dx}{x}[/tex] {By separation of variables}
Now, integrating both sides we get,
[tex]\int {\frac{2v}{ 1- v^{2}}} \, dv = \int {\frac{dx}{x} } \, dx[/tex]
⇒ [tex]- \int {\frac{d(1 - v^{2} )}{1 - v^{2}}} = \int {\frac{dx}{x} }[/tex]
⇒ [tex]- ln (1 - v^{2}) = ln x + c[/tex] {Where c is the integration constant}
⇒ [tex]ln [1 - (\frac{y}{x} )^{2} ] + ln x + c = 0[/tex] (Answer)
