Initial conditions on hypergeometric equation Normally, when dealing with linear second order differential equations of the type:
$$a y''(x)+by'(x)+cy(x)=0$$
The solution is given by:
$$Ae^{mx}+Be^{nx} \quad \text{or} \quad e^{mx}(A+Bx) \quad \text{or} \quad Ae^{px}\cos(qx)+Be^{px}\sin(qx) $$ 
depending on whether the roots of the characteristic equation are real and distinct, real and equal or complex.
The coefficients $A$ and $B$ can be easily found by matching the initial conditions on $y$ and $y'$.
However with an hypergeometric equation of the type:
$$xy''(x)+(b-x)y'(x)+ay(x)=0$$
The solution is given by the hypergeometric function:
$$y(x)={}_{1}F_1(a,b,x)$$
What about the initial conditions on the equation? How can we match them?
(Do you have any reference article for looking more deeply into this kind of equations?)
Also, what are the derivatives of ${}_{1}F_1(a,b,x)$ with respect to $x$?
 A: The equation you have is known as the confluent hypergeometric equation! The general solution is
$$y(z)=cM(a, b, z)+c'U(a, b, z)$$
Given
$$y(0)=A$$
and
$$y'(0)=B$$ applying the first condition you have
$$cM(a, b, z)+c'U(a, b, 0)=A$$
For the second condition you use the fact that
$$\frac{d}{dz}M(a, b, z)=\frac{a}{b}M(a+1, b+1, z)$$
and
$$\frac{d}{dz}U(a, b, z)=-aU(a+1, b+1, z)$$
So,
$$c\frac{a}{b}M(a+1, b+1, 0)-ac'U(a+1, b+1, 0)=B$$
Then solve for $c$ and $c'$. As simple as that! Look at the Abramovitz book on the special functions, it is all time classics! Or in any textbook with the name like special functions of mathematical physics!
A: The equation you state as the hypergeometric equation is rather Kummer's equation and its solutions are Kummer functions or confluent hypergeometric functions. Hypergeometric functions have an additional $c$ parameter $F(a,b;c;z)$
You can check this reference for a description of the properties of the differential equation, the Kummer functions and derivatives.
