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

## 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.