Find the special solution of the differential equation y '' '= e ^ (- x) satisfying the initial condition y | (x = 1) = y' | (x = 1) = y '' | (x = 1) = 0

Find the special solution of the differential equation y '' '= e ^ (- x) satisfying the initial condition y | (x = 1) = y' | (x = 1) = y '' | (x = 1) = 0

Integral: Y "= - e ^ (- x) + C1, substituting Y" (1) = 0, C1 = e ^ (- 1),
That is Y "= - e ^ (- x) + 1 / E
Re integration: y '= e ^ (- x) + X / E + C2, substituting y' (1) = 0, C2 = - 2 / E
That is y '= e ^ (- x) + X / E-2 / E
Re integration: y = - e ^ (- x) + x ^ 2 / (2e) - 2x / E + C3, substituting y (1) = 0, C3 = 5 / (2e)
So y = - e ^ (- x) + x ^ 2 / (2e) - 2x / E + 5 / (2e)