What is the solution for y^{''}-4y^'-5y=0,y(0)=2,y^'(0)=3 ?

The solution for y^{''}-4y^'-5y=0,y(0)=2,y^'(0)=3 is y=(2-7/6)e^{5t}+7/6 e^{-t}

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y^{''}-4y^'-5y=0,y(0)=2,y^'(0)=3

y^{^{''}} - 4 y^{^{'}} - 5 y = 0, y (0) = 2, y^{^{'}} (0) = 3

y = (2 - \frac{7}{6}) e^{5 t} + \frac{7}{6} e^{- t}

Solution steps

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

Solve linear ODE:

y = (2 - \frac{7}{6}) e^{5 t} + \frac{7}{6} e^{- t}

y = (2 - \frac{7}{6}) e^{5 t} + \frac{7}{6} e^{- t}

(d^{5} + 8 d^{3} + 16 d) y = 0

y^{^{''}} - 6 y^{^{'}} + 9 y = 0, y (0) = 1, y (1) = 0

\frac{d ^{2} x}{d t ^{2}} + 81 x = 0

\frac{1}{4} y^{^{''}} + y^{^{'}} + 4 y = 0

y^{^{'''}} + 5 y^{^{''}} + 3 y^{^{'}} - 9 y = 0

What is the solution for y^{''}-4y^'-5y=0,y(0)=2,y^'(0)=3 ?
The solution for y^{''}-4y^'-5y=0,y(0)=2,y^'(0)=3 is y=(2-7/6)e^{5t}+7/6 e^{-t}