Variation of Parameters — Question 5

Question 5 – Solution

Strategy. Use variation of parameters where the leading coefficient is nonzero, then examine the original equation separately at its singular point.

Step 1: Normalize and integrate. The normalized forcing is $g=t$, and $W=(t+1)e^t-e^t=te^t$. Hence $$u_1'=-1,\qquad u_2'=(t+1)e^{-t}.$$ Choose $u_1=-t$, $u_2=-(t+2)e^{-t}$, giving $$y_p=-t(t+1)-(t+2)=-t^2-2t-2.$$ Thus $y=C(t+1)+De^t-t^2-2t-2$ on $(0,\infty)$.

Step 2: Fit and verify. The conditions are $2C+De=5$ and $C+De=4$, so $C=1$, $D=3/e$ and $$\boxed{y=3e^{t-1}-t^2-t-1.}$$ Its derivatives are $y'=3e^{t-1}-2t-1$, $y''=3e^{t-1}-2$. Substitution cancels the exponential terms; the polynomial terms yield $$-2t+(t+1)(2t+1)-t^2-t-1=t^2.$$ Both data at one vanish. The normalized problem is regular on $(0,\infty)$, where its solution is unique.

Step 3: Continue the selected function. The formula is analytic on $\mathbb R$, and the same residual identity holds there. At zero, $$y(0)=y'(0)=3/e-1.$$ The undivided equation reduces to $-y'(0)+y(0)=0$, which these values satisfy. This extension does not require dividing by $t$ at zero.

Step 4: Demonstrate nonuniqueness at zero. The nonzero homogeneous function $h=e^t-(t+1)$ is analytic, with $h(0)=h'(0)=0$. For every real $k$, $y+kh$ solves the undivided equation and has the same two data at zero. These are distinct functions since $h''(0)=1$. The Wronskian vanishes at zero precisely where the leading coefficient vanishes; the regular normalized uniqueness theorem does not apply there.