Substitutions — Question 1

Question 1 – Solution

Strategy. The ratio substitution removes the repeated $y/x$, but differentiating $y=xv(x)$ requires both product-rule terms.

Step 1: Transform the equation. Since $y'=v+xv'$, substitution gives $$v+xv'=v+\sqrt{1+v^2},\qquad \frac{dv}{\sqrt{1+v^2}}=\frac{dx}{x}.$$ The initial value is $v(1)=0$. An antiderivative on the left is $\ln(v+\sqrt{1+v^2})$, whose logarithm has a strictly positive argument for every real $v$. Hence $$\ln(v+\sqrt{1+v^2})=\ln x.$$

Step 2: Recover the original unknown. Exponentiating gives $v+\sqrt{1+v^2}=x$. Squaring $\sqrt{1+v^2}=x-v$ and simplifying gives $$v=\frac{x^2-1}{2x},\qquad \boxed{y=\frac{x^2-1}{2}}.$$ This step is valid: for the recovered expression, $x-v=(x^2+1)/(2x)>0$ on $x>0$, so the unsquared relation also holds.

Step 3: Verify the branch and interval. For this $y$, $$\sqrt{1+(y/x)^2}=\frac{x^2+1}{2x},\qquad \frac yx+\sqrt{1+(y/x)^2}=x=y'.$$ Also $y(1)=0$. The formula satisfies the original equation for every $x>0$, so its maximal interval there is $\boxed{(0,\infty)}$. The polynomial has a value at $0$, but the given equation does not; an algebraic extension alone does not extend this solution through the excluded point.

Step 4: Diagnose the missing derivative term. The omitted term is $v$. Writing only $xv'$ would produce $xv'=v+\sqrt{1+v^2}$ instead of $xv'=\sqrt{1+v^2}$. The two equations describe different ratio functions, so the product rule is essential to the reduction.