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

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

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y^{''}-6y^'+8y=2e^{3t},y(0)=0,y^'(0)=2

y^{^{''}} - 6 y^{^{'}} + 8 y = 2 e^{3 t}, y (0) = 0, y^{^{'}} (0) = 2

y = 2 e^{4 t} - 2 e^{3 t}

Solution steps

y^{''} (t) - 6 y^{'} (t) + 8 y = 2 e^{3 t}

Solve linear ODE:

y = 2 e^{4 t} - 2 e^{3 t}

y = 2 e^{4 t} - 2 e^{3 t}

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

y^{^{''}} + y = cos (- t + \frac{π}{2}), y (0) = 0, y^{^{'}} (0) = 1

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

y^{^{''}} + 2 y^{^{'}} + y = 2 cos (2 x) + 3 x + 2 + 3 e^{x}

y^{^{''}} + y = t^{2} sin (t), y (0) = 0, y^{^{'}} (0) = 0

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