Stuck while using Laplace transform to solve delayed DE.y′(t)=y(t−1)y[−1,0]=y0where y0 is a constant by using the method of steps to gety(t)=∑y0tnn!+c on [n,n+1].

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sorrisi7yny · Accepted Answer

Step 1I have to make a necessary assumption in order to solve your differential equation using Laplace transform, that being y(t) is a causal function, meaning we don&#039;t care about its history&amp;nbsp;(y=0&amp;nbsp;if&amp;nbsp;t&amp;lt;0). That being said, if we shift the function 1 unit to the right, it would still be zero for&amp;nbsp;t&amp;lt;1, and the equation turns to:y′(t)=y(t−1)H(t−1)where H(t) is the Heaviside or unit step function.Taking Laplace transforms:sY(s)−y0=Y(s)e−ssY(s)−Y(s)e−s=y0(s−e−s)Y(s)=y0Y(s)=y0s−e−s=y0s;11−e−ss=y0s∑n=0+∞(e−ss)n=y0∑n=0+∞e−nssn+1if&amp;nbsp;e-ss&amp;lt;1Step 2Now there&#039;s another assumption to make, and that is the inverse Laplace of a series is the series of the inverse Laplace (operator and series commute). It&#039;s not obvious, hard to prove, but widely used to solve for example, RLC circuits with periodic voltage sources.Then,&amp;nbsp;L-1e-nssn+1=L-11n!n!sn+1e-ns=1n!L-1n!sn+1t-nH(t-n)=(t-n)nn!H(t-n)Resulting iny(t)=y0∑n=0+∞(t-n)nn!H(t-n)=y0H(t)+(t-1)H(t-1)+12(t-2)2H(t-2)+...Step 3Where you can compute by hand or with GeoGebra that its derivative is a shifted version of itself. Here we&#039;ve found a solution of the delayed differential equation restricting the solution to be causal, a requirement to use Laplace transforms in the first place.

Stuck while using Laplace transform to solve delayed

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