What is the solution for y^{''}-2y^'+y=e^{t-1},y(1)=0,y^'(1)=5 ?

The solution for y^{''}-2y^'+y=e^{t-1},y(1)=0,y^'(1)=5 is y=-(9e^{t-1})/2+4e^{t-1}t+(e^{t-1}t^2)/2

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y^{''}-2y^'+y=e^{t-1},y(1)=0,y^'(1)=5

y^{^{''}} - 2 y^{^{'}} + y = e^{t - 1}, y (1) = 0, y^{^{'}} (1) = 5

y = - \frac{9 e ^{t - 1}}{2} + 4 e^{t - 1} t + \frac{e ^{t - 1} t ^{2}}{2}

Solution steps

y^{''} (t) - 2 y^{'} (t) + y = e^{t - 1}

Solve linear ODE:

y = - \frac{9 e ^{t - 1}}{2} + 4 e^{t - 1} t + \frac{e ^{t - 1} t ^{2}}{2}

y = - \frac{9 e ^{t - 1}}{2} + 4 e^{t - 1} t + \frac{e ^{t - 1} t ^{2}}{2}

y^{^{''}} - y^{^{'}} = x e^{2 x}

y^{^{''}} + 4 y = - 8 sin (2 t)

y^{^{''}} - 2 y^{^{'}} + 2 y = e^{2} x (cos (x) - 9 sin (x))

\frac{d ^{2} y}{d t ^{2}} - 2 \frac{d y}{d t} + y = 5 e^{t}

y^{^{''}} - x e^{- x} = 2 y^{^{'}}

What is the solution for y^{''}-2y^'+y=e^{t-1},y(1)=0,y^'(1)=5 ?
The solution for y^{''}-2y^'+y=e^{t-1},y(1)=0,y^'(1)=5 is y=-(9e^{t-1})/2+4e^{t-1}t+(e^{t-1}t^2)/2