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An integrating factor is $\mu (x)=\cosh x$, since ${\mu }^{\prime }/\mu =\tanh x$. ThusFor negative $x$ the integral vanishes; for positive $x$ it is $\sinh x$. ConsequentlyBoth 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,so its unique maximum is $y=\sqrt{2}$ at $x=\mathrm{arsinh}1$, after which it decreases to the horizontal asymptote $y=1$ from above.