Fundamental Sets of Solutions — Question 3

Question 3

For $y''-3y'+2y=0$ on $\mathbb R$, let $$f=e^t,\quad g=e^{2t},\quad h=f+g,\quad k=2f-g.$$ A student writes $y=Af+Bg+Ch+Dk$ and claims this gives four independent solution freedoms.

Tasks

1. Verify that all four functions solve the equation and that $f,g$ form a fundamental set.

2. Determine which of the six unordered pairs chosen from $f,g,h,k$ are fundamental. Justify every pair.

3. Find every quadruple $(A,B,C,D)$ that represents the zero function, and explain the error in the student’s claim.

4. Find all quadruples representing the solution with $y(0)=1$, $y\prime(0)=3$. Give its unique representation using only $h,k$.

Question 3 – Solution

Strategy. Express every candidate in the already verified pair $f,g$, then count effective coefficients.

Step 1: Verify the underlying pair. Substitution of $e^{rt}$ gives residual $(r^2-3r+2)e^{rt}$, which vanishes for $r=1,2$. Linearity verifies $h,k$. The data determinant for $f,g$ at zero is $2-1=1$, so this pair is fundamental.

Step 2: Check all six pairs. The coefficient columns relative to $f,g$ are $(1,0)$, $(0,1)$, $(1,1)$ and $(2,-1)$. Their determinants, in the listed order, are $$\begin{array}{c|rrrrrr} \text{pair}&(f,g)&(f,h)&(f,k)&(g,h)&(g,k)&(h,k)\\\hline \text{determinant}&1&1&-1&-1&-2&-3 \end{array}$$ Every determinant is nonzero, so every pair is an invertible change of the fundamental pair $f,g$ and is itself fundamental.

Step 3: Identify redundancy. The proposed expression is $$y=(A+C+2D)f+(B+C-D)g.$$ It is zero exactly when $A=-C-2D$ and $B=-C+D$, where $C,D$ are arbitrary. These nontrivial constant relations show that the four functions are not independent. Only two effective coefficients are present.

Step 4: Fit the specified data. Writing $y=af+bg$ gives $a+b=1$, $a+2b=3$, hence $a=-1,b=2$. All four-coefficient representations are $$\boxed{(A,B,C,D)=(-1-C-2D,\ 2-C+D,\ C,\ D).}$$ Since $h-k=-f+2g$, the unique representation in the pair $h,k$ is $\boxed{y=h-k}$. Its data are $-1+2=1$ and $-1+4=3$.