First-order linear differential equation ($y^{\prime }+p(x)y=q(x)$)

A first-order linear differential equation has the form $y^{\prime }+p(x)y=q(x)$. Multiplication by an integrating factor $\mu$ satisfying ${\mu }^{\prime }=p\mu$ turns it into $(\mu y{)}^{\prime }=\mu q$.

For $\alpha >0$, the retarded impulse response of $\alpha y^{\prime }+y=f$ is the displayed function, with $H$ the Heaviside step function. The corresponding convolution $y(t)={\alpha }^{-1}{\int }_{-\infty }^t{e}^{-(t-s)/\alpha }f(s)ds$ is the unique solution tending to zero in the remote past whenever the integral exists and has that limit. For example, bounded forcing that tends to zero as $t\to -\infty$ satisfies these conditions. A general forcing need not: $f\equiv 1$ admits no such solution. A unit-duration pulse with $\alpha =1$ gives $y=0$ before zero, $y=1-e^{-t}$ between zero and one, and $y=(1-e^{-1})e^{-(t-1)}$ thereafter.

For $y\prime -ay=e^{\lambda x}$, subtracting an appropriate homogeneous solution from the particular solution removes its apparent singularity as $\lambda \to a$. The resulting derivative in the parameter produces the resonant particular solution $x{e}^{ax}$. The arbitrary constant must be reparametrized before taking the limit.