What is the solution for y^{''}-7y^'+12y=2e^{4t},y(0)=1,y^'(0)=0 ?

The solution for y^{''}-7y^'+12y=2e^{4t},y(0)=1,y^'(0)=0 is y=-5e^{4t}+6e^{3t}+2e^{4t}t

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y^{''}-7y^'+12y=2e^{4t},y(0)=1,y^'(0)=0

y^{^{''}} - 7 y^{^{'}} + 12 y = 2 e^{4 t}, y (0) = 1, y^{^{'}} (0) = 0

y = - 5 e^{4 t} + 6 e^{3 t} + 2 e^{4 t} t

Solution steps

y^{''} (t) - 7 y^{'} (t) + 12 y = 2 e^{4 t}

Solve linear ODE:

y = - 5 e^{4 t} + 6 e^{3 t} + 2 e^{4 t} t

y = - 5 e^{4 t} + 6 e^{3 t} + 2 e^{4 t} t

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

y^{^{''}} - 5 y^{^{'}} + 6 y = sin (4 x)

y^{^{''}} - 2 y^{^{'}} + 4 y = e^{4 x}

y^{^{''}} - 7 y^{^{'}} + 10 y = x^{2} - 4 + e^{x}

y^{^{''}} + 8 y^{^{'}} + 16 y = t^{- 2} e^{- 4 t}

What is the solution for y^{''}-7y^'+12y=2e^{4t},y(0)=1,y^'(0)=0 ?
The solution for y^{''}-7y^'+12y=2e^{4t},y(0)=1,y^'(0)=0 is y=-5e^{4t}+6e^{3t}+2e^{4t}t