Substitutions — Question 4

Question 4 – Solution

Strategy. Replace the repeated linear combination by one variable. The extra derivative of $x$ cancels the constant term in the equation.

Step 1: Reduce to a separable IVP. For $u=x+y$, $u'=1+y'=u^2$, and $u(0)=1$. Separating on the nonzero branch gives $$-\frac 1u=x+C,\qquad C=-1,\qquad u=\frac 1{1-x}.$$ Thus $$\boxed{y=-x+\frac 1{1-x},\qquad I=(-\infty,1)}.$$ The only finite obstruction is the pole at $x=1$, where $y\to+\infty$ from the left.

Step 2: Find the minimum and verify. Differentiation gives $y'=-1+(1-x)^{-2}$. On $I$ this is negative for $x<0$, zero at $0$, and positive for $0<x<1$. Hence the unique global minimum on $I$ is $\boxed{y(0)=1}$.

Also $x+y=1/(1-x)$, so $(x+y)^2-1=(1-x)^{-2}-1=y'$, and the initial value checks directly.

Step 3: Restore the constant transformed branch. The equation $u'=u^2$ also has $u\equiv 0$. In the original variables it gives $\boxed{y=-x}$ on all of $\mathbb R$. Its derivative is $-1$ and its right-hand side is $0^2-1=-1$, so it is valid. Division by $u^2$ excludes it. A constant value of a combination of $x$ and $y$ need not represent a constant value of $y$.