The solution of the differential equation $x^3\frac{dy}{dx}=y^3+y^2\sqrt{y^2-x^2}$ is:
Answer & explanation
Correct answer: option 1
$\frac{dy}{dx}=\frac{y^3}{x^3}+\frac{y^2}{x^2}\sqrt{\frac{y^2}{x^2}-1}$
Let $y=tx⇒\frac{dy}{dx}=t+x\frac{dt}{dx};t+x\frac{dt}{dx}=t^3+t^2\sqrt{t^2-1}$
$\frac{dt}{t\sqrt{t^2-1}[t+\sqrt{t^2-1}]}=\frac{dx}{x};\int\frac{[t-\sqrt{t^2-1}]}{t\sqrt{t^2-1}}dt=\int\frac{dx}{x};\int\frac{dt}{\sqrt{t^2-1}}-\int\frac{dt}{t}=\int\frac{dx}{x};ln|\frac{t+\sqrt{t^2-1}}{t}|=ln\,cx$
Substitute $t = \frac{y}{x}$ to get: $y+\sqrt{y^2-x^2}=cxy$