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

The solution for y^{''}-2y^'-3y=3e^{2t},y(0)=1,y^'(0)=0 is y=e^{3t}+e^{-t}-e^{2t}

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

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

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

Solution steps

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

Solve linear ODE:

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

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

x \to \infty lim (\frac{1}{3})

\frac{d y}{d x} + 2 y - 8 e^{- x} = 0, y (0) = 7

\int_{0}^{1} (e^{x}) - (x e^{x^{2}}) d x

\int \frac{1 + x ^{2}}{x - x ^{3}} d x

\int \frac{1}{196 - x ^{2}} d x

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