What is the solution for y^{''}-2y^'+y=6t-2,y(-1)=3,y^'(-1)=7 ?

The solution for y^{''}-2y^'+y=6t-2,y(-1)=3,y^'(-1)=7 is y=e^{1+t}+2e^{1+t}t+6t+10

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y^{''}-2y^'+y=6t-2,y(-1)=3,y^'(-1)=7

y^{^{''}} - 2 y^{^{'}} + y = 6 t - 2, y (- 1) = 3, y^{^{'}} (- 1) = 7

y = e^{1 + t} + 2 e^{1 + t} t + 6 t + 10

Solution steps

y^{''} (t) - 2 y^{'} (t) + y = 6 t - 2

Solve linear ODE:

y = e^{1 + t} + 2 e^{1 + t} t + 6 t + 10

y = e^{1 + t} + 2 e^{1 + t} t + 6 t + 10

y^{^{''}} + 4 y^{^{'}} + y = 4, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} - 4 y^{^{'}} + 5 y = e^{2 x} (sin (x) + 2 cos (x))

y^{^{''}} - 3 y^{^{'}} + 5 y = 24 e^{- 2 x}

x^{^{''}} + 5 x^{^{'}} + 6 x = 2 e^{- t}

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

What is the solution for y^{''}-2y^'+y=6t-2,y(-1)=3,y^'(-1)=7 ?
The solution for y^{''}-2y^'+y=6t-2,y(-1)=3,y^'(-1)=7 is y=e^{1+t}+2e^{1+t}t+6t+10