Variation of Parameters — Question 3

Question 3 – Solution

Strategy. Normalize the Euler operator before solving the derivative equations, and retain its positive-domain restriction.

Step 1: Verify the basis and normalization. For $y=x^m$, the left side is $(m-1)(m-2)(m-3)x^m$. Thus the proposed functions are homogeneous. Their Wronskian is $2x^3\ne 0$ for $x>0$. Division by $x^3$ gives normalized forcing $x\ln x$, so $$\begin{pmatrix}x&x^2&x^3\\1&2x&3x^2\\0&2&6x\end{pmatrix} \begin{pmatrix}u_1'\\u_2'\\u_3'\end{pmatrix} =\begin{pmatrix}0\\0\\x\ln x\end{pmatrix}.$$

Step 2: Integrate the solved parameter derivatives. Elimination gives $u_1'=x^2\ln x/2$, $u_2'=-x\ln x$, $u_3'=\ln x/2$. Convenient antiderivatives are $$u_1=\frac{x^3\ln x}{6}-\frac{x^3}{18},\quad u_2=-\frac{x^2\ln x}{2}+\frac{x^2}{4},\quad u_3=\frac{x\ln x}{2}-\frac{x}{2}.$$ Combining $xu_1+x^2u_2+x^3u_3$ gives $$\boxed{y_p=x^4\left(\frac{\ln x}{6}-\frac{11}{36}\right),\qquad y=y_p+c_1x+c_2x^2+c_3x^3.}$$ Integration constants are already represented by the homogeneous terms.

Step 3: Fit and check the initial vector. The particular solution has vector $(-11/36,-19/18,-5/2)$ at $1$. The three linear equations for its cancellation give $$\boxed{y=\frac{x}{18}-\frac{x^2}{4}+\frac{x^3}{2} +x^4\left(\frac{\ln x}{6}-\frac{11}{36}\right).}$$ The added terms have vector $(11/36,19/18,5/2)$, verifying the data. For a direct residual check, let $\theta=xD$. The operator is $(\theta-1)(\theta-2)(\theta-3)$; on $x^4(a\ln x+b)$ it gives $x^4(6a\ln x+11a+6b)$. With $a=1/6,b=-11/36$, this is exactly $x^4\ln x$.

Step 4: Respect the regular interval. The normalized coefficients, forcing and nonsingular derivative matrix are valid throughout $(0,\infty)$, the largest such interval containing $1$. At zero the leading coefficient and Wronskian vanish and $\ln x$ is undefined. Finite limits of selected terms do not make that parameter system invertible or extend the theorem across zero.