Example 17.6.1 Solve the differential equation $\ds \ddot y-\dot y-6y=18t^2+5$. The general solution of the homogeneous equation is $\ds Ae^{3t}+Be^{-2t}$. We guess that a solution to the non-homogeneous equation might look like $f(t)$ itself, namely, a quadratic $\ds y=at^2+bt+c$. Substituting this guess into the differential equation we get $$\ddot y-\dot y-6y = 2a-(2at+b)-6(at^2+bt+c) = -6at^2+(-2a-6b)t+(2a-b-6c).$$ We want this to equal $18t^2+5$, so we need $$\eqalign{ -6a&=18\cr -2a-6b&=0\cr 2a-b-6c&=5\cr}$$ This is a system of three equations in three unknowns and is not hard to solve: $a=-3$, $b=1$, $c=-2$. Thus the general solution to the differential equation is $\ds Ae^{3t}+Be^{-2t}-3t^2+t-2$. $\square$

Example 17.6.2 Consider the initial value problem $\ds m\ddot y +ky=-mg$, $y(0)=2$, $\ds\dot y(0)=50$. The left hand side represents a mass-spring system with no damping, i.e., $b=0$. Unlike the homogeneous case, we now consider the force due to gravity, $-mg$, assuming the spring is vertical at the surface of the earth, so that $g=980$. To be specific, let us take $m=1$ and $k=100$. The general solution to the homogeneous equation is $\ds A\cos(10t)+B\sin(10t)$. For the solution to the non-homogeneous equation we guess simply a constant $y=a$, since $-mg=-980$ is a constant. Then $\ds \ddot y+100y= 100a$ so $a=-980/100=-9.8$. The desired general solution is then $\ds A\cos(10t)+B\sin(10t)-9.8$. Substituting the initial conditions we get $$\eqalign{ 2&=A-9.8\cr 50&=10B\cr}$$ so $A=11.8$ and $B=5$ and the solution is $\ds 11.8\cos(10t)+5\sin(10t)-9.8$. $\square$

Example 17.6.3 Find the general solution to $\ds\ddot y+7\dot y+10y=e^{3t}$. The characteristic equation is $r^2+7r+10=(r+5)(r+2)$, so the solution to the homogeneous equation is $Ae^{-5t}+Be^{-2t}$. For a particular solution to the inhomogeneous equation we guess $Ce^{3t}$. Substituting we get $$9Ce^{3t}+21Ce^{3t}+10Ce^{3t}=e^{3t}40C.$$ When $C=1/40$ this is equal to $f(t)=e^{3t}$, so the solution is $Ae^{-5t}+Be^{-2t}+(1/40)e^{3t}$. $\square$

Example 17.6.4 Find the general solution to $\ds\ddot y+7\dot y+10y=e^{-2t}$. Following the last example we might guess $Ce^{-2t}$, but since this is a solution to the homogeneous equation it cannot work. Instead we guess $Cte^{-2t}$. Then $$(-2Ce^{-2t}-2Ce^{-2t}+4Cte^{-2t})+7(Ce^{-2t}-2Cte^{-2t})+10Cte^{-2t} =e^{-2t}(3C).$$ Then $C=1/3$ and the solution is $Ae^{-5t}+Be^{-2t}+(1/3)te^{-2t}$. $\square$

Example 17.6.5 Find the general solution to $\ds\ddot y-6\dot y+9y=e^{3t}$. The characteristic equation is $\ds r^2-6r+9=(r-3)^2$, so the general solution to the homogeneous equation is $Ae^{3t}+Bte^{3t}$. Guessing $Ct^2e^{3t}$ for the particular solution, we get $$(9Ct^2e^{3t}+6Cte^{3t}+6Cte^{3t}+2Ce^{3t})-6(3Ct^2e^{3t}+2Cte^{3t})+9Ct^2e^{3t} =e^{3t}2C.$$ The solution is thus $\ds Ae^{3t}+Bte^{3t}+(1/2)t^2e^{3t}$. $\square$

Example 17.6.6 Find the general solution to $\ds\ddot y+6\dot y+25y=\cos(4t)$. The roots of the characteristic equation are $-3\pm 4i$, so the solution to the homogeneous equation is $\ds e^{-3t}(A\cos(4t)+B\sin(4t))$. For a particular solution, we guess $C\cos(4t)+D\sin(4t)$. Substituting as usual: $$\eqalign{ (-16C\cos(4t)&+-16D\sin(4t))+6(-4C\sin(4t)+4D\cos(4t))+25(C\cos(4t)+D\sin(4t))\cr &=(24D+9C)\cos(4t)+(-24C+9D)\sin(4t).\cr}$$ To make this equal to $\cos(4t)$ we need $$\eqalign{ 24D+9C&=1\cr 9D-24C&=0\cr}$$ which gives $C=1/73$ and $D=8/219$. The full solution is then $\ds e^{-3t}(A\cos(4t)+B\sin(4t))+(1/73)\cos(4t)+(8/219)\sin(4t)$.

The function $\ds e^{-3t}(A\cos(4t)+B\sin(4t))$ is a damped oscillation as in example 17.5.3, while $\ds(1/73)\cos(4t)+(8/219)\sin(4t)$ is a simple undamped oscillation. As $t$ increases, the sum $\ds e^{-3t}(A\cos(4t)+B\sin(4t))$ approaches zero, so the solution $$e^{-3t}(A\cos(4t)+B\sin(4t))+(1/73)\cos(4t)+(8/219)\sin(4t)$$ becomes more and more like the simple oscillation $\ds(1/73)\cos(4t)+(8/219)\sin(4t)$—notice that the initial conditions don't matter to this long term behavior. The damped portion is called the transient part of the solution, and the simple oscillation is called the steady state part of the solution. A physical example is a mass-spring system. If the only force on the mass is due to the spring, then the behavior of the system is a damped oscillation. If in addition an external force is applied to the mass, and if the force varies according to a function of the form $\ds a\cos(\omega t)+b\sin(\omega t)$, then the long term behavior will be a simple oscillation determined by the steady state part of the general solution; the initial position of the mass will not matter. $\square$

Example 17.6.7 Find the general solution to $\ds\ddot y+16y=-\sin(4t)$. The roots of the characteristic equation are $\pm4i$, so the solution to the homogeneous equation is $A\cos(4t)+B\sin(4t)$. Since both $\cos(4t)$ and $\sin(4t)$ are solutions to the homogeneous equation, $C\cos(4t)+D\sin(4t)$ is also, so it cannot be a solution to the non-homogeneous equation. Instead, we guess $Ct\cos(4t)+Dt\sin(4t)$. Then substituting: $$\eqalign{ (-16Ct\cos(4t)&-16Dt\sin(4t)+8D\cos(4t)-8C\sin(4t)))+16(Ct\cos(4t)+Dt\sin(4t))\cr &=8D\cos(4t)-8C\sin(4t).\cr}$$ Thus $C=1/8$, $D=0$, and the solution is $\ds A\cos(4t)+B\sin(4t)+(1/8)t\cos(4t)$. $\square$