Variation of Parameters — Question 4

Question 4 – Solution

Strategy. Track the constant coordinate transformation and isolate the effect of the integration base point.

Step 1: Identify the basis transformation. The coefficient columns give $$M=\begin{pmatrix}1&0&1\\1&1&0\\0&1&1\end{pmatrix},\qquad \det M=2.$$ All six displayed functions are quadratic or lower and solve $y'''=0$. The Wronskian of $f$ is $1$, and that of $\widetilde f$ is $\det M=2$, so both triples are fundamental.

Step 2: Transform the coordinates. The standard derivative system gives $u'=g(x)(x^2/2,-x,1)^T$. Since the derivative matrices satisfy $\widetilde F=FM$, their parameter equations imply $Mv'=u'$, hence $$\boxed{v'=g(x)\begin{pmatrix}x^2/4-x/2-1/2\\-x^2/4-x/2+1/2\\x^2/4+x/2+1/2\end{pmatrix}.}$$ With $u(0)=v(0)=0$, integration gives $u=Mv$ at every $x$. Thus $fu=fMv=\widetilde f\,v$ exactly, not merely up to a homogeneous term. Both equal $$Y_0(x)=\frac 12\int_0^x(x-t)^2g(t)\,dt.$$

Step 3: Change the integration base point. The construction based at $b$ gives $Y_b(x)=\frac 12\int_b^x(x-t)^2g(t)\,dt$. Therefore $$\boxed{Y_b-Y_0=-\frac{x^2}{2}\int_0^b g(t)\,dt +x\int_0^b t g(t)\,dt-\frac 12\int_0^b t^2g(t)\,dt.}$$ The three integrals are constants with respect to $x$, even when $b<0$. The difference is a quadratic polynomial and hence homogeneous.

Step 4: Count genuine solution freedom. Once a basis and lower limit are fixed, the invertible derivative matrix uniquely fixes the parameter derivatives, and the specified parameter values fix their integrals. Allowing arbitrary integration constants adds a constant linear combination of the three homogeneous basis functions. Adding a separate homogeneous triple afterward repeats that same freedom rather than creating six independent constants. Particular solutions need not coincide, but their differences are homogeneous; the full solution space of $y'''=g$ is one particular solution plus the three-dimensional homogeneous space.