true">y={12(sint−cost+e−t) if 0<t<π12(3sint−cost+e−t)+e−π2(e−(t−π)+cost+sint) if t>π_.

The given data is L=1H,C=1F,I(0)=0 and v(t)={1−e−t if 0<t<π0 if t>π.

The equation for LC-circuit is 1C∫idt+Ldidt=v(t).

Substitute the value of L=1H,C=1F,I(0)=0 and v(t)={1−e−t if 0<t<π0 if t>π in 1C∫idt+Ldidt=v(t).

11∫idt+1didt=(1−e−t)(1−u(t−π))∫idt+didt=1−e−t−(1−e−t)u(t−π)

Using the Laplace transformation.

L[∫idt]+L[didt]=L[1]−L[e−t]−L[(1−e−tu(t−π))]I[s]s+(sI[s]−I(0))=1s−1s+1−e−πsL(1−e−(t+π))I(s)=1(1+s2)(1−e−πs)−s(s+1)(1+s2)(1−e−π(s+1))

The Laplace inverse of the above equation is as follows:

L−1[I(s)]=12L−1[11+s2]−L−1[e−πs1+s2]+12L−1[1s+1]−12L−1[s1+s2]+e−π2L−1[e−πs1s+1]−L−1[e−πss1+s2]+L−1[e−πs11+s2]

Using L−1[F(s)e−as]=f(t−a)u(t−a) in the above equation.

i(t)=12sint−sin(t−π)u(t−π)+12e−t−12cost+e−π2(e−(t−π)u(t−π)−cos(t−π)u(t−π)+sin(t−π)u(t−π))=12(sint−cost+e−t)+u(t−π)[sint+e−π2(e−(t−π)−cost+sint)]

If 0<t<π then u(t−π)=0 and hence y=12(sint−cost+e−t).

If t>1 then u(t−4)=1 and hence y=12(3sint−cost+e−t)+e−π2(e−(t−π)+cost+sint).

The graph of the current is as shown below:

ADV.ENG.MATH (LL) W/WILEYPLUS BUNDLE, Chapter 6, Problem 42RQ

Hence, the solution of the given system is y={12(sint−cost+e−t) if 0<t<π12(3sint−cost+e−t)+e−π2(e−(t−π)+cost+sint) if t>π_.

Chapter 6 Solutions