What is the solution for y^{''}-2y^'-3y=1,y(0)=1,y^'(0)=-1 ?

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

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y^{''}-2y^'-3y=1,y(0)=1,y^'(0)=-1

y^{^{''}} - 2 y^{^{'}} - 3 y = 1, y (0) = 1, y^{^{'}} (0) = - 1

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

Solution steps

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

Solve linear ODE:

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

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

(D^{2} + 5 D + 4) Y = x^{2} + 7 x + 9

(d^{2} + 4) y = cos (x)

y^{^{''}} + 2 y^{^{'}} - 15 y = 7 e^{3 t} + (1 - 2 t) sin (6 t)

y^{^{''}} + 2 y^{^{'}} + 2 y = g (t), y (0) = 8, y^{^{'}} (0) = 6

y^{^{''}} - 4 y = 1, y^{^{'}} (0) = 2

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