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Given the differential equation $$\ddot{y} + y + y^3 = 0$$

where $y = y(t)$, $\dot{y} = \frac{dy}{dt}$. By multiplying this equation by $\dot{y}$, assuming $\dot{y}>0$ and integrating, find $\frac{dy}{dt}$.

$$I = \int{\dot{y}\ddot{y} + \dot{y}y+\dot{y}y^3}\ \rm dt = 0$$

I don't really know how to integrate this, I understand we can do the following $$I = \int \left(\ddot{y}+y+y^3\right)\dot{y}\ \text{dt} = \int\ddot{y}+y+y^3 \, \, \text{dy} = 0$$ but since $y=y(t)$ and $\dot{y} = \dot{y}(t)$, I don't think this is integrable in this form.

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  • $\begingroup$ Just use $\displaystyle \int u'u^{\alpha}=\frac {u^{\alpha +1}}{\alpha +1}$, for appropriate $u$'s and $\alpha$ on each term of the sum. $\endgroup$ – Git Gud Oct 26 '14 at 17:42
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Hint: with $u = \dot y$, apply $$ \frac d{dt} \left(\frac 12 u^2\right) = u\dot u $$and also the chain rule.

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