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}

\frac{d}{d x} (30 e^{x})

\frac{d}{d x} (22 x)

\frac{d}{d x} (\frac{x}{e ^{- x}})

x \to 2 - lim (x^{2} - 4)

n = 0 \sum \infty (\frac{5}{3})^{n}

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}