What is the solution for y^{''}-2y^'-3y=4,y^'(0)=-1 ?

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

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y^{''}-2y^'-3y=4,y^'(0)=-1

y^{^{''}} - 2 y^{^{'}} - 3 y = 4, y^{^{'}} (0) = - 1

y = \frac{e ^{3 t} ( - 1 + c _{2} )}{3} + c_{2} e^{- t} - \frac{4}{3}

Solution steps

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

Solve linear ODE:

y = \frac{e ^{3 t} ( - 1 + c _{2} )}{3} + c_{2} e^{- t} - \frac{4}{3}

y = \frac{e ^{3 t} ( - 1 + c _{2} )}{3} + c_{2} e^{- t} - \frac{4}{3}

y^{^{''}} + 9 y = cos (3 t), y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + 9 y = - 18

y^{^{''}} - y = e^{3 x}

1 q^{^{''}} + 4 q^{^{'}} + 5 q = 5 cos (40 t)

y^{^{''}} - 2 y^{^{'}} - 3 y = 12 e x - 3

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