Solution check of differential equation.

Question

Are these steps correct? If not what step is wrong?

1. $$ \frac{dt}{dZ} = \frac{N}{P}\left (\frac{1}{Z} + \frac{1}{P-Z} \right ) $$
2. $$ \frac{dZ}{dt} = \frac{1}{\frac{N}{P}\left (\frac{1}{Z} + \frac{1}{P-Z} \right )} $$
3. $$ \frac{dZ}{dt} \left( \frac{1}{Z} + \frac{1}{P-Z} \right ) = \frac{P}{N} $$
4. $$ \left( \frac{1}{Z} + \frac{1}{P-Z} \right ) dZ \frac{1}{dt}= \frac{P}{N} $$
5. $$ \left( \frac{1}{Z} + \frac{1}{P-Z} \right ) dZ = \frac{P}{N}dt $$
6. $$ \int \left( \frac{1}{Z} + \frac{1}{P-Z} \right ) dZ = \int \frac{P}{N}dt $$
7. $$ ln(|Z|) - ln (| Z-P|)= \frac{P}{N}t + C $$
8. $$ ln \left( \left |\frac{Z}{Z-P} \right | \right) = \frac{P}{N}t + C $$
9. $$ e^{\frac{P}{N}t + C} = \frac{Z}{Z-P} $$

Answer 1

Every step is right

The only "mistake" that I found is in row 9.

from 9: we apply the exponent rules

$$e^{\left[\left(\frac{P}{N}\right)t\ +C\ \right]}=e^{\left(\frac{P}{N}\right)t}e^C=Ce^{\left(\frac{P}{N}\right)t}$$

Then: $$\mathbf{Ce^{\left(\frac{P}{N}\right)t}=\frac{Z}{Z-P}\qquad} or\mathbf{\qquad CZ=(Z-P)e^{\left(\frac{P}{N}\right)t}}$$