Problems solving a DE I have the following DE $$\frac{\mathrm{dn} }{\mathrm{d} \theta } = \frac{f'(\theta ) - \alpha }{c'(n)}.$$
By taking the integral on both sides and letting the constants $= 0$, I get $$\theta = \frac{nc'(n)}{f'(\theta) - \alpha}.$$. 
This is what I've done: $\int c'(n) dn =  \int \left [ f'(\theta ) - \alpha \right ] d\theta \Rightarrow nc'(n) + K_{n} = \theta f'(\theta) - \alpha \theta + K_{\theta } \Rightarrow nc'(n) = \theta f'(\theta) - \alpha \theta + K_{\theta } - K_{n} \Rightarrow \frac{nc'(n)- K}{f'(\theta)-\alpha} = \theta$.   
But I want to get $\theta = \frac{-nc'(n)}{f'(\theta)-\alpha}$.
Where am I going wrong with this? 
 A: I do not quite know how you can get what you got.
Start from
$$\frac{\mathrm{dn} }{\mathrm{d} \theta } = \frac{f'(\theta ) - \alpha }{c'(n)} \tag{1}$$
We can get 
$$c'(n)\frac{\mathrm{dn} }{\mathrm{d} \theta } = f'(\theta ) - \alpha  \tag{2}$$
or
$$\frac{\mathrm{dc}}{\mathrm{dn}}\frac{\mathrm{dn} }{\mathrm{d} \theta } = f'(\theta ) - \alpha  \tag{3}$$
$$\frac{\mathrm{dc}}{\mathrm{d} \theta } = f'(\theta ) - \alpha  \tag{4}$$
So integration of (4) leads to
$$c(\theta) = f(\theta) - \alpha \theta +\beta  \tag{5}$$
Setting $\beta=0$, we get
$$\alpha \theta = f(\theta) -c(\theta) \tag{6}$$
A: What you should do is put all $n$'s on one side and all $\theta$'s on the other.  Then you can integrate both sides.
A: Your differential equation can be solved writing 
$$
 c'(n) dn = (f'(\theta) -\alpha ) d\theta 
$$
Now you can integrate both sides between $n_1, n_2$ and $\theta_1, \theta_2 $, to obtain
$$
\theta_2 - \theta_1 = \frac{1}{\alpha} \left [ f(\theta_2) - f(\theta_1) - (c(n_2)-c(n_1) \right ]
$$
