Solving a differential equation with composite functions If $c=f(a+e^b)+g(a-e^b)$ where $f$ and $g$ are functions of $a+b^2$ and $a-b^2$ respectively, find $c$ such that when $b=0$, you find that $c=0$ and $\frac{\partial c}{\partial b}=1+a$.
 A: You say, that
$$
c(a,b) = f(a+e^b) + g(a-e^b)
$$
where $f, g : \mathbb R \to \mathbb R$, and you need to find some particular form of $c(a,b)$ so that
\begin{align}
c(a,0) &= 0\\
\left . \frac {\partial c}{\partial b} \right |_{b = 0} &= 1 + a
\end{align}
Use first restriction.
$$
c(a,0) = f(a+1) + g(a-1) = 0 \tag 1
$$
Now, let's find the form of of $c_b(a,b)$
$$
c_b(a,b) = f'(a+e^b)e^b - g'(a-e^b)e^b = e^b \left ( f'(a+e^b) - g'(a+e^b)\right) \tag 2
$$
so
$$
c_b(a,0) = f'(a+1)-g'(a-1) = 1 + a \tag 3
$$
Now, differentiate $(1)$
$$
f'(a+1) + g'(a-1) = 0 \tag 4
$$
and therefore
$$
g'(a-1) = -f'(a+1) \tag 5
$$
substitute $(5)$ to $(3)$
$$
2f'(a+1) = a+1
$$
or
$$
f'(x) = \frac x2
$$
which has obvious solution
$$f(x) = \frac {x^2}4 + C$$
Now, find $g(x)$
$$
g(a-1) = -f(a+1) = -\frac {(a+1)^2}4 - C
$$
or, after some trivial manipulations
$$
g(x) = -\frac {(x+2)^2}4 - C
$$
So, final answer is
$$
c(a,b) = f(a+e^b) + g(a-e^b) = \frac {\left ( a+e^b\right )^2}4 - \frac {\left ( a-e^b+2\right )^2}4
$$
