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Differential equation theory

This page is about solving a second-order linear differential equation with real and coincident roots. The equation being solved was given earlier but it is almost there halfway down following the word "thus". It is homogeneous, i.e. it equals 0. I have followed the writing down to the line following "So". Can someone explain the rest?

Re: Differential equation theory

Re: Differential equation theory

Another way of seeing that is to use the "reduction of order" method. If we have an nth order d.e, and know one solution, u(x), then looking for a solution of the form y(x)= xu(x).

For example, if a second order differential equation has characteristice equation with at double root, that is, of the form , then the differential equation is of the form . Knowing that [tex]e^{ax}[/itex] is a solution ( ), we try a solution of the form y= . Then and . Putting those into the equation, [tex]a^2e^{ax}u+ 2ae^{ax}u'+ e^{ax}u''- 2a(ae^{ax}u+ e^{ax}u')+ a^2(e^{ax}u)= e^{ax}(u''+ (2a- 2a)u'+ (a^2- 2a^2+ a^2)u)= e^{ax}u''= 0[tex]. The terms involving u only have canceled because is a solution to the differential equation. The terms involving u' only have canceled because a is a **double** root. Since is never 0, we must have u''= 0. Integrating twice, we have u(x)= C+ Dx so that . The " " we already had but now we know that is also a solution.