Variation of Parameters — Question 1

Question 1 – Solution

Strategy. Normalize first, then combine the four parameter integrals into one positive kernel.

Step 1: Form the normalized parameter system. Put $g(x)=1/(1+x^2)$. The derivative matrix of the given basis is triangular with unit diagonal, so its Wronskian is $1$. The parameter derivatives satisfy $$\begin{pmatrix} 1&x&x^2/2&x^3/6\\0&1&x&x^2/2\\0&0&1&x\\0&0&0&1 \end{pmatrix} \begin{pmatrix}u_0'\\u_1'\\u_2'\\u_3'\end{pmatrix} =\begin{pmatrix}0\\0\\0\\g\end{pmatrix}.$$ The first three zero rows remove derivative terms introduced by varying the constants; the last row supplies the normalized forcing.

Step 2: Solve and combine. Back substitution gives $(u_0',u_1',u_2',u_3')=g(-x^3/6,x^2/2,-x,1)$. Choose each $u_j(0)=0$. In $y=\sum u_jf_j$, the combined numerator is $(x-t)^3$, yielding $$\boxed{y(x)=\frac 16\int_0^x\frac{(x-t)^3}{1+t^2}\,dt.}$$ Oriented integration also defines the solution for negative $x$.

Step 3: Verify the initial state and residual. Successive derivatives have kernels $(x-t)^2/2$, $x-t$, and $1$, all divided by $1+t^2$. The fourth derivative is $1/(1+x^2)$. The first four state components vanish at zero, and multiplication by $1+x^2$ gives the original forcing. If one incorrectly uses $g=1$, the same construction gives $x^4/24$, whose original residual is $1+x^2$, not $1$.

Step 4: Bound the positive integral. For $0\le t\le x$, $1/(1+x^2)\le 1/(1+t^2)\le 1$. Hence $$\boxed{\frac{x^4}{24(1+x^2)}\le y(x)\le\frac{x^4}{24}\quad(x\ge 0).}$$ Both inequalities are strict for $x>0$, since their integrands differ on an interval of positive length where $(x-t)^3>0$. Both become equalities at $x=0$. The figure labels the lower and upper bounds $L$ and $U$, respectively.