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-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} - 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} - e^{2 t}

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

y^{^{''}} = 2 y + tan^{3} (x), y^{1} (x) = \frac{1}{2} tan (x)

y^{^{''}} - 2 y + y = \frac{e ^{x}}{1 + x ^{2}}

y^{^{''}} - 11 y^{^{'}} + 28 y = 108 e^{t}

y^{^{''}} - 5 y^{^{'}} + 6 y = sin (3 x)

y^{^{''}} - 2 y^{^{'}} + y = x - 2 - 2 cos (8)

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-e^{2t}