What is the solution for (d^2-4d+3)y=e^{-2t} ?

The solution for (d^2-4d+3)y=e^{-2t} is y=c_{1}e^{3t}+c_{2}e^t+1/15 e^{-2t}

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(d^2-4d+3)y=e^{-2t}

(d^{2} - 4 d + 3) y = e^{- 2 t}

y = c_{1} e^{3 t} + c_{2} e^{t} + \frac{1}{15} e^{- 2 t}

Solution steps

(d^{2} - 4 d + 3) y = e^{- 2 t}

Solve linear ODE:

y = c_{1} e^{3 t} + c_{2} e^{t} + \frac{1}{15} e^{- 2 t}

y = c_{1} e^{3 t} + c_{2} e^{t} + \frac{1}{15} e^{- 2 t}

y^{^{''}} + y^{^{'}} - 12 = 0

y^{^{''}} + 15 y^{^{'}} + 50 y = 1296 t e^{- t}

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

y^{^{''}} + 9 y = cos (3 t) + sin (3 t)

50 x^{^{''}} + 1 0^{6} x = 10

What is the solution for (d^2-4d+3)y=e^{-2t} ?
The solution for (d^2-4d+3)y=e^{-2t} is y=c_{1}e^{3t}+c_{2}e^t+1/15 e^{-2t}