What is the solution for y^{''}-2y^'-3y=6,y(0)=6,y^'(0)=17 ?

The solution for y^{''}-2y^'-3y=6,y(0)=6,y^'(0)=17 is y=(-7/4+8)e^{3t}+7/4 e^{-t}-2

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y^{''}-2y^'-3y=6,y(0)=6,y^'(0)=17

y^{^{''}} - 2 y^{^{'}} - 3 y = 6, y (0) = 6, y^{^{'}} (0) = 17

y = (- \frac{7}{4} + 8) e^{3 t} + \frac{7}{4} e^{- t} - 2

Solution steps

y^{''} (t) - 2 y^{'} (t) - 3 y = 6

Solve linear ODE:

y = (- \frac{7}{4} + 8) e^{3 t} + \frac{7}{4} e^{- t} - 2

y = (- \frac{7}{4} + 8) e^{3 t} + \frac{7}{4} e^{- t} - 2

y^{^{''}} - 8 y^{^{'}} + 16 y = (7 t^{4} - 2 t^{3}) e^{4 t}

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

y^{^{''}} - 4 y^{^{'}} - 5 y = 2 e^{(3 x)}

y^{^{''}} + 5 y^{^{'}} + 6 y = 19 t e^{5 t}

y^{^{''}} - 2 y^{^{'}} + y = 15 x e^{x}

What is the solution for y^{''}-2y^'-3y=6,y(0)=6,y^'(0)=17 ?
The solution for y^{''}-2y^'-3y=6,y(0)=6,y^'(0)=17 is y=(-7/4+8)e^{3t}+7/4 e^{-t}-2