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

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

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y^{''}-4y^'+5y=4e^{3t},y(0)=2,y^'(0)=7

y^{^{''}} - 4 y^{^{'}} + 5 y = 4 e^{3 t}, y (0) = 2, y^{^{'}} (0) = 7

y = e^{2 t} sin (t) + 2 e^{3 t}

Solution steps

y^{''} (t) - 4 y^{'} (t) + 5 y = 4 e^{3 t}

Solve linear ODE:

y = e^{2 t} sin (t) + 2 e^{3 t}

y = e^{2 t} sin (t) + 2 e^{3 t}

y^{^{'''}} - 2 y^{^{''}} - 5 y^{^{'}} + 6 y = e^{x} + x^{2}

x^{^{''}} + 3 x^{^{'}} + 2 x = sin (3 t)

x^{^{''}} + 8 x^{^{'}} + 16 = 0

y^{^{''}} + 5 y^{^{'}} + 6 y = e^{- t}

x^{^{''''}} - x^{^{''}} = 1

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