What is the solution for y^{''}-4y^'-5y=0,y(1)=0,y^'(1)=2 ?

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

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y^{''}-4y^'-5y=0,y(1)=0,y^'(1)=2

y^{^{''}} - 4 y^{^{'}} - 5 y = 0, y (1) = 0, y^{^{'}} (1) = 2

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

Solution steps

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

Solve linear ODE:

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

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

n = 1 \sum \infty 8^{n}

g (x) = 9 - x

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

\frac{d y}{d x} = \frac{10 x}{3 y}

\int x^{3} e^{- x} d x

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