Q$.$ For the given circuit, if $v_g(t)=50tu(t),$ find $v_o(t)."$

Answer:

$4.$

$V_g(s)=\frac{50}{s^2}.$

$V_o(s)=1000(s+5000)**\frac{\frac{50}{s^2}}{s^2+6000(s)+25{10}^6}$

$=\frac{K_1}{s+3000-j4000}+\frac{K_1^{**}}{s+3000+j4000}+\frac{K_2}{s^2}+\frac{K_3}{s}$

$-v_o(t)=[10\sqrt[\phantom{2}]{?}5{10}^{-4}{e}^{-3000t}cos(4000t+79\mathrm{.}7)+10t-4{10}^{-4}]u(t)[V]$