Euler Equations — Question 5

Question 5 – Solution

Strategy. Remove the repeated characteristic exponential before integrating the forcing.

Step 1: Factor the logarithmic-time operator. The equation is $(D-2)^2Y=e^{2t}t$, where $D=d/dt$. Since $(D-2)(e^{2t}Z)=e^{2t}Z'$, a second application gives $$\boxed{Z''=t.}$$

Step 2: Integrate through the resonance. Two integrations yield $Z=t^3/6+C_1+C_2t$, hence $$\boxed{y=x^2\left(C_1+C_2\ln x+\frac{(\ln x)^3}{6}\right).}$$ The proposed trial has $Z=A+Bt$ and therefore $Z''=0$: it belongs entirely to the homogeneous solution space. Integrating the degree-one forcing twice produces degree three, accounting for the cubic logarithm and its factor $1/6$.

Step 3: Impose and verify the initial data. At $x=1$, $t=0$, so $y(1)=C_1$ and $y'(1)=2C_1+C_2$. Thus both constants vanish and $$\boxed{y=\frac{x^2(\ln x)^3}{6}.}$$ For this solution, $Z=t^3/6$ satisfies $Z''=t$ exactly; therefore the original left-hand side is $e^{2t}t=x^2\ln x$. The two initial values are zero.

Step 4: Inspect regularity at the singular point. Writing $t=\ln x$, $$y'=x\left(\frac{t^3}{3}+\frac{t^2}{2}\right),\qquad y''=\frac{t^3}{3}+\frac{3t^2}{2}+t.$$ As $x\to 0^+$, the power of $x$ dominates every logarithmic power, so $y\to 0$, $y'\to 0$, and $y(x)/x\to 0$. Assigning $y(0)=0$ therefore gives a $C^1$ right extension, which may be joined to zero on the left as a $C^1$ function. But $y''\to-\infty$, so no $C^2$ extension is possible.