add e^x to both sides
dN/dt + (1/10) N(t) = 9e^x
Glad to help. What you want to do is calculate an integrating factor, which is in this case:
$\displaystyle \mu(t)=e^{\frac{1}{10}\int\,dt}=e^{\frac{t}{10}}$
You want to multiply the equation in standard linear form by this factor:
$\displaystyle e^{\frac{t}{10}}\frac{dN}{dt}+\frac{1}{10}e^{\frac {t}{10}}N(t)=9e^{\frac{t}{10}}$
Do you see now that the left side is $\displaystyle \frac{d}{dt}\left(e^{\frac{t}{10}}N(t) \right)$ ?
after the left hand side above, I have RHS=9e^1/10t
Integrate both sides(is it ok to have 0.1t instead of 1/10t?) e^0.1t N(t) = 9e^0.1t + C
Then I'm stuck, I think I have to divide both sides by e^0.1t but not sure
Any help gratefully accepted
You have integrated the right-hand side incorrectly. Recall:
$\displaystyle a\int e^{bx}\,dx=\frac{a}{b}e^{bx}+C$ where $\displaystyle a,b$ are constants, and $\displaystyle b\ne0$.
edit: yes, it is fine to replace 1/10 with 0.1. It's a matter of preference, so use which you prefer.
Integrating both sides with respect to $\displaystyle t$, you should have:
$\displaystyle \int\,d\left(e^{\frac{t}{10}}N(t) \right)=9\int e^{\frac{t}{10}}\,dt$
You may choose to use definite integrals instead, using the boundaries as the limits, but I suspect you may prefer to use the initial conditions to find the parameter (constant of integration) instead.
Your notes are telling you that after you integrate, on the left you will have the product of the integrating factor and the dependent variable, which in our case is:
$\displaystyle e^{\frac{t}{10}}N(t)$
and on the right you will have the integral of the product of the integrating factor and what was on the right before we multiplied through by the integrating factor, which in our case was the constant 9. And this is what we have, you are not missing anything.
So, use the rule I gave above to integrate the right side.