Show that y as a function of x, given parametrically by x
Show that $y$ as a function of $x$, given parametrically by $x = 2t+3t^{2}$, $y = t^{2}+2t^{3}$, satisfies $2y'^{3} + y'^{2} - y = 0$, where $y' = \dfrac{dy}{dx}$.
Explanation of Method
The parameter turns out to be the slope itself.
Step 1: Compute $y' = \dfrac{dy}{dx}$
Differentiate both coordinates with respect to $t$ and factor: \[x'_{t} = 2+6t = 2\bigl(1+3t\bigr), \qquad y'_{t} = 2t+6t^{2} = 2t\bigl(1+3t\bigr).\]
The common factor $2(1+3t)$ cancels, and the result is strikingly simple: $y' = \frac{2t(1+3t)}{2(1+3t)} = t, \qquad t \neq -\frac13$.
So the parameter is the slope of the curve at the corresponding point.
Step 2: Write $y$ in terms of $t$
\[ y = t^{2}+2t^{3}. \]
Step 3: Substitute into the equation
Replacing $y'$ by $t$ and $y$ by $t^{2}+2t^{3}$, everything cancels: \[2y'^{3}+y'^{2}-y = 2t^{3}+t^{2}-\bigl(t^{2}+2t^{3}\bigr) = 2t^{3}+t^{2}-t^{2}-2t^{3} = 0.\]
Step 4: Verification
\[t = 1:\quad x = 5,\ y = 3,\ y' = 1; \qquad 2(1)+1-3 = 0 \;\;\checkmark\]
\[t = 2:\quad x = 16,\ y = 20,\ y' = 2; \qquad 2(8)+4-20 = 0 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,2y'^{3}+y'^{2}-y \equiv 0 \qquad\text{for all } t \neq -\tfrac13\,}\]
Remark
Remark. An equation relating $y$ and $y'$ with no $x$ in it is called a Clairaut- or Lagrange-type equation, and the standard way to solve one is exactly the reverse of this computation: set $y' = t$, solve for $y$, then recover $x$ by integrating $dx = dy/t$. The excluded value $t = -\tfrac13$ is where $x'_{t} = y'_{t} = 0$ — a cusp of the parametrisation, at the point $\left(-\tfrac13,\,\tfrac{1}{27}\right)$.