What is the solution for 2y^{''}-6y^'+4y=6e^{2t} ?

The solution for 2y^{''}-6y^'+4y=6e^{2t} is y=c_{1}e^{2t}+c_{2}e^t+3te^{2t}

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2y^{''}-6y^'+4y=6e^{2t}

2 y^{^{''}} - 6 y^{^{'}} + 4 y = 6 e^{2 t}

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

Solution steps

2 y^{''} (t) - 6 y^{'} (t) + 4 y = 6 e^{2 t}

Solve linear ODE:

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

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

\int e^{2 x} (3 + e^{2 x})^{5} d x

\int_{- 1}^{2} \frac{5 x}{x ^{2} - x - 6} d x

y^{^{''}} + 3 y^{^{'}} + 2 y = 4

\int csc^{3} (x) cot^{2} (x) d x

e^{θ}

What is the solution for 2y^{''}-6y^'+4y=6e^{2t} ?
The solution for 2y^{''}-6y^'+4y=6e^{2t} is y=c_{1}e^{2t}+c_{2}e^t+3te^{2t}