What is the solution for y^{''}-2y^'-3y=e^t,y(0)=2,y^'(0)=4 ?

The solution for y^{''}-2y^'-3y=e^t,y(0)=2,y^'(0)=4 is y=(-5/8+9/4)e^{3t}+5/8 e^{-t}-1/4 e^t

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y^{''}-2y^'-3y=e^t,y(0)=2,y^'(0)=4

y^{^{''}} - 2 y^{^{'}} - 3 y = e^{t}, y (0) = 2, y^{^{'}} (0) = 4

y = (- \frac{5}{8} + \frac{9}{4}) e^{3 t} + \frac{5}{8} e^{- t} - \frac{1}{4} e^{t}

Solution steps

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

Solve linear ODE:

y = (- \frac{5}{8} + \frac{9}{4}) e^{3 t} + \frac{5}{8} e^{- t} - \frac{1}{4} e^{t}

y = (- \frac{5}{8} + \frac{9}{4}) e^{3 t} + \frac{5}{8} e^{- t} - \frac{1}{4} e^{t}

y^{^{''}} - y = 2 t - 4

y^{^{'''}} - 2 y^{^{''}} - 4 y^{^{'}} + 8 y = e^{- 3 x} + 8 x^{2}

\frac{1}{6} y^{^{''}} + 6 y = tan (6 t) - \frac{1}{3} e^{3 t}

y^{^{''}} + 4 y^{^{'}} + 5 y = e^{- 2 x} sec (x)

y^{^{''}} - 5 y^{^{'}} + 4 y = 1

What is the solution for y^{''}-2y^'-3y=e^t,y(0)=2,y^'(0)=4 ?
The solution for y^{''}-2y^'-3y=e^t,y(0)=2,y^'(0)=4 is y=(-5/8+9/4)e^{3t}+5/8 e^{-t}-1/4 e^t