Variation of Parameters — Question 2

Question 2 – Solution

Strategy. Use parameter integrals to isolate the growing homogeneous mode and the initial curvature needed to cancel it.

Step 1: Solve the derivative system. The functions solve $y'''-y'=0$ and have Wronskian $2$. With $g=e^{-x^2}$, the equations are $$u_0'+e^xu_+'+e^{-x}u_-'=0,\quad e^xu_+'-e^{-x}u_-'=0,\quad e^xu_+'+e^{-x}u_-'=g.$$ Thus $u_0'=-g$, $u_+'=e^{-x}g/2$, and $u_-'=e^xg/2$.

Step 2: Construct the zero-data response. Integrating from zero and combining gives $$\boxed{y_0(x)=\int_0^x[\cosh(x-t)-1]e^{-t^2}\,dt.}$$ For $G(s)=\cosh s-1$, we have $G(0)=G'(0)=0$, $G''(0)=1$, and $G'''-G'=0$. Leibniz differentiation therefore gives $y_0'''-y_0'=g$ and the three zero initial data.

Step 3: Identify the growing coefficient. Define $$A(x)=\int_0^xe^{-t^2-t}\,dt,\quad B(x)=\int_0^xe^{-t^2+t}\,dt,\quad C(x)=\int_0^xe^{-t^2}\,dt.$$ Their limits $A_\infty,B_\infty,C_\infty$ are finite: for $t\ge 2$, each integrand is at most $e^{-t^2/2}\le e^{-t}$. All are positive. Since $y_0=e^xA(x)/2+e^{-x}B(x)/2-C(x)$, $$\boxed{\lim_{x\to\infty}e^{-x}y_0(x)=A_\infty/2>0.}$$

Step 4: Cancel growth without losing the first two data. Any other solution with the same value and slope is $y=y_0+k(\cosh x-1)$, where $k=y''(0)$. Boundedness requires $k=-A_\infty$, and no other choice cancels the growing coefficient. For this choice, $$y=-\tfrac 12e^x(A_\infty-A(x)) +\tfrac 12e^{-x}(B(x)-A_\infty)+A_\infty-C(x).$$ The first term tends to zero: for $x>0$, $e^x\int_x^\infty e^{-t^2-t}\,dt\le e^{-x^2}$. The second also tends to zero. Therefore the choice is sufficient, and $$\boxed{y''(0)=-A_\infty,\qquad \lim_{x\to\infty}y=A_\infty-C_\infty.}$$