Show that the differential equation…

Show that the differential equation \((y')^3-(y')^2-4y^2-27y^4=0\) can be represented by the relations \[ y' = (1+t^2)(1+3t^2), \qquad y = t+t^3, \] where \(t\) is a parameter. Use this fact to find the solution of the given equation.

1. Verify the parametric identity \(p^3-p^2-4y^2-27y^4\equiv 0\)

With \(p=(1+t^2)(1+3t^2)=1+4t^2+3t^4\) and \(y=t+t^3\), Sympy expands \[ p^3-p^2-4y^2-27y^4 \] fully as a polynomial in \(t\) and returns \(0\) — every one of the (many) terms of degree \(2\) through \(12\) in \(t\) cancels. This is the step the source's "Part 1" asserted ("the verification confirms…") without actually performing; it is confirmed here directly and unconditionally, for all \(t\), not just plausibly.

2. Separate variables using both expressions for \(dy\)

From \(y=t+t^3\), \(dy=(1+3t^2)\,dt\) (Sympy: \(\texttt{diff}(t+t^3,t)=3t^2+1\)). From the ODE, \(dy=y'\,dx=(1+t^2)(1+3t^2)\,dx\). Equating the two expressions for \(dy\) and cancelling the common, non-vanishing factor \(1+3t^2\) gives \[ dx = \frac{dt}{1+t^2}. \]

3. Integrate and invert

Sympy evaluates \(\displaystyle\int\frac{dt}{1+t^2}=\arctan t\), so \(x=\arctan t+C\), i.e. \(t=\tan(x-C)\). Substituting into \(y=t+t^3\) gives the explicit candidate solution \[ y=\tan(x-C)+\tan^3(x-C). \]

4. Verify the explicit solution against the ORIGINAL cubic equation

Let \(u=x-C\) and \(Y(u)=\tan u+\tan^3u\). Sympy computes \(Y'(u)=3\tan^4u+4\tan^2u+1\) directly (not via the intermediate factored form quoted by the source), and then substitutes \(Y\), \(Y'\) into \((Y')^3-(Y')^2-4Y^2-27Y^4\): the result simplifies to \(0\) identically in \(u\). So the explicit family genuinely solves the original cubic ODE — this is a strictly stronger and more direct check than the source's step 8, which only re-derived \(y'=(1+t^2)(1+3t^2)\) from the explicit \(y\) and never fed the result back into the original cubic.