Couchy's problem

Couchy's problem

Postby Guest » Sat Jan 30, 2021 2:07 pm


It is given that u''(x)+1/3*u'(x) + 3*u(x), u'(0)=1, u(0)=3/2.

How should I go about solving this?

Re: Couchy's problem

Postby HallsofIvy » Sat Feb 06, 2021 8:05 pm

First it is "Cauchy". Second, what you have written is NOT a differential equation because it is not an equation! I assume you mean u''(x)+ (1/3)u'+ 3u= 0.

This is a "homogeneous linear equation with constant coefficients", about the easiest kind of differential equation.

In Calculus you should have learned that "exponentials", functions of the form [tex]y= e^{ax}[/tex] have derivative [tex]y'= ae^{ax}[/tex], the same "kind" of function as y. And because in u''(x)+ (1/3)u'(x)+ 3u(x)= 0 the various derivatives have to cancel, they must be the same "kind". So it makes sense to try [tex]y= e^{ax}[/tex].

If [tex]y= e^{ax}[/tex] then [tex]y'= ae^{ax}[/tex] and [tex]y''= a(ae^{ax})= a^2e^{ax}[/tex]. With [tex]y= a^{ax}[/tex] the equation becomes [tex]y''+ (1/3)y'+ 3y= a^2e^{ax}+ (1/3)ae^{ax}+ 3e^{ax}= (a^2+ (1/3)a+ 3)e^{ax}= 0[/tex]. Since [tex]e^{ax}[/tex] is never 0 we must have [tex]a^2+ (1/3)a+ 3= 0[/tex].

That is called the "characteristic equation" for the differential equation and its roots, [tex]a_1[/tex] and [tex]a_2[/tex], are the "characteristic roots". The general solution to the differential equation is [tex]u(x)= C_1e^{a_1x}+ C_2e^{a_2x}[/tex].

So back to you- what are the roots to the equation [tex]a^2+ (1/3)a+ 3= 0[/tex], what is [tex]u(x)= C_1e^{a_1x}+ C_2e^{a_2x}[/tex], and what must [tex]C_1[/tex] and [tex]C_2[/tex] be so that [tex]u'(0)= 1[/tex] and [tex]u(0)= 3/2[/tex]?

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