get first ODE general solution using integrating factor and laplace transform

Question

$y' + ay = h(t) , y(0)= b $ is the question. I get $e^{at}$ as integrating factor, but I don't know where it uses while doing laplace transform.

Answer 1

The "associated homogeneous equation" is $y'+ y= \frac{dy}{dx}+ ay= 0$ which can be separated as $\frac{dy}{dx}= -ay$ and then as $\frac{dy}{y}= --a dx$. Integrating, $ln|y|= ax+ C$ where C is an arbirtrary constant. Taking the exponential of both sides, $y= C'e^{-ax}$ where $C'= e^C$.

Now to find a single function that satisfies the entire equation, using "variation of parameters", we look for a solution of the form $y= u(x)e^{-ax}$ (allowing the "parameter", C', to "vary").

$y'= u'(x)e^{-ax}- au(x)e^{ax}$ so $y'+ y= u'e^{ax}- au(x)e^{-x}+ au(x)e^{-ax}= u'e^{ax}= h(x)$. Then $u'e^{-ax}= h(x)$.

$u'= h(x)e^{ax}$ and we just need to integrate. Using "integration by parts", let u= h and $dv= e^{ax}dx$. $du= h'(x)dx$ and $v= \frac{1}{a}e^{ax}$. We have $\int h(x)e^{ax}dx= \frac{1}{a}h(x)e^{ax}- \frac{1}{a}\int h'(x)e^{ax}dx$.

Putting those together the general solution to y'+ ay= h is $y(x)= Ce^{-ax}+ e^{-ax}\left(\frac{1}{a}h(x)e^{ax}- \frac{1}{a}\int h(t)e^{at}dt\right)= Ce^{-ax}+ \frac{1}{a}h(x)- \frac{1}{a}e^{-ax}\int h(t)e^{at} dt$.