What is the solution for y^{''}-y^'-y=0 ?

The solution for y^{''}-y^'-y=0 is y=c_{1}e^{((1+sqrt(5))t)/2}+c_{2}e^{((1-sqrt(5))t)/2}

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y^{''}-y^'-y=0

y^{^{''}} - y^{^{'}} - y = 0

y = c_{1} e^{\frac{( 1 + 5 ) t}{2}} + c_{2} e^{\frac{( 1 - 5 ) t}{2}}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{\frac{( 1 + 5 ) t}{2}} + c_{2} e^{\frac{( 1 - 5 ) t}{2}}

y = c_{1} e^{\frac{( 1 + 5 ) t}{2}} + c_{2} e^{\frac{( 1 - 5 ) t}{2}}

x \to 4 lim (2 x - 1)

\frac{d}{d x} (2 e^{x} ln (x))

x \to - 2 lim (- x^{2} + 8 x - 4)

\int \frac{28}{( 1 - x ^{2} ) ^{\frac{3}{2}}} d x

(1 x + y e^{\frac{y}{x}}) d x - x e^{\frac{y}{x}} d y = 0, y (1) = 3

What is the solution for y^{''}-y^'-y=0 ?
The solution for y^{''}-y^'-y=0 is y=c_{1}e^{((1+sqrt(5))t)/2}+c_{2}e^{((1-sqrt(5))t)/2}