Systems of Differential Equations — Question 4

Question 4 – Solution

Strategy. Differentiate one row, substitute the other, and retain the algebraic formula that recovers the missing coordinate.

Step 1: Eliminate $y$ while retaining initial information. The first row gives $y=x'-x$. Differentiating it and using the second gives $x''=x'+2x-y+t=3x+t$. Thus $$\boxed{x''-3x=t,\qquad x(0)=1,\quad x'(0)=3.}$$ The initial slope is $x(0)+y(0)$, not the initial value of $y$ alone.

Step 2: Reconstruct and verify. Given a twice differentiable solution of this scalar IVP, define $y=x'-x$. Then $x'=x+y$ identically, and $$y'=x''-x'=3x+t-x'=2x-(x'-x)+t=2x-y+t.$$ Also $y(0)=3-1=2$. This proves the converse as well as the forward implication: no extra scalar solutions remain after reconstruction and data matching.

Step 3: Eliminate in the other direction. The second row gives $x=(y'+y-t)/2$. Differentiating that row yields $$y''=2x'-y'+1=2x+2y-y'+1=3y+1-t.$$ Therefore the equivalent alternative is $$\boxed{y''-3y=1-t,\quad y(0)=2,\quad y'(0)=0, \qquad x=\frac{y'+y-t}{2}.}$$ Indeed $x'=(y''+y'-1)/2=(3y+y'-t)/2=x+y$; the reconstruction also gives $x(0)=1$ and the second row directly.

Step 4: Diagnose the mismatched data. If $x(0)=1$ and $x'(0)=2$, reconstruction forces $y(0)=2-1=1$. The calculation solves a different original IVP, with initial state $(1,1)$. Insisting simultaneously on $y(0)=2$ would require $x'(0)=1+2=3$, contradicting the imposed slope. Correct differential equations alone do not make two initial-value problems equivalent.