Use integration to find the general solution to the differential equation .

Question

Use integration to find the general solution to the differential equation  . ​

Quizplus · Accepted Answer

Taking the antiderivative of both sides gives:
$$\int \frac{\mathrm{d}y}{\sqrt{1-y^2}} = \int \frac{x^2}{\sqrt{1-x^6}}\,\mathrm{d}x$$
The left-hand side can be evaluated by the substitution $y=\sin u$, $\mathrm{d}y = \cos u\,\mathrm{d}u$:
$$\int \frac{\mathrm{d}y}{\sqrt{1-y^2}} = \int \frac{\cos u}{\sqrt{1-\sin^2 u}}\,\mathrm{d}u = \int \mathrm{d}u = u + C_1$$
where $C_1$ is the constant of integration. The right-hand side can be evaluated by the substitution $x^6 = \frac{1}{t^2}$, $6x^5\,\mathrm{d}x = -\frac{2}{t^3}\,\mathrm{d}t$:
\begin{align*}
\int \frac{x^2}{\sqrt{1-x^6}}\,\mathrm{d}x &= -\frac{1}{3}\int \frac{\mathrm{d}t}{t\sqrt{t^4-1}} \
&= -\frac{\mathrm{arsinh}\left(t^2ight)}{3} + C_2 \
&= -\frac{\mathrm{arsinh}\left(\frac{1}{x^3}ight)}{3} + C_2
\end{align*}
where $\mathrm{arsinh}$ denotes the inverse hyperbolic sine function, and $C_2$ is another constant of integration. Therefore, the general solution is given by:
$$\mathrm{arsinh}\left(\frac{1}{x^3}ight) = -3u + C$$
where $C=C_1+C_2$ is the combined constant of integration.

Use Integration to Find the General Solution to the Differential