Now we consider second order equations of the form $\ds a\ddot y+b\dot y+cy=f(t)$, with $a$, $b$, and $c$ constant. Of course, if $a=0$ this is really a first order equation, so we assume $a\not=0$. Also, much as in exercise 20 of section 17.5, if $c=0$ we can solve the related first order equation $\ds a\dot h+bh=f(t)$, and then solve $\ds h=\dot y$ for $y$. So we will only examine examples in which $c\not=0$.

Suppose that $\ds y_1(t)$ and $\ds y_2(t)$ are solutions to $\ds a\ddot y+b\dot y+cy=f(t)$, and consider the function $\ds h=y_1-y_2$. We substitute this function into the left hand side of the differential equation and simplify: $$a(y_1-y_2)''+b(y_1-y_2)'+c(y_1-y_2)=ay_1''+by_1'+cy_1 - (ay_2''+by_2'+cy_2)=f(t)-f(t)=0.$$ So $h$ is a solution to the homogeneous equation $\ds a\ddot y+b\dot y+cy=0$. Since we know how to find all such $h$, then with just one particular solution $\ds y_2$ we can express all possible solutions $\ds y_1$, namely, $\ds y_1=h+y_2$, where now $h$ is the general solution to the homogeneous equation. Of course, this is exactly how we approached the first order linear equation.

To make use of this observation we need a method to find a single solution $y_2$. This turns out to be somewhat more difficult than the first order case, but if $f(t)$ is of a certain simple form, we can find a solution using the method of undetermined coefficients, sometimes more whimsically called the method of judicious guessing.

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 $\ds Ae^{3t}+Be^{-2t}-3t^2+t-2$. $\square$

So the "judicious guess'' is a function with the same form as $f(t)$ but with undetermined (or better, yet to be determined) coefficients. This works whenever $f(t)$ is a polynomial.

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 $\ds 11.8\cos(10t)+5\sin(10t)-9.8$. $\square$

More generally, this method can be used when a function similar to $f(t)$ has derivatives that are also similar to $f(t)$; in the examples so far, since $f(t)$ was a polynomial, so were its derivatives. The method will work if $f(t)$ has the form $p(t)e^{\alpha t}\cos(\beta t)+ q(t)e^{\alpha t}\sin(\beta t)$, where $p(t)$ and $q(t)$ are polynomials; when $\alpha=\beta=0$ this is simply $p(t)$, a polynomial. In the most general form it is not simple to describe the appropriate judicious guess; we content ourselves with some examples to illustrate the process.

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 $Ae^{-5t}+Be^{-2t}-(1/3)te^{-2t}$. $\square$

In general, if $f(t)=e^{kt}$ and $k$ is one of the roots of the characteristic equation, then we guess $Cte^{kt}$ instead of $Ce^{kt}$. If $k$ is the only root of the characteristic equation, then $Cte^{kt}$ will not work, and we must guess $Ct^2e^{kt}$.

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 $\ds Ae^{3t}+Bte^{3t}+(1/2)t^2e^{3t}$. $\square$

It is common in various physical systems to encounter an $f(t)$ of the form $\ds a\cos(\omega t)+b\sin(\omega t)$.

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. $\square$

As with the exponential form, such a simple guess may not work.

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)&-16D\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 $\ds C\cos(4t)+D\sin(4t)+(1/8)t\cos(4t)$. $\square$

In general, if $f(t)=a\cos(\omega t)+b\sin(\omega t)$, and $\pm \omega i$ are the roots of the characteristic equation, then instead of $C\cos(\omega t)+D\sin(\omega t)$ we guess $Ct\cos(\omega t)+Dt\sin(\omega t)$.

Exercises 17.6

Find the general solution to the differential equation.

Ex 17.6.1 $\ds\ddot y -10\dot y+25y=\cos t$ (answer)

Ex 17.6.2 $\ds\ddot y+2\sqrt2\dot y+2y=10$ (answer)

Ex 17.6.3 $\ds\ddot y+16y=8t^2+3t-4$ (answer)

Ex 17.6.4 $\ds\ddot y+2y=\cos(5t)+\sin(5t)$ (answer)

Ex 17.6.5 $\ds\ddot y-2\dot y+2y=e^{2t}$ (answer)

Ex 17.6.6 $\ds\ddot y-6y+13=1+2t+e^{-t}$ (answer)

Ex 17.6.7 $\ds\ddot y+\dot y-6y=e^{-3t}$ (answer)

Ex 17.6.8 $\ds\ddot y-4\dot y+3y=e^{3t}$ (answer)

Ex 17.6.9 $\ds\ddot y+16y=\cos(4t)$ (answer)

Ex 17.6.10 $\ds\ddot y +9y=3\sin(3t)$ (answer)

Ex 17.6.11 $\ds\ddot y+12\dot y+36y=6e^{-6t}$ (answer)

Ex 17.6.12 $\ds\ddot y-8\dot y+16y=-2e^{4t}$ (answer)

Ex 17.6.13 $\ds\ddot y+6\dot y+5y=4$ (answer)

Ex 17.6.14 $\ds\ddot y-\dot y-12y=t$ (answer)

Ex 17.6.15 $\ds\ddot y+5y=8\sin(2t)$ (answer)

Ex 17.6.16 $\ds\ddot y-4y=4e^{2t}$ (answer)

Solve the initial value problem.

Ex 17.6.17 $\ds\ddot y-y=3t+5$, $y(0)=0$, $\ds\dot y(0)=0$ (answer)

Ex 17.6.18 $\ds\ddot y+9y=4t$, $y(0)=0$, $\ds\dot y(0)=0$ (answer)

Ex 17.6.19 $\ds\ddot y +12\dot y +37y=10e^{-4t}$, $y(0)=4$, $\ds\dot y(0)=0$ (answer)

Ex 17.6.20 $\ds\ddot y+6\dot y+18y=\cos t-\sin t$, $y(0)=0$, $\ds\dot y(0)=2$ (answer)

Ex 17.6.21 Find the solution for the mass-spring equation $\ds\ddot y+4\dot y+29y=689\cos(2t)$. (answer)

Ex 17.6.22 Find the solution for the mass-spring equation $\ds3\ddot y+12\dot y+24y=2\sin t$. (answer)

Ex 17.6.23 Consider the differential equation $\ds m\ddot y+b\dot y+ky=\cos(\omega t)$, with $m$, $b$, and $k$ all positive and $\ds b^2< 2mk$; this equation is a model for a damped mass-spring system with external driving force $\cos(\omega t)$. Show that the steady state part of the solution has amplitude $${1\over \sqrt{(k-m\omega^2)^2+\omega^2b^2}}.$$ Show that this amplitude is largest when $\ds \omega={\sqrt{4mk-2b^2}\over 2m}$. This is the resonant frequency of the system.