Past exam of the algebra 1 and differential equations 1 course of the University of Oxford / 2014 / 4 / b

Dividing by $x^2$, which is allowed precisely away from zero, givesThis function is smooth on each of the two half-planes $x>0$ and $x<0$, and is affine in the dependent variable $y$. Its derivative with respect to $y$ is $(1-2x)/x^2$.

For fixed $a\ne 0$ and $B>0$, the rectangleis compact and stays away from $x=0$. The function $f$ in part (i) is continuous there. The extreme value theorem makes $|f|$ attain a finite maximum $M_0$ on $K$; choosing $\boxed{M=M_0+1}$ gives the requested strict bound.

The absolute values make this a positive-$x$ rectangle even when $a<0$. Its boundedness is still true, but for a negative initial point the Picard-Lindelöf theorem must instead be applied to a rectangle centered at that negative point, as in part (iii).

Take any $a\ne 0$ and any $b\in \mathbb{R}$. Choose $0<h_0<|a|$, so $[a-h_0,a+h_0]$ avoids zero. Onboth $f$ and $f_y=(1-2x)/x^2$ are continuous. They are bounded by compactness, and the bound on $f_y$ gives the uniform Lipschitz continuity in $y$ required by the Picard-Lindelöf theorem.

If $M$ is a bound for $|f|$, shrink $h$ so $0<h<h_0$ and $hM<1$. Then the solution has $|y(x)-b|\le hM<1$, which supplies even strict inequalities at both edges of the requested vertical strip.

At $a=0$, neither $f$ nor the original exponential expression is defined. Thus the theorem applies exactly for $a\ne 0$, with no restriction on $b$. Failure of the hypotheses alone would not prove nonexistence; the explicit solutions below settle that issue.