How to solve the following differential equation.....?
\begin{equation} x^3\frac{dy}{dx}=y^3+y^2 (y^2-x^2)^{1/2} \end{equation}
It is a homogeneous equation. So after substituting $y=vx$ I get
\begin{equation} \frac{dv}{v^3 + v^2 (v^2 -1)^{1/2}-v}=\frac{dx}{x} \end{equation}
and what next?
On
To evalute the integral on the left side of $$\int \begin{equation} \frac{dv}{v^3 + v^2 (v^2 -1)^{1/2}-v}= \int\frac{dx}{x} \end{equation}$$ You may let $v=\sec t$ and your integral simplifies to $$ \int \frac {\sec t \tan t dt}{\sec ^3 t + \sec ^2 t \tan t -\sec t}$$ which in turn simplifies to $$ \int \frac {\cos t dt}{\sin t +1}$$
You can finish the problem easily.
Next is you have to integrate both sides. $$\begin{equation} \int\frac{dv}{v^3 + v^2 (v^2 -1)^{1/2}-v}=\int\frac{dx}{x} \end{equation}$$
substitue $v=\sec u$
$${\displaystyle\int}\dfrac{\tan\left(u\right)}{\sec\left(u\right)\sqrt{\sec^2\left(u\right)-1}+\sec^2\left(u\right)-1}\,\mathrm{d}u$$ $${\displaystyle\int}\dfrac{\tan\left(u\right)}{\sec\left(u\right)\tan\left(u\right)+\sec^2\left(u\right)-1}\,\mathrm{d}u$$ $${\displaystyle\int}\dfrac{\tan\left(u\right)}{\tan^2\left(u\right)+\sec\left(u\right)\tan\left(u\right)}\,\mathrm{d}u$$ $${\displaystyle\int}\dfrac{1}{\tan\left(u\right)+\sec\left(u\right)}\,\mathrm{d}u$$ $${\displaystyle\int}\class{steps-node}{\cssId{steps-node-1}{\cos\left(u\right)}}\cdot\class{steps-node}{\cssId{steps-node-2}{\dfrac{1}{\sin\left(u\right)+1}}}\,\mathrm{d}u$$
substitute $w=\sin u+1$ thus $dw=\cos u du $ $$\int \frac{1}{w} dw$$
can you complete the solution from here