1. ## A+b=a'/b'

Find two real functions whose sum is the ratio of their first derivatives, i.e. solve:

$\displaystyle A(x)+B(x)=\frac{A'(x)}{B'(x)}$

Or prove no solutions exist.

(This is a problem that cropped up in another thread: http://www.mathhelpforum.com/math-he...tml#post345253)

2. There are an infinite number of solutions aren't there ?

Just solve as a first order ode in $\displaystyle A(x)$.

Rewrite as

$\displaystyle A'(x) - B'(x)A(x) = B(x)B'(x)$,

multiply by the integrating factor

$\displaystyle e^{-B(x) }$

and the equation can be written as

$\displaystyle (A(x)e^{-B(x)})' = B(x)B'(x)e^{-B(x)}$

so

$\displaystyle A(x)e^{-B(x)} = \int B(x)B'(x)e^{-B(x)} dx$

and

$\displaystyle A(x) = e^{B(x)}\int B(x)B'(x)e^{-B(x)} dx$.

Now just choose something convenient for $\displaystyle B(x)$.

An easy choice would be $\displaystyle B(x) = x$ after which an integration by parts would lead to $\displaystyle A(x) = Ce^{x} - x - 1$.

3. Originally Posted by BobP
There are an infinite number of solutions aren't there ?

Just solve as a first order ode in $\displaystyle A(x)$.

Rewrite as

$\displaystyle A'(x) - B'(x)A(x) = B(x)B'(x)$,

multiply by the integrating factor

$\displaystyle e^{-B(x) }$

and the equation can be written as

$\displaystyle (A(x)e^{-B(x)})' = B(x)B'(x)e^{-B(x)}$

so

$\displaystyle A(x)e^{-B(x)} = \int B(x)B'(x)e^{-B(x)} dx$

and

$\displaystyle A(x) = e^{B(x)}\int B(x)B'(x)e^{-B(x)} dx$.

Now just choose something convenient for $\displaystyle B(x)$.

An easy choice would be $\displaystyle B(x) = x$ after which an integration by parts would lead to $\displaystyle A(x) = Ce^{x} - x - 1$.
We can actually go further and integrate fully giving

$\displaystyle A = c_1 e^{B} - B - 1$