# How to calculate step response for $y''(t) - y(t) = x'(t) - x(t)$ in time domain?

How to calculate step response for $y''(t) - y(t) = x'(t) - x(t)$ in time domain? So, without Laplace or Fourier Transforms.

This is what I tried:

The Homogeneous solution of the differential equation would be: $y_{homogeneous} = A*e^t + B*e^{-t}$.

u(t) is the unit step function. When I try to fill in the stepfunction for x: x(t)=u(t). So x'(t) = u'(t) = $\delta(t)$

And I try a step response $y= s= y_{homogeneous}*u(t) = (A*e^t + B*e^{-t})*u(t)$

$\frac{\partial ^2 ( \left(A e^t+B e^{-t} \right)u(t) )}{\partial t^2}-\left(A e^t+B e^{-t}\right) u(t)=\delta(t)-u(t)$

So, I will get:

$2 (A-B) \delta(t)-A \delta(t)+A \left( \delta ' (t) \right )+B \delta(t)+B \left( \delta ' (t) \right)=\delta(t)-u(t)$

But there's no u(t) at the left hand side.

Multiply with $e^{-t}$. Then the ODE can be rewritten as $$(e^{-t}y')'+(e^{-t}y)'=(e^{-t}x)'$$ which integrates to $$y'+y=x+Ce^t$$ This removes the need to introduce distributions. Now integrate to the left and to the right of the jump and adjust the two integration constants on both sides in such a way that the combined function is continuous.