What is the solution for y^{''}-4y^'-5y=0,y(-1)=3,y^'(-1)=9 ?

The solution for y^{''}-4y^'-5y=0,y(-1)=3,y^'(-1)=9 is y=2e^{5+5t}+e^{-t-1}

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y^{''}-4y^'-5y=0,y(-1)=3,y^'(-1)=9

y^{^{''}} - 4 y^{^{'}} - 5 y = 0, y (- 1) = 3, y^{^{'}} (- 1) = 9

y = 2 e^{5 + 5 t} + e^{- t - 1}

Solution steps

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

Solve linear ODE:

y = 2 e^{5 + 5 t} + e^{- t - 1}

y = 2 e^{5 + 5 t} + e^{- t - 1}

\frac{d}{d x} (4 e^{4 e^{x} + x})

f (x) = \frac{- x ^{2} - 4}{( x ^{2} - 4 ) ^{2}}

\frac{d ^{2}}{d x ^{2}} (x^{3})

x^{2} arctan (x)

\int 3 sin^{3} (x co) s^{5} x d x

What is the solution for y^{''}-4y^'-5y=0,y(-1)=3,y^'(-1)=9 ?
The solution for y^{''}-4y^'-5y=0,y(-1)=3,y^'(-1)=9 is y=2e^{5+5t}+e^{-t-1}