Undetermined Coefficients — Question 9

Question 9 – Solution

Strategy. Use acceleration and its derivative to identify the forcing, then use value and slope to select the trajectory.

Step 1: Identify the forcing parameters. At zero, the equation gives $1=A$. Differentiating once gives $y^{(3)}-3y''+2y'=3Ae^{3t}+B$, so at zero $-3=3A+B$. Therefore $$\boxed{A=1,\qquad B=-6.}$$

Step 2: Match a particular solution. Neither $3$ nor $0$ is a root of $(r-1)(r-2)$, so use $y_p=ce^{3t}+dt+k$. The residual is $2ce^{3t}+2dt+(2k-3d)$. Matching $e^{3t}-6t$ gives $$\boxed{y_p=\tfrac 12e^{3t}-3t-\tfrac 92.}$$

Step 3: Fit and verify. Write $y=y_p+Ce^t+De^{2t}$. Since $y_p(0)=-4$ and $y_p'(0)=-3/2$, the zero data give $C+D=4$, $C+2D=3/2$. Hence $$\boxed{y=\tfrac 12e^{3t}+\tfrac{13}{2}e^t-\tfrac 52e^{2t}-3t-\tfrac 92.}$$ At zero its value and slope are $1/2+13/2-5/2-9/2=0$ and $3/2+13/2-5-3=0$. Its second and third derivatives there are $9/2+13/2-10=1$ and $27/2+13/2-20=0$. The coefficient matching already verifies the forcing everywhere.

Step 4: Interpret the information count. Before calibration there are two unknown forcing parameters and two trajectory constants. The four measurements determine those four quantities here. Once the forcing is fixed, only the value and slope are freely assignable: the equation and its derivative determine the higher derivatives. Arbitrary additional acceleration measurements would be compatibility tests, not extra solution freedoms.