= Solution
An <integrating factor> is $\mu(x)=\cosh x$, since $\mu'/\mu=\tanh x$. Thus
$$
(\cosh x\,y)'=\cosh x\,H(x),\qquad
\cosh x\,y(x)=1+\int_0^x\cosh t\,H(t)\,dt.
$$
For negative $x$ the integral vanishes; for positive $x$ it is $\sinh x$. Consequently
$$
\boxed{y(x)=\begin{cases}
\operatorname{sech}x,&x\le0,\\
\operatorname{sech}x+\tanh x,&x\ge0.
\end{cases}}
$$
Both branches give $y(0)=1$. This is <Heaviside forcing in a first-order integrating-factor equation>: the solution is continuous and piecewise differentiable, while its one-sided derivatives at zero are $0$ and $1$. The <Heaviside step function>'s value at zero does not affect the integral solution. The equation holds classically on the two open half-lines and in the weak sense across zero.
The left branch rises from zero at $-\infty$ to one. On the right,
$$
y'(x)=\frac{1-\sinh x}{\cosh^2x},
$$
so its unique maximum is \b[$y=\sqrt2$ at $x=\operatorname{arsinh}1$], after which it decreases to the horizontal asymptote $y=1$ from above.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2001/ia/paper-2-step-solution.png]
{title=Continuous step-forced solution with its derivative jump, maximum and asymptotes}
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