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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

f^{^{'}} (x) = x^{2} - x

\int (4 - 3 x)^{3} d x

\int \frac{arccot ( e ^{x} )}{e ^{x}} d x

2 + \frac{4 ( s - 4 )}{s ^{2} + 16}

\frac{d}{d x} (- 3 sin (2 x))

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