D solve initial and boundary value problems ivps with piecewise coefficients

PiecewiseExpand[UnitStep[x] + Max[x, x ^ 2]]

Max[x, x^2] + UnitStep[x]

DSolve[y'[x] == UnitStep[x], y, x]

{{y -> Function[{x}, C[1] + Integrate[UnitStep[x], x]]}}

eqn = {y'[x] == UnitStep[x], y[0] == 1};

sol = DSolve[eqn, y, x]

{{y -> Function[{x}, 1 - Integrate[1, 0] + Integrate[UnitStep[x], x]]}}

Plot[y[x] /. sol, {x, -3, 3}]

-Graphics-

Simplify[eqn /. sol[[1]], x  > 0 || x  < 0]

{Derivative[1, Function[{x}, 1 - Integrate[1, 0] + Integrate[UnitStep[x], x]], x] == UnitStep[x], True}

sol = DSolve[ {y'[x] + Max[x, 1] y[x] == 0, y[0] == 1}, y[x], x ]

{{y[x] -> E^(Integrate[1, 0] - Integrate[Max[1, x], x])}}

Plot[y[x] /. sol, {x, -3, 3}]

-Graphics-

FinalSol  = DSolve[ {y'[t] == If[ t ≤ 2, y[t], -y[t] / 2], y[0] == 1}, y, t]

DSolve[{Derivative[1][y][t] == If[t <= 2, y[t], -(y[t]/2)], y[0] == 1}, y, t]

SolFromMinusInfinityToTwo = DSolve[{y'[t] == y[t], y[0] == 1}, y, t]

{{y -> Function[{t}, E^t]}}

SolFromTwoToInfinity = DSolve[{y'[t] == -y[t] / 2, y[2] == E ^ 2}, y, t]

{{y -> Function[{t}, E^(3 - t/2)]}}

eqn = {y''[t] + y[t] == Piecewise[{{-1, t < 0}, {1, t < 1}, {Sin[t], t < 2}}], y[0] == 1, y'[0] == 1};

sol = DSolve[ eqn, y, t ]

DSolve[{y[t] + Derivative[2][y][t] == Piecewise[{{-1, t < 0}, {1, t < 1}, {Sin[t], t < 2}}, 0], y[0] == 1, Derivative[1][y][0] == 1}, y, t]

eqn = {y''[x]  - Clip[x] * y[x] == 0, y[0] == 0, y'[0] == -1};

DSolve[eqn, y, x]

DSolve[{-(Clip[x]*y[x]) + Derivative[2][y][x] == 0, y[0] == 0, Derivative[1][y][0] == -1}, y, x]