# Solving a first order linear ODE via the method of integrating factors

Use the integrating factor method to find the general solution of the following IVP

$$y'+p(t)y=q(t), y(t_{o})=y_{o}$$

What i tried

While i understand and know how to use the integrating factor method to solve a linear ODE. Previously all the questions i did were those that involves values and number. Now my challenge is to do the same but this time involving expressions that does not contain .any values.

What i did first was to find the integrating factor, $u(t)=e^{ \int^{}p(t)dt}$. Then i multiply both the LHS and the RHS by the integrating factor to get $$ye^{ \int^{}p(t)dt}=\int^{}q(t)e^{ \int^{}p(t)dt}$$. However i am unable to proceed from here. Could anyone explain. Thanks

suppose $A$ is the unique solution of $A' = p(t)A, A(t_0) = 1.$ you can write down the solution
$$A = exp \left({\int_{t0}^t p(s) \, ds}\right), \dfrac{1}{A}= exp \left(-{\int_{t0}^t p(s) \, ds}\right)$$
multiplying by $A$ simplifies the equation to $(Ay)' = Aq.$ integrate this from $t_0$ to $t$ gets you, $A(t)y(t) - y_0 = \int_{t_0}^t A(s)y(s) \, ds.$ therefore your $$y = \frac{1}{A}\left( y_0 + \int_{t_0}^t A(s)y(s) \, ds \right) .$$