It might seem counterintuitive, but even though y'(t) appears in the ODE, you don't actually need to calculate it. No DIFF, no GRADIENT, no analytical derivatives, no anything. First let me show you that this is true, then I will show you why it is true.
Just as an example, say, m = c = g = k = 1. We'll say y(t) = cos(t), and we'll take some arbitrary initial conditions x(0) = 2 and x'(0) = 3.
% Method 1, your current method, explicitly differentiating to get -sin(t):
F = @(t,x) [0 1; -1 -1]*x + [0;-1 + cos(t) - sin(t)];
x0 = 2; xp0 = 3; IC = [x0; xp0];
[tout, xout1] = ode45(F,[0:0.1:10],IC);
figure, plot(tout,xout1(:,1));
hold all
% Method 2, without differentiating y in any way:
F = @(t,z) [-1 1; -1 0]*z + [1;1]*[cos(t)] + [-t;0];
IC = [x0; x0 + xp0 - cos(0)];
[tout, xout2] = ode45(F,[0:0.1:10],IC);
plot(tout,xout1(:,1),'r:'), legend('Method 1', 'Method 2');
You can see they give exactly the same answer, and it is true in general, you will never have to differentiate y for this sort of linear ODE.
Now, I will will show you how I derived this. Start with your equation:
m*x''(t) + c*x'(t) + k*x(t) + m*g = k*y(t) + c*y'(t)
Step 1. Divide through by m. I will call k/m = K and c/m = C.
x'' + C*x' + K*x + g = K*y + C*y'
Step 2. Rearrange to get x'' by itself.
x'' = -C*x' - K*x + K*y + C*y' - g
Step 3. Integrate this up once with respect to time, assuming zero intial conditions for everything (don't worry, we'll deal with them later). I will call int(y) = Y and int(x) = X.
x' = -C*x - K*X + K*Y + C*y - g*t
Step 4. Define z1 = x and rewrite.
z1' = -C*z1 - K*X + K*Y + C*y - g*t
Step 5. Define -K*X + K*Y = z2, then differentiate z2.
z1' = -C*z1 + z2 + C*y - g*t
z2' = -K*x + K*y = -K*z1 + K*y
Step 6. If you let z = [z1;z2], the final equation is:
z' = [-C 1; -K 0]*z + [C;K]*y + [-g*t; 0]
And, as we defined in step 4, your final answer x(t) is given by the first state, z1.
Step 7. How are z(0) and z'(0) related to x(0) and x'(0)? Figure out how to map initial conditions.
z1(0) is simply equal to x(0).
z2(0) is a bit more tricky. Plug in t=0 into the equation for z1'.
z1'(0) = -C*z1(0) + z2(0) + C*y(0) - g*0 = -C*z1(0) + z2(0) + C*y(0)
Rearrange this to solve for z2(0).
z2(0) = z1'(0) + C*z1(0) -C*y(0)
z2(0) = x'(0) + C*x(0) -C*y(0)
DONE
There you have it. This is a very general method, and it is used in particular to generate state space equations from linear ODEs. See http://en.wikipedia.org/wiki/State_space_representation#Canonical_realizations
