Good day everyone,

Studying for the first time 2nd order differential equations, we focused on the case of linear homogeneous equations. However, it comes out that the general integral, or general solution (or structure theorem) of the equation, well this theorem, can be proved in more advanced courses.

Given AY''+BY' +CY=0 where A, B, C are reals and A is different from 0,the general solution is the linear combination of two linearly independent solutions Y(x) = C1*Y1+C2*Y2, where C1 and C2 are arbitrary numbers.

However, it turns out as well that it is possible to give an elementary demonstration by constructing Cauchy problems and demonstrating that any particular solution can be represented as a linear combination of 2 solutions with specific coefficients.

Could someone please explain to me the intuition which lies behind this demonstration and its steps?

Thank you in advance!

  • $\begingroup$ Do you know and understand what linear independence of two functions means? $\endgroup$ – Lutz Lehmann Oct 29 '19 at 17:34
  • $\begingroup$ @LutzL yes, their ratio is not constant $\endgroup$ – Shootforthemoon Oct 29 '19 at 17:38
  • $\begingroup$ What does this tell you then about the derivatives? $\endgroup$ – Lutz Lehmann Oct 29 '19 at 17:47
  • $\begingroup$ @LutzL they are also linearly independent? $\endgroup$ – Shootforthemoon Oct 29 '19 at 18:08
  • $\begingroup$ That is not necessarily true. What about the derivative of the ratio? $\endgroup$ – Lutz Lehmann Oct 29 '19 at 23:14

By your definition, $Y_1/Y_2$ is not a constant. This means that its derivative $$ \frac{Y_1'Y_2-Y_1Y_2'}{Y_2^2} $$ is almost nowhere zero. The numerator is the determinant of the Wronskian. This in turn means that the system $$ Y(x_0)=C_1Y_1(x_0)+C_2Y_2(x_0)\\ Y'(x_0)=C_1Y_1'(x_0)+C_2Y_2'(x_0) $$ has a solution for almost any $x_0$. Pick one $x_0$ and apply the existence and uniqueness theorem to find that $Y=C_1Y_1+C_2Y_2$ everywhere.

Thus the solution space is spanned by two linearly independent functions, thus has dimension $2$. You can add arbitrary initial conditions on the basis solutions, the only restriction is that the Wronskian determinant has to be non-zero in the initial point.

  • $\begingroup$ Thank you very much. I suspected that the Wronskian was behind the discussion on the derivatives, but we did not study it. However I'll look for its precise meaning. As for the rest, your answer is simple and clear. (I up voted and the site recorded but saying it does not show votes cast by those with less than 15 reputation). Thanks again! $\endgroup$ – Shootforthemoon Oct 31 '19 at 15:19

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