Proof. The function \(\mu (t) = e^{P(t)}\), the integrating factor, is positive and differentiable with \(\mu ' = p\mu \). For any differentiable \(y\) on \(I\), \[ (\mu y)' = \mu y' + p\mu y = \mu (y' + py) . \] If \(y\) solves the problem, then \((\mu y)' = \mu q\), and integrating from \(t_0\), using \(\mu (t_0) = 1\), gives \(\mu (t)y(t) = y_0 + \int _{t_0}^t\mu q\), which is the formula; so any solution must be this function, which proves uniqueness. Conversely, the function defined by the formula is differentiable, by the fundamental theorem of calculus, and satisfies \((\mu y)' = \mu q\), hence \(y' + py = q\), with \(y(t_0) = y_0\). □