What is the solution for y^{''}-y^'-6y=e^{2t},y^'(0)=1,y(0)=3 ?

The solution for y^{''}-y^'-6y=e^{2t},y^'(0)=1,y(0)=3 is y=(8e^{3t})/5+33/20 e^{-2t}-1/4 e^{2t}

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y^{''}-y^'-6y=e^{2t},y^'(0)=1,y(0)=3

y^{^{''}} - y^{^{'}} - 6 y = e^{2 t}, y^{^{'}} (0) = 1, y (0) = 3

y = \frac{8 e ^{3 t}}{5} + \frac{33}{20} e^{- 2 t} - \frac{1}{4} e^{2 t}

Solution steps

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

Solve linear ODE:

y = \frac{8 e ^{3 t}}{5} + \frac{33}{20} e^{- 2 t} - \frac{1}{4} e^{2 t}

y = \frac{8 e ^{3 t}}{5} + \frac{33}{20} e^{- 2 t} - \frac{1}{4} e^{2 t}

y^{^{''}} - 4 y = 14 cos (2 x) + 21 sin (2 x)

y^{^{''}} + 5 y^{^{'}} + 6 y = - 5 - e^{x}

y^{^{''}} + 3 y^{^{'}} + 9 y = 8 sin (6 t)

y^{^{''}} - y^{^{'}} + 0.25 y = 3 + e^{\frac{x}{2}}

y^{^{''}} + 2 y^{^{'}} - 3 y = 8 x e^{x}

What is the solution for y^{''}-y^'-6y=e^{2t},y^'(0)=1,y(0)=3 ?
The solution for y^{''}-y^'-6y=e^{2t},y^'(0)=1,y(0)=3 is y=(8e^{3t})/5+33/20 e^{-2t}-1/4 e^{2t}