Question

Solve
6y^2dx-x(2x^3+y)dy=0

Answer 1

Given DE is

$$6y^2dx-x(2x^3+y)dy=0$$

After rearranging the above equation ,we can write

$$\frac{dx}{dy}-\frac{x}{6y}=\frac{x^4}{3y^2}\\ \implies -\frac{3\frac{dx}{dy}}{x^4}+\frac{1}{2yx^3}=-\frac{1}{y^2}$$

Put

$$v=\frac{1}{x^3}\implies \frac{dv}{dy}= -\frac{3\frac{dx}{dy}}{x^4}$$

Thus, we get

$$\frac{dv}{dy}+\frac{1}{2y}v=-\frac{1}{y^2}$$

Hence, Integrating Factor is

$$I.F=e^{\int 1/2ydy}=\sqrt{y}$$

So it follows

$$\frac{d}{dy}(\sqrt{y}v)=-y^{-3/2}\\ \implies \sqrt{y}v=-\int y^{-3/2}dy=2y^{-1/2}+C_1\\ \implies v=C_1y^{-1/2}+2y^{-1}\\ \implies \frac{1}{x^3}=C_1y^{-1/2}+2y^{-1}\\$$

After simplification we get,

$$y{\left(x \right)} = \frac{3 C_{1} \sqrt{x^{9} \left(9 C_{1}^{2} x^{3} + 8\right)}}{2} + \frac{x^{3} \left(9 C_{1}^{2} x^{3} + 4\right)}{2}\\ y{\left(x \right)} = - \frac{3 C_{1} \sqrt{x^{9} \left(9 C_{1}^{2} x^{3} + 8\right)}}{2} + \frac{x^{3} \left(9 C_{1}^{2} x^{3} + 4\right)}{2}$$