Is this theorem true?

1. Sep 27, 2004

quasar987

In my physics textbook there is a theorem that goes "If x1(t) and x2(t) are both solutions of a linear homogeneous d.e., then x(t) = x1(t) + x2(t) is also a solution."

I need to know if the inverse is true, i.e. If x(t) = x1(t) + x2(t) is a solution, does it implies that x1(t) and x2(t) are also solutions separately.

I tried to do a proof similar to that of the first theorem but I come to (for the case of a second order d.e. and using f(x) = g(x) + h(x) instead of x(t) = x1(t) + x2(t)):

[a2*d²g/dx² + a1*dg/dx + a0*g] + [a2*d²h/dx² + a1*dh/dx + a0*h] = 0

of course [a2*d²g/dx² + a1*dg/dx + a0*g] = 0 and [a2*d²h/dx² + a1*dh/dx + a0*h] = 0 is a solution but maybe it's [a2*d²g/dx² + a1*dg/dx + a0*g] = -[a2*d²h/dx² + a1*dh/dx + a0*h] too, right?

2. Sep 27, 2004

StatusX

no, the converse isn't true. for example, y'' = 0 has a solution y = a*x + b, but that doesn't mean y = cosh(e^(1-tan(x))) and y = a*t +b - cosh(e^(1-tan(x))) are also solutions. is that what you were asking?

3. Sep 27, 2004

quasar987

I'm not sure I follow your reasoning involving cosh, e, t and x, but if you say that the converse of the theorem "If x1(t) and x2(t) are both solutions of a linear homogeneous d.e., then x(t) = x1(t) + x2(t) is also a solution." isn't true, that is indeed what I wanted to know!

Thank you.

4. Sep 27, 2004

quasar987

It is true for a complex function though!

If f(x) = u(x) + i*w(x) is the solution of a lin. homo. d.e. WITH REAL coefficients, then u(x) and i*w(x) are solutions separately.

5. Sep 30, 2004

StatusX

sorry, i was being a smart ass. its just not true cause if f(t) is a solution, then your theorem would mean g(t) and f(t)-g(t) would both be solutions for ANY g(t), which obviously isn't true.