What is the solution for y^{''}-8y^'+16y=8e^{2t},y(0)=3,y^'(0)=2 ?

The solution for y^{''}-8y^'+16y=8e^{2t},y(0)=3,y^'(0)=2 is y=e^{4t}-6e^{4t}t+2e^{2t}

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y^{''}-8y^'+16y=8e^{2t},y(0)=3,y^'(0)=2

y^{^{''}} - 8 y^{^{'}} + 16 y = 8 e^{2 t}, y (0) = 3, y^{^{'}} (0) = 2

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + y^{^{'}} = x, y (0) = 5, y^{^{'}} (0) = 0

y^{^{''}} + 9 y = x^{2}

y^{^{''}} - 2 y^{^{'}} - 3 y = sin (x)

y^{^{''}} + 2 y^{^{'}} + 2 y = cos (t)

y^{^{''}} - 6 y^{^{'}} + 9 y = sin (3 t)

What is the solution for y^{''}-8y^'+16y=8e^{2t},y(0)=3,y^'(0)=2 ?
The solution for y^{''}-8y^'+16y=8e^{2t},y(0)=3,y^'(0)=2 is y=e^{4t}-6e^{4t}t+2e^{2t}