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

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

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

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

y = \frac{3 e ^{5 t - 5}}{2} - \frac{3 e}{2} e^{- t}

Solution steps

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

Solve linear ODE:

y = \frac{3 e ^{5 t - 5}}{2} - \frac{3 e}{2} e^{- t}

y = \frac{3 e ^{5 t - 5}}{2} - \frac{3 e}{2} e^{- t}

f (x) = 7 - 6 x, at x = 1

y^{2} \frac{d y}{d x} = y^{3} - x^{3}

x \to - 2 lim (x^{2} + 5)

f (x) = x^{2}, g (x) = x + 2

(x^{2} + 1) y^{^{'}} + 3 x (y - 1) = 0, y (0) = 9

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