Solve the differential equation y'' + y = 12.

Step-by-step solution

1. 1

   Split the problem in two. For a linear equation with constant coefficients, the general solution is $y=y_h+y_p$: the general solution of the homogeneous equation plus any one particular solution of the full equation. Both parts are needed — omitting $y_p$ or dropping the arbitrary constants both give wrong answers.

2. 2

   Solve the homogeneous equation. Setting the right side to zero and trying $y=e^{rx}$ gives the characteristic equation

   $r^2+1=0\qquad\Longrightarrow\qquad r=\pm i$

   A purely imaginary conjugate pair $\pm i\beta$ corresponds to oscillation with no decay, so

   $y_h=C_1\cos x+C_2\sin x$

3. 3

   Choose a trial particular solution. The right side is the constant $12$, so try a constant, $y_p=A$. Since a constant has zero derivatives, $y_p''=0$ and substitution gives

   $0+A=12\qquad\Longrightarrow\qquad A=12$

   so $y_p=12$. (This works because $r=0$ is not a root of the characteristic equation — otherwise the trial would have needed an extra factor of $x$.)

4. 4

   Assemble the general solution.

   $y(x)=C_1\cos x+C_2\sin x+12$

5. 5

   Interpret the answer. The solution oscillates with period $2\pi$ and amplitude $\sqrt{C_1^2+C_2^2}$ about the equilibrium level $y=12$. That $12$ is the steady state the constant forcing imposes: the homogeneous part supplies the wobble, the particular part sets the centre line.

6. 6

   Verify by substitution. With $y=C_1\cos x+C_2\sin x+12$: $y''=-C_1\cos x-C_2\sin x$, so

   $y''+y=\left(-C_1\cos x-C_2\sin x\right)+\left(C_1\cos x+C_2\sin x+12\right)=12\ \checkmark$

   Numerically with $C_1=1.3$, $C_2=-0.7$, a central second difference of step $10^{-4}$ gives $y''+y=12.000$ at $x=0.2,\ 1.5,\ -1.1$ ✓.