Nonhomogeneous Differential Equations — Question 10

Question 10 – Solution

Strategy. Prescribe a trajectory and compute its forcing directly; then use a homogeneous difference equation to analyze initial-data error.

Step 1: Fit the cubic trajectory. Write $y=c_0+c_1t+c_2t^2+c_3t^3$. The initial conditions give $c_0=c_1=0$. The final conditions give $c_2+c_3=1$ and $2c_2+3c_3=0$, so $$\boxed{y_*(t)=3t^2-2t^3.}$$ The coefficient equations have a unique solution, including among all polynomials of lower degree.

Step 2: Compute the forcing. The derivatives are $y_*'=6t-6t^2$ and $y_*''=6-12t$. Thus $$\boxed{g=y_*''+2y_*'+y_*=6-9t^2-2t^3.}$$ The derivative formulas give zero initial value and slope, and final value $3-2=1$ with slope $6-6=0$. The computed forcing verifies the equation identically.

Step 3: Locate the sign change. We have $g(0)=6$, $g(1)=-5$, and $g'=-18t-6t^2<0$ for $t>0$. Therefore there is exactly one zero. More precisely, $$g(3/4)=3/32>0,\qquad g(4/5)=-98/125<0,$$ so the zero lies strictly between $3/4$ and $4/5$. This same cubic fixes $g$ pointwise, so a nonnegative forcing cannot realize it on the whole interval. This conclusion concerns the specified cubic, not every possible trajectory.

Step 4: Propagate the initial error. The difference $h=y-y_*$ satisfies $h''+2h'+h=0$, with $h(0)=\varepsilon$, $h'(0)=0$. Hence $h=\varepsilon(1+t)e^{-t}$ and $h'=-\varepsilon te^{-t}$. At the endpoint, $$\boxed{y(1)-1=2\varepsilon/e,\qquad y'(1)=-\varepsilon/e.}$$ For nonzero $\varepsilon$ the target state is missed. The forcing is unchanged; the error evolves entirely through the homogeneous correction.