Undetermined Coefficients — Question 2

Question 2 – Solution

Strategy. Shift the exponential factor out of the operator; the repeated characteristic root reveals the necessary power of $t$.

Step 1: Identify the repeated root. The characteristic polynomial is $(r-2)^2$. Product differentiation gives $$(e^{2t}F)''-4(e^{2t}F)'+4e^{2t}F=e^{2t}F''.$$ Thus the transformed equation is $F''=6t+2$.

Step 2: Reject incomplete shifts. The first trial has $F=At+B$, so $F''=0$. The second has $F=At^2+Bt$, so $F''=2A$, which cannot supply the linear term $6t$. Multiplying the usual degree-one trial by only one power of $t$ is still insufficient at a double root.

Step 3: Use the correct trial. With $F=At^3+Bt^2$, coefficient matching gives $6At+2B=6t+2$, hence $A=B=1$. The complete solution is $$y=e^{2t}(C+Dt+t^2+t^3).$$ The initial value sets $C=1$, and the slope at zero is $2C+D$, so $D=-2$. Therefore $$\boxed{y=e^{2t}(1-2t+t^2+t^3).}$$

Step 4: Account for every constant. Constant and linear terms in $F$ have zero second derivative and correspond exactly to $e^{2t}$ and $te^{2t}$, the homogeneous modes. Leaving them out of the particular trial avoids redundant coefficients; the constants $C,D$ restore them in the general solution. For the final polynomial, $F''=2+6t$, so the equation holds. Also $F(0)=1$ and $2F(0)+F'(0)=2-2=0$, verifying both data.