What is the solution for y^{''}-4y^'-5y=0,y(1)=0,y^'(1)=5 ?

The solution for y^{''}-4y^'-5y=0,y(1)=0,y^'(1)=5 is y=(5e^{5t-5})/6-(5e)/6 e^{-t}

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y^{''}-4y^'-5y=0,y(1)=0,y^'(1)=5

y^{^{''}} - 4 y^{^{'}} - 5 y = 0, y (1) = 0, y^{^{'}} (1) = 5

y = \frac{5 e ^{5 t - 5}}{6} - \frac{5 e}{6} e^{- t}

Solution steps

y^{''} (t) - 4 y^{'} (t) - 5 y = 0

Solve linear ODE:

y = \frac{5 e ^{5 t - 5}}{6} - \frac{5 e}{6} e^{- t}

y = \frac{5 e ^{5 t - 5}}{6} - \frac{5 e}{6} e^{- t}

x^{^{'''}} - 12 x^{^{''}} + 48 x^{^{'}} - 64 x = 0

y^{^{''}} = 4 y^{^{'}} - 3 y

(D^{2} + 3 D - 10) y = 0

y^{^{''}} + 8 y^{^{'}} + 8 y = 0

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

What is the solution for y^{''}-4y^'-5y=0,y(1)=0,y^'(1)=5 ?
The solution for y^{''}-4y^'-5y=0,y(1)=0,y^'(1)=5 is y=(5e^{5t-5})/6-(5e)/6 e^{-t}