# Notes on Differential Equations (1) —— $\frac{dy}{dx}=ay+e^{st}$

This series will be my personal notes on differential equations (including ordinary differential equation (ODE), partial differential equation (PDE) and stochastic differential equation (SDE)). These posts may contain mistakes.

The first a few posts will be different solutions for different forms of first order ODE.

The first form is $\frac{dy}{dx}=ay+e^{st}$.

The first form is $\frac{dy}{dx}=ay+e^{st}$ and $y=y(0)$ at $t=0$. In this case, $e^{st}$ is the input, or the source term, and $y(t)$ is called "Exponential Response". Here we look for the particular solution $y_p=Ye^{st}$.

We can substitute $y_p$ into $\frac{dy}{dx}=ay+e^{st}$ and we get the particular solution as below.

\begin{align*}\frac{dY_p}{dt}&=ay_p+e^{st}\\ \frac{d}{dt}Ye^{st}&=aYe^{st}+e^{st}\\ Yse^{st}&=aYe^{st}+e^{st}\\ Ys&=aY+1\\ Y(s-a)&=1\\ Y&=\frac{1}{s-a} \end{align*}

Now we need to get the homogeneous solution, or the full solution, by adding $y_p$ with any null solution $y_n$ (which is when $t=0$, and the subscript $n$ denotes the 'null' solution). And the null solution of $\frac{dy}{dt}=ay$ is $Ce^{at}$

Then we have

\begin{align*}y(t)&=y_p+y_n\\ &=\frac{e^{st}}{s-a}+Ce^{at} \end{align*}

Now consider the initial condition, $y=y(0)$ at $t=0$, we have

\begin{align*}y(0)&=\frac{1}{s-a}+C\\ C&=y(0)-\frac{1}{s-a} \end{align*}

The complete solution is

\begin{align*}y(t)&=\frac{e^{st}}{s-a}-\left[y(0)-\frac{1}{s-a}\right]e^{at}\\ &=\frac{e^{st}-e^{at}}{s-a}+y(0)e^{at}\\ y_p+y_n&=\frac{e^{st}-e^{at}}{s-a}+y(0)e^{at} \end{align*}

But there is an exception when $s=a$ because the denominator will be $0$. This is the case called resonance. Therefore our formula has to change. However, you may notice that the nominator is also $0$. That's to say we have a situation where $\frac{0}{0}$ appears. The formula when $s=a$ is

$$y_p+y_n=te^{at}+y(0)e^{at}$$

Why? Because we can actually apply L'Hopital's rule here when $s\rightarrow a$:

\begin{align*}&\lim_{s\rightarrow a}\frac{\frac{d}{ds}(e^{st}-e^{at})}{\frac{d}{ds}(s-a)}\\ =&\lim_{s\rightarrow a}\frac{te^{st}}{1}\\ =&\lim_{s\rightarrow a}te^{st}\\ =&te^{at} \end{align*}