My teacher said that for first order non homogeneous IVP $$y'+a(x)y=f(x) , \ \ y(0)=y_o$$ we can get the solution directly using the following formula : $$\mu y|_{x=0}^x=\int_{\zeta=0}^{\zeta=x} f(\zeta)\mu(\zeta)d\zeta$$ I would like to know what is the proof ? and whether it is related to green function ? (because my teacher explained the above formulas for first order IVP while talking about green function to solve second order IVP but I could not understand the relation of the above formula to green's method)
Note :
here is my trial to prove the above formula , is this a right proof and is there another way to prove it ? can we prove using green's method?
$$\mu(x)=e^{\int a(x)dx}$$
$$\mu(x)y'+\mu(x)a(x)y'=\mu(x)f(x)$$
$$\frac{d}{dx}\mu(x) y=f(x)\mu(x)$$
integrating both sides
$$\mu(x) y=\int f(x)\mu(x) dx+ C$$
Applying the initial condition
$$\mu(x=0) y(x=0)={\int \mu(x) dx}\ \ |_{x=0}+ C$$
$$C=\mu(x=0) y(x=0)-{\int \mu(x) dx}\ \ |_{x=0} $$
$$\mu(x) y=\int f(x)\mu(x) dx+ C$$
$$\mu(x) y=\int f(x)\mu(x) dx+ \mu(x=0) y(x=0)-{\int \mu(x) dx}\ \ |_{x=0}$$
$$\mu(x) y-\mu(x=0) y(x=0)=\int f(x)\mu(x) dx-{\int \mu(x) dx}\ \ |_{x=0}$$
$$\mu y|_{x=0}^x=\int_{\zeta=0}^{\zeta=x} f(\zeta)\mu(\zeta)d\zeta$$
I tried to work it out by my own but I was slightly unsuccessful at satisfying the boundary condition. Of course if I found a slot I will proceed. However, I found this note useful. Also, there is a paper which can shed some light.