Nonhomogeneous Differential Equations — Question 3

Question 3 – Solution

Strategy. Changing the particular solution transfers a homogeneous term between the two pieces without changing the full solution.

Step 1: Verify efficiently. For any twice differentiable $F$, differentiation gives $$(Fe^{-t})''+2(Fe^{-t})'+Fe^{-t}=e^{-t}F''.$$ Both supplied polynomials have second derivative $1$, so both are particular solutions. Their difference $(3-2t)e^{-t}$ has zero residual, either by this identity or by subtracting the two forced equations.

Step 2: Match coefficients. Equality of the two representations is equivalent to $$\boxed{A=\alpha+3,\qquad B=\beta-2.}$$ This is an invertible reassignment of constants, so the two formulas describe the same complete solution family.

Step 3: Impose the data. In the first representation, $y(0)=A$ and $y'(0)=B-A$, because $p(0)=p'(0)=0$. Thus $A=2$, $B=1$, while the second representation has $\alpha=-1$, $\beta=3$. Both give $$\boxed{y=e^{-t}(2+t+\tfrac 12t^2).}$$ Its initial value is $2$. Differentiation gives $y'=-e^{-t}(1+\tfrac 12t^2)$, whose value at zero is $-1$, as required.

Step 4: Interpret the complete trajectory. The derivative follows from subtracting the full polynomial when differentiating its exponential factor: $$y'=e^{-t}[(1+t)-(2+t+\tfrac 12t^2)]=-e^{-t}(1+\tfrac 12t^2).$$ For $t\ge 0$, $y>0$ and $y'<0$; the polynomial times $e^{-t}$ tends to zero. The same trajectory admits both decompositions above, so neither piece is unique without a chosen particular solution.