# Linear first order differential equations - just a question

1. Jun 20, 2005

I thought the general solution of the linear first order differential equation

$$y ^{\prime} + p(t)y = g(t), \qquad y\left( t_0 \right) = y_0$$

were

$$y = \frac{1}{\mu (t)} \left[ \int \mu (s) g(s) \: ds + \mathrm{C} \right],$$

where

$$\mu (t) = \exp \int p(s) \: ds$$

However, I have found this

$$y = \frac{1}{\mu (t)} \left[ \int _{t_0} ^t \mu (s) g(s) \: ds + y_0 \right],$$

where

$$\mu (t) = \exp \int _{t_0} ^t p(s) \: ds$$

Can anybody please clarify that? I don't see why those integral limits should be there. Any help is highly appreciated.

Last edited: Jun 20, 2005
2. Jun 20, 2005

### geosonel

general solution correct.

however, in this case, integration limits needed for initial condition:

$$y\left( t_0 \right) = y_0$$

that is:

$$\mu (t_0) = \exp \int _{t_0} ^{t_0} p(s) \: ds = \exp(0) = 1$$

$$y(t_0) = \frac{1}{\mu (t_0)} \left[ \int _{t_0} ^{t_0} \mu (s) g(s) \: ds + y_0 \right] = \frac {1} {1} \left [(0) \ + \ y_0 \right ] \ = \ y_0$$

nothing really complicated here