Chapter 8 - Linear Differential Equations of Order n - 8.5 Oscillations of a Mechanical System - Problems - Page 540: 14

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$\frac{d^2y}{dt^2}+3\frac{dy}{dt}+2t=0$ Rewrite as: $D^2y+3Dy+2y=0$ The complementary function is $y_c(x)=c_1e^{-t}+c_2e^{-2t}$ Since $y(0)=1\\ \frac{dy}{dt}(0)=-3$ We found $c_1=-1,c_2=2$ Hence, the general solution is $y(x)=-e^{-t}+2e^{-2t}$ b) When the mass passes through the equilibrium position: $y(x)=0$ then $-e^{-t}+2e^{-2t}=0 \\ 2e^{-2t}=e^{-t} \\ 2e^{-t}=1\\ e^t=2\\ t=\ln 2$