Variation of Parameters — Question 5

Question 5 – Solution

Strategy. Separate parameter divergence, solution limits and the regularity needed for initial data.

Step 1: Integrate from the regular point. The triangular parameter system for $y'''=g$ gives $(u_0',u_1',u_2')=(x^2g/2,-xg,g)$. Here $g=1/x$, so $$u_0=(x^2-1)/4,\qquad u_1=1-x,\qquad u_2=\ln x.$$ Combining $u_0+xu_1+x^2u_2/2$ yields $$\boxed{y=\frac{x^2\ln x}{2}-\frac{3x^2}{4}+x-\frac 14\quad(x>0).}$$

Step 2: Check derivatives and endpoint behavior. Direct differentiation gives $$y'=x\ln x-x+1,\qquad y''=\ln x,\qquad y'''=1/x.$$ These formulas and $\ln 1=0$ verify all three data at $1$. Since $x\ln x$ and $x^2\ln x$ tend to zero at the positive endpoint, $$\boxed{\lim_{x\to 0^+}y=-1/4,\qquad \lim_{x\to 0^+}y'=1,\qquad \lim_{x\to 0^+}y''=-\infty.}$$

Step 3: Classify the extension. Defining $y(0)=-1/4$ gives right derivative $1$, because $(y(x)+1/4)/x\to 1$. This derivative is continuous from the right, so the solution extends as a $C^1$ function to $[0,\infty)$; it may also be continued to the left with matching value and slope. No $C^2$ extension is possible because $y''$ diverges. Although $u_2=\ln x$ diverges, it is multiplied by $x^2/2$ in $y$, whose contribution tends to zero.

Step 4: Reject the singular base point. For fixed $x>0$ and $0<t<x/2$, $(x-t)^2/(2t)\ge x^2/(8t)$. The proposed improper integral therefore diverges to $+\infty$. Its lower limit cannot simply be moved to zero. The equation is singular there, and even the solution’s second derivative lacks finite initial data. Two finite endpoint limits do not supply a regular third-order initial state.