Here's the problem:
(1) (a) Show that the solution of the linear equation $y'+p(t)y=g(t)$ can be written in the form y=cy1(t)+Y(t), where c is an arbitrary constant. Identify the functions y1(t) and Y(t).
So I'm trying to solve the general DE $y'+p(t)y=g(t)$ for y... I'm trying to use the integrating factor μ(t) to do so, where $μ(t)p(t)=μ'(t)$. I just plugged back in μ(t) to $(μ(t)y)'=g(t)μ(t)$ but now I'm having trouble putting it into the form they're looking for, which is $y=cy1(t)+Y(t)$.
Right now I have:
$$y=\dfrac {\int \mu(s)g(s)ds+c} {\mu(t)} $$
How would you put it into the form above?
You have already proved that $$y=(1/μ(t))[∫μ(s)g(s)ds+c]$$
Note that $$y=(1/μ(t))[∫μ(s)g(s)ds] +c/μ(t)$$
Let $$y_p = (1/μ(t))[∫μ(s)g(s)ds]$$
and $$y_h =c/μ(t)$$
Then $$ y= y_p + y_h$$