If p(x) is a solution to a first-order differential equation in the form of dfdx=g(f), then p(x+c), with c-constant, is a solution as well. I know the idea of the proof, but I am having trouble expressing it in a rigorous proof. Namely, my idea is that if we shift p(x) to the left by c, then both the LHS and RHS will shift by cP (invoking the chain rule), so they are equal. Should I define u=x+c and do a change of variable?

Question

If p(x) is a solution to a first-order differential equation in the form of dfdx=g(f), then p(x+c), with c-constant, is a solution as well.  I know the idea of the proof, but I am having trouble expressing it in a rigorous proof. Namely, my idea is that if we shift p(x) to the left by c, then both the LHS and RHS will shift by cP (invoking the chain rule), so they are equal. Should I define u=x+c and do a change of variable?

Goodwin2ug · Accepted Answer

Sippose p(x) is a solution of  dydx=g(y).    (1)  that means  p;(a)=g(p(a)) for all a in the domain of p.    (2)  in particular putting a=x+c in (2),we have  (p(x+c))′=p′(x+c)=g(p(x+c))  which says that y=p(x+c) is a solution of (1). we have used the chain rule in the first equality of the last equation.  the geometric reason is that the slope field g(y) is translation invariant with respect to x.

If p(x) is a solution to a first-order differential equation in the form of \frac{df}{dx}=g

Answered question

Answer & Explanation

New Questions in Differential Equations