very hard differential equations Hi I would need help with solving the two following differential equations:


*

*$\dot{N} =\left(a\frac{\dot{x}(t)}{x(t)} -  b\frac{\dot{y}(t)}{y(t)} - c \frac{\dot{z}(t)}{z(t)} \right) \alpha N$

*$\dot{N} =\left(a\frac{\dot{x}(t)}{x(t)} -  b\frac{\dot{y}(t)}{y(t)} - c \frac{\dot{z}(t)}{z(t)} \right) \alpha(t)N$


I also know that $N = (px(t) - ry(t) - sz(t))\frac{1}{\alpha+\kappa} $ and $N = (px(t) - ry(t) - sz(t))\frac{1}{\alpha(t)+\kappa}$ respectively - although I am not sure if this information is crucial or not.  
The way I was thinking to approach this problem is that I know there is a class of separable differential equations:
$\frac{dx}{dt}= F(t)g(x)$ 
And I know how to solve these, but I have no idea if 1. or 2. can be expressed in such form. I think that this should be possible since everything in brackets is a function of t, but I don't think that it is rigorous to just say that $F(t)=\left(a\frac{\dot{x}(t)}{x(t)} -  b\frac{\dot{y}(t)}{y(t)} - c \frac{\dot{z}(t)}{z(t)} \right)$. But I don't know if then the rule for separable differential equation still applies as this would make F a compound function.  
I also tried to work with inverse functions $t(x)=x^{-1}(t)$, $t(y)=y^{-1}(t)$ and $t(z)=z^{-1}(t)$ and their derivatives, but I feel that just confused me even more.
I have never encountered a complex problem such as this so any help would be much appreciated.
 A: Well, when $\text{N}\left(t\right)\ne0$:
$$\text{N}'\left(t\right)=\alpha\cdot\left\{\text{a}\cdot\frac{\text{x}'\left(t\right)}{\text{x}\left(t\right)}+\text{b}\cdot\frac{\text{y}'\left(t\right)}{\text{y}\left(t\right)}+\text{c}\cdot\frac{\text{z}'\left(t\right)}{\text{z}\left(t\right)}\right\}\cdot\text{N}\left(t\right)\space\Longleftrightarrow\space$$
$$\int\frac{\text{N}'\left(t\right)}{\text{N}\left(t\right)}\space\text{d}t=\alpha\cdot\left\{\text{a}\cdot\int\frac{\text{x}'\left(t\right)}{\text{x}\left(t\right)}\space\text{d}t+\text{b}\cdot\int\frac{\text{y}'\left(t\right)}{\text{y}\left(t\right)}\space\text{d}t+\text{c}\cdot\int\frac{\text{z}'\left(t\right)}{\text{z}\left(t\right)}\space\text{d}t\right\}$$
$$\ln\left|\text{N}\left(t\right)\right|+\text{C}_1=\alpha\cdot\left(\text{a}\cdot\ln\left|\text{x}\left(t\right)\right|+\text{b}\cdot\ln\left|\text{y}\left(t\right)\right|+\text{c}\cdot\ln\left|\text{z}\left(t\right)\right|+\text{C}_2\right)\tag1$$
Was too long for a comment.
Take the $\exp$ of both sides:
$$\text{K}_1\cdot\left|\text{N}\left(t\right)\right|=e^{\alpha\cdot\text{a}\cdot\ln\left|\text{x}\left(t\right)\right|}\cdot e^{\alpha\cdot\text{b}\cdot\ln\left|\text{y}\left(t\right)\right|}\cdot e^{\alpha\cdot\text{c}\cdot\ln\left|\text{z}\left(t\right)\right|}\cdot e^{\text{C}_2}=$$
$$\text{K}_2\cdot\left|\text{x}\left(t\right)\right|^{\alpha\cdot\text{a}}\cdot\left|\text{y}\left(t\right)\right|^{\alpha\cdot\text{b}}\cdot\left|\text{z}\left(t\right)\right|^{\alpha\cdot\text{c}}\tag2$$
So, we get:
$$\text{K}\cdot\left|\text{N}\left(t\right)\right|=\left(\left|\text{x}\left(t\right)\right|^\text{a}\cdot\left|\text{y}\left(t\right)\right|^\text{b}\cdot\left|\text{z}\left(t\right)\right|^\text{c}\right)^\alpha\tag3$$
