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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d}{d x} (x - ln (e^{x} cos (x)))

\frac{d}{d x} (\frac{1}{2} x)

\frac{d}{d x} (\frac{x ^{2}}{2 x + 1})

\frac{d}{d y} (x^{2} - y^{2})

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