What is the solution for y^{''}-4y^'-5y=0,y(1)=0,y^'(1)=8 ?

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

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y^{''}-4y^'-5y=0,y(1)=0,y^'(1)=8

y^{^{''}} - 4 y^{^{'}} - 5 y = 0, y (1) = 0, y^{^{'}} (1) = 8

y = \frac{4 e ^{5 t - 5}}{3} - \frac{4 e}{3} e^{- t}

Solution steps

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

Solve linear ODE:

y = \frac{4 e ^{5 t - 5}}{3} - \frac{4 e}{3} e^{- t}

y = \frac{4 e ^{5 t - 5}}{3} - \frac{4 e}{3} e^{- t}

- cot (x)

y^{^{''''}} + 2 y^{^{''}} = 0

\int \frac{x + 2}{x ^{2} + 4 x} d x

\int \frac{1}{1 + 3 x} d x

y = sin (tan (3 x))

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