\( y ^ { 3 } y ^ { \prime \prime } = 1 \)

To solve the equation \( y^3 y'' = 1 \), we begin by rewriting it in a more manageable form. This can be expressed as \( y'' = \frac{1}{y^3} \). It's a second-order differential equation, and observing that the right-hand side is a function of \( y \) suggests a potential use of substitution. You can let \( p = y' \), and thus \( y'' = p \frac{dp}{dy} \). This transforms your equation into \( p \frac{dp}{dy} = \frac{1}{y^3} \), which you can integrate by separating variables. The journey doesn't end there, though! After integrating, you'll need to revisit \( p \) and express the solution in terms of \( y \) to find the general solution, or specific solutions, depending on your initial conditions. Happy solving! To really dig into the essence of solving differential equations like this one, knowing the relationship between functions and their derivatives can have profound implications in physics, engineering, and beyond. For instance, oscillation problems in mechanical systems often lead to similar types of equations, showcasing just how intertwined mathematics and the physical world truly are. In tackling differential equations, one common mistake is rushing to solve without considering the nature of the solutions. Always check if your function is defined over the relevant interval, and remember to verify any constants of integration with initial or boundary conditions. This can save you from pursuing paths that led you astray!