- 373

- 0

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

[tex]y ^{\prime} + p(t)y = g(t), \qquad y\left( t_0 \right) = y_0[/tex]

were

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

where

[tex]\mu (t) = \exp \int p(s) \: ds[/tex]

However, I have found this

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

where

[tex]\mu (t) = \exp \int _{t_0} ^t p(s) \: ds[/tex]

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

[tex]y ^{\prime} + p(t)y = g(t), \qquad y\left( t_0 \right) = y_0[/tex]

were

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

where

[tex]\mu (t) = \exp \int p(s) \: ds[/tex]

However, I have found this

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

where

[tex]\mu (t) = \exp \int _{t_0} ^t p(s) \: ds[/tex]

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

Last edited: