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

The solution for y^{''}-4y^'+3y=t^2e^{2t} is y=c_{1}e^{3t}+c_{2}e^t-e^{2t}t^2-2e^{2t}

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y^{''}-4y^'+3y=t^2e^{2t}

y^{^{''}} - 4 y^{^{'}} + 3 y = t^{2} e^{2 t}

y = c_{1} e^{3 t} + c_{2} e^{t} - e^{2 t} t^{2} - 2 e^{2 t}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{3 t} + c_{2} e^{t} - e^{2 t} t^{2} - 2 e^{2 t}

y = c_{1} e^{3 t} + c_{2} e^{t} - e^{2 t} t^{2} - 2 e^{2 t}

y^{^{''}} + 2 y^{^{'}} + 2 y = e^{- x} csc (x)

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

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

y^{^{''}} - y = x^{2} e^{x} + e^{3 x}

\frac{d ^{2} y}{d x ^{2}} - 3 \frac{d y}{d x} = 2 e^{2 x} sin (x)

What is the solution for y^{''}-4y^'+3y=t^2e^{2t} ?
The solution for y^{''}-4y^'+3y=t^2e^{2t} is y=c_{1}e^{3t}+c_{2}e^t-e^{2t}t^2-2e^{2t}