Fundamental Sets of Solutions — Question 1

Question 1 – Solution

Strategy. A fundamental pair must supply every value-and-slope pair, so test the two-by-two initial-data system.

Step 1: Prove the criterion. If $D(t_0)\ne 0$, the equations $$A u(t_0)+Bv(t_0)=a,\qquad A u'(t_0)+Bv'(t_0)=b$$ have unique constants $A,B$ for every $a,b$. The resulting combination solves the equation; uniqueness makes it equal to every solution with those data. Conversely, if $D(t_0)=0$, some nonzero constant pair $(A,B)$ gives zero value and slope. By uniqueness, $Au+Bv$ is identically zero, contradicting unique representation. Thus the criterion is necessary and sufficient.

Step 2: Verify the candidates. Their derivatives are $$u'=e^{-t}(-\cos 2t-2\sin 2t),\quad u''=e^{-t}(-3\cos 2t+4\sin 2t),$$ $$v'=e^{-t}(-\sin 2t+2\cos 2t),\quad v''=e^{-t}(-3\sin 2t-4\cos 2t).$$ Both residuals vanish. At zero their data are $(1,-1)$ and $(0,2)$, so $D(0)=2\ne 0$ and the pair is fundamental on $\mathbb R$.

Step 3: Normalize the data. The required choices are $$\boxed{\phi=u+\tfrac 12v,\qquad\psi=\tfrac 12v.}$$ Their data are $(1,0)$ and $(0,1)$ respectively, so they too form a fundamental set.

Step 4: Fit the solution. The normalized coefficients equal the data: $$\boxed{y=2\phi-3\psi=e^{-t}(2\cos 2t-\tfrac 12\sin 2t).}$$ At zero its value is $2$; its slope is $-2-1=-3$. It satisfies the equation by linearity and is unique on $\mathbb R$.