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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

x cos (x)

\int \frac{2 + x}{4 x ^{2} - 4 x - 3} d x

f (x) = ln (x - 1)

\int (12 x^{2} - 8 x + 2) d x

\frac{d y}{d x} = \frac{( x - 1 )}{y}

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