What is the solution for y^{''}-4y^'-5y=0,y(0)=3,y^'(0)=4 ?

The solution for y^{''}-4y^'-5y=0,y(0)=3,y^'(0)=4 is y=(3-11/6)e^{5t}+11/6 e^{-t}

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y^{''}-4y^'-5y=0,y(0)=3,y^'(0)=4

y^{^{''}} - 4 y^{^{'}} - 5 y = 0, y (0) = 3, y^{^{'}} (0) = 4

y = (3 - \frac{11}{6}) e^{5 t} + \frac{11}{6} e^{- t}

Solution steps

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

Solve linear ODE:

y = (3 - \frac{11}{6}) e^{5 t} + \frac{11}{6} e^{- t}

y = (3 - \frac{11}{6}) e^{5 t} + \frac{11}{6} e^{- t}

\frac{d}{d x} (x - ln (8 x))

(7 x + 4)^{2}

\frac{d y}{d x} = 3 x y

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

\frac{d}{d x} (x^{2} arctan (7 x))

What is the solution for y^{''}-4y^'-5y=0,y(0)=3,y^'(0)=4 ?
The solution for y^{''}-4y^'-5y=0,y(0)=3,y^'(0)=4 is y=(3-11/6)e^{5t}+11/6 e^{-t}