Posted by: Steven Sam | February 3, 2009

Some simple differential equations

Here is a short one about two simple differential equations. They both have “standard” solutions that appear in textbooks, but here is a method that treats them in similar ways.

First, an easy one. If f is a (infinitely) differentiable function such that f' = f, then f(x) = re^x for some constant r. Proof: Write \displaystyle g(x) = \frac{f(x)}{e^x}. Then g'(x) = -f(x)e^{-x} + f'(x)e^{-x} = 0 and hence \displaystyle \frac{f(x)}{e^x} is a constant function r.

Here is another one. Let f be a (infinitely) differentiable function defined on [0,2\pi) such that f'' = -f. Then f(x) = A\cos(x) + B\sin(x) for some constants A and B. First make the substitution x = \arcsin(t), and set g(t) = f(\arcsin(t)). Consider also the function h(t) = \cos(\arcsin(t)). Then

\displaystyle h'(t) = \frac{t}{\sqrt{1-t^2}}

and

\displaystyle h''(t) = \frac{1}{(1-t^2)^{3/2}}.

Now we do the same for g:

\displaystyle g'(t) = \frac{f'(\arcsin(t))}{\sqrt{1-t^2}}

and

\begin{array}{ll} \displaystyle g''(t) & \displaystyle = \frac{f''(\arcsin(t)) + tf'(\arcsin(t))(1-t^2)^{-1/2}}{1-t^2}\\&\displaystyle = \frac{f''(\arcsin(t))\sqrt{1-t^2} + tf'(\arcsin(t))}{(1-t^2)^{3/2}} \end{array}.

The numerator of the last term is actually a constant: its derivative is

\displaystyle f'''(\arcsin(t)) - \frac{tf''(\arcsin(t))}{\sqrt{1-t^2}} + f'(\arcsin(t)) + \frac{tf''(\arcsin(t))}{\sqrt{1-t^2}},

which is 0 by the fact that f''' = -f'.

Hence Ah''(t) = g''(t) for some constant A. Integrating twice, we get g(t) = Ah(t) + Bt + C for some constants B and C. Now substituting back t = \sin(x), we get f(x) = A\cos(x) + B\sin(x) + C. Since f'' = -A\cos(x) - B\sin(x), we conclude that C = 0.

-Steven

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Responses

  1. for the second problem ,

    If m is a positive real number , the only solution to differential equations :

    f”(x) + mf(x) = 0

    is of the form :

    f(x) = a cos(nx) + b sin(nx) , where n = (m)^1/2

    Consider the function :

    g(x) = f(x)cos(nx) – (1/n)*f'(x)sin(nx)

    h(x) = f(x)sin(nx) + (1/n)*f'(x)cos(nx)

    By differentiating g(x) and h(x) , we get :

    g'(x) = 0 = h'(x)

    so they’re just constant function.Hence , there exist a , b real numbers such that :

    a = g(x) = f(x)cos(nx) – (1/n)*f'(x)sin(nx)

    b = h(x) = f(x)sin(nx) + (1/n)*f'(x)cos(nx)

    and we get :

    f(x) = acos(nx) + bsin(nx)

    Q.E.D.


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