# Couchy's problem

### Couchy's problem

Hello.

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?
Guest

### Re: Couchy's problem

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 $$y= e^{ax}$$ have derivative $$y'= ae^{ax}$$, 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 $$y= e^{ax}$$.

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

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

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

HallsofIvy

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