Solve the differential equation (y')^7 - (y')^5 + y' + 3 = 0
Solve the differential equation $(y')^7 - (y')^5 + y' + 3 = 0$.
Explanation of Method
The equation involves only \(y'\) with no explicit \(x\) or \(y\) dependence. Setting \(p=y'\) reduces the problem to finding the real roots of a degree-7 polynomial \(p^7-p^5+p+3=0\); each real root \(p_i\) gives a family of straight-line solutions \(y=p_ix+C\).
Set up the polynomial in \(p=y'\)
Let \(p = y' = \dfrac{dy}{dx}\). The equation becomes
\[p^7 - p^5 + p + 3 = 0.\]
This is a seventh-degree polynomial in \(p\), so by the Abel–Ruffini theorem no general closed-form (radical) expression for its roots exists.
Rule out rational roots
By the Rational Root Theorem, any rational root must divide \(3\), i.e. lie in \(\{\pm1,\pm3\}\). Direct substitution gives \(f(-1)=2\), \(f(1)=4\), \(f(-3)=-1944\), \(f(3)=1950\), none zero. So the polynomial has no rational roots.
Locate the real roots
Let \(f(p)=p^7-p^5+p+3\). A sign check gives \(f(-2)=-95<0\) and \(f(0)=3>0\), so by the Intermediate Value Theorem there is a real root between \(-2\) and \(0\). Applying Sturm's theorem (or a numerical root solver such as sympy.Poly.nroots) to the full degree-7 polynomial shows this is the only real root; the other six roots form three complex-conjugate pairs. High-precision root finding gives
\[p_0 \approx -1.2521412990911069,\]
which satisfies \(f(p_0)=0\) to the precision shown. (Note: a coarse Newton-Raphson estimate can easily land on a nearby but slightly inaccurate value such as \(-1.2207\); the value above was verified numerically to 20 digits.)
Integrate each real root
For any root \(p_i\) of \(f(p)=0\) (real or complex), \(y'=p_i\) is a constant, so integrating directly gives \(y = p_i x + C_i\). Since the ODE has no \(x,y\) dependence, this constant-slope family is always a valid solution branch whenever \(f(p)=0\) has a real solution.
Result
Final answer:
The general solution consists of straight lines \(y = p_i x + C\), one family for each real root \(p_i\) of \(p^7-p^5+p+3=0\). This polynomial has exactly one real root (confirmed by Sturm's theorem), so there is exactly one real solution family:
\[\boxed{y = p_0\,x + C, \qquad p_0 \approx -1.2521412990911069 \ \text{(the unique real root of } p^7-p^5+p+3=0\text{)}}\]
The remaining six roots of the polynomial are complex (three conjugate pairs) and do not yield additional real solution families.