This is sometimes called the Heaviside expansion theorem.(a) Use this theorem to write the solution of \( y'' + 3y' + 2y = f(t) \), \( y(0) = y'(0) = 0 \).(b) Give an explicit evaluation of the solution in (a) for the cases \( f(t) = e^{3t} \) and \( f(t) = t \).(c) Find the solutions in (b) by using the superposition principle (13). **5.** When the polynomial \( z(p) \) has distinct real zeros \( a \) and \( b \), so that\[\frac{1}{z(p)} = \frac{1}{(p-a)(p-b)} = \frac{A}{p-a} + \frac{B}{p-b}\]for suitable constants \( A \) and \( B \), then\[h(t) = Ae^{at} + Be^{bt}\]and (15) takes the form\[y(t) = \int_{0}^{t} f(\tau) \left[ A e^{a(t-\tau)} + B e^{b(t-\tau)} \right] d\tau.\]

Given (a) y" + 3y' + 2y = f(t), y(0)=y'(O)=0. (b)f(t) = e3t and f(t) = t.(a) Taking laplace transform for the given equation L(y'')= s2γ(s)-sy(0)-y'(0) =s2γ(s)-0-0 =s2γ(s)L(y')= sγ(s)-y(0)=sγ(s)L(f(t))=F(s)then,F(s)=s2γ(s)+3sγ(s)+2γ(s) =γ(s)(s2+3s+2)F(s)(s2+3s+2)=γ(s)F(s)(s2+2s+s+2)=γ(s)F(s)s(s+2)+(s+2)=γ(s)F(s)(s+2)(s+1)=γ(s)so,y(t)=L-1(γ(s)) =L-1F(s)(s+2)(s+1) =L-1F(s)1s+1-1s+2Now based on the F(s),the value will depend (b) For the given equation f(t)=e3t Then taking the laplace transform for the given equation Lf(t)=F(s)=1s-3y(t)=L-1F(s1s+1-1s+2 =L-11(s-3)(s+1)-1(s+2)(s-3) =L-1141(s-3)(s+1)-151(s+2)(s-3) =14e3t-14e-t -15e3t +15e-2t Similarly for f(t)=t, L(f(t))=L(t)=1s2y(t)=L-11s21s+1-1s+2 =t +e-t-1-t2+e-2t4-14 y(t) =t2+e-t-e-2t4-34