ENGR 213 Practice: Forced Second-Order Equations

order comes from the highest derivative. A linear equation contains the unknown function and its derivatives only to the first power, without multiplying them together. A coefficient can depend on $t$ while the equation remains linear.

A is first-order and linear. Use a first-order method, such as an integrating factor. B is second-order and linear with constant coefficients and exponential forcing; it fits this practice.

C is second-order and linear, but the coefficient of $y'$ varies with $t$. It is not a constant-coefficient equation. D is second-order and nonlinear because of $(y')^2$. Superposition and the characteristic-polynomial recipe cannot simply be carried over to it.

the characteristic roots are $-1$ and $-2$. Compare the forcing exponent $+1$ with those roots.

The homogeneous solution is $C_1e^{-t}+C_2e^{-2t}$. The forcing exponential $e^t$ is not part of that family, so try $y_p=Ae^t$. Substitution gives $6Ae^t=e^t$, hence $A=1/6$.

For

$$\begin{aligned}y={}&C_1e^{-t}+C_2e^{-2t}\\&+\tfrac16e^t,\end{aligned}$$

the conditions give $C_1+C_2+1/6=0$ and $-C_1-2C_2+1/6=0$. Therefore

$$\boxed{\begin{aligned}y={}&-\tfrac12e^{-t}+\tfrac13e^{-2t}\\&+\tfrac16e^t.\end{aligned}}$$

Each homogeneous term contributes zero to the left-hand side. The particular term contributes $e^t$. At zero, $-1/2+1/3+1/6=0$ and $1/2-2/3+1/6=0$. No factor $t$ was needed.

determine whether $e^{-t}$ is already a homogeneous solution. Test $t$ times that exponential.

The equation is solvable; the first trial belongs to the homogeneous family. The root $-1$ is simple, so use $y_p=Ate^{-t}$. The left-hand side becomes $Ae^{-t}$, giving $A=2$.

The full solution is $C_1e^{-t}+C_2e^{-2t}+2te^{-t}$. Its conditions give

$$\begin{gathered}C_1+C_2=1,\\ -C_1-2C_2+2=0.\end{gathered}$$

Thus $C_1=0$, $C_2=1$ and

$$\boxed{y=2te^{-t}+e^{-2t}.}$$

For the first term, the left-hand side gives $2e^{-t}$; for the second, it gives zero. Also $y(0)=1$. Differentiation gives $y'=2(1-t)e^{-t}-2e^{-2t}$, so $y'(0)=0$.

factor $r^2-2r+1$. You can also substitute $y=e^t u$ to see what the left-hand side becomes.

The characteristic polynomial is $(r-1)^2$. The homogeneous family is $(C_1+C_2t)e^t$, so both suggested trials are homogeneous. The next trial is $y_p=At^2e^t$.

With $y=e^t u$, the left-hand side reduces to $e^t u''$. Taking $u=At^2$ gives $2Ae^t=6e^t$, hence $A=3$. Therefore

$$y=(C_1+C_2t+3t^2)e^t.$$

At zero, $C_1=0$. Differentiating gives $y'(0)=C_1+C_2$, so $C_2=1$:

$$\boxed{y=(t+3t^2)e^t.}$$

The same substitution verifies the equation: $u=t+3t^2$ has $u''=6$. Also $y(0)=0$ and $y'(0)=1$. Multiplying by $t$ only once would still leave the trial inside the homogeneous family.

differentiating $t^2$ twice produces a constant. Allow the particular solution to cancel it.

Use $y_p=At^2+Bt+C$. Then

$$y_p''+y_p=At^2+Bt+(C+2A).$$

Matching coefficients gives $A=1$, $B=0$ and $C=-2$. The homogeneous solution is $C_1\cos t+C_2\sin t$, so

$$y=C_1\cos t+C_2\sin t+t^2-2.$$

The initial values give $C_1=2$ and $C_2=0$:

$$\boxed{y=t^2-2+2\cos t.}$$

Now $y'=2t-2\sin t$ and $y''=2-2\cos t$, so $y''+y=t^2$ and both initial values are zero. The reduced guess $At^2$ would leave an unwanted constant $2A$ in the equation.

try $y_p=t[A\cos (3t)+B\sin (3t)]$. Substitute before solving for $A$ and $B$.

The homogeneous family is $C_1\cos (3t)+C_2\sin (3t)$. The proposed trial gives

$$\begin{aligned}y_p''+9y_p={}&-6A\sin(3t)\\&+6B\cos(3t).\end{aligned}$$

Matching the forcing gives $A=-1/6$ and $B=0$, so $y_p=-\tfrac{t}{6}\cos(3t)$. It satisfies $y_p(0)=0$ but has $y_p'(0)=-1/6$, which is not the required initial derivative.

Add the homogeneous solution. The first condition gives $C_1=0$. The second gives $3C_2-1/6=0$, so $C_2=1/18$:

$$\boxed{y=\frac1{18}\sin(3t)-\frac{t}{6}\cos(3t).}$$

Its first derivative simplifies to $y'=\tfrac{t}{2}\sin(3t)$. Then

$$y''=\frac12\sin(3t)+\frac32t\cos(3t),$$

and adding $9y$ gives $\sin (3t)$. Both conditions are zero. For $t \ge 0$, the oscillatory particular solution has the linearly growing envelope $t/6$ in this ideal undamped resonant model.