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

The solution for y^'+y=y^{''} 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^{''}

y^{^{'}} + y = y^{^{''}}

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

Solution steps

y^{'} (t) + y = y^{''} (t)

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}}

\frac{5}{6} y^{^{''}} + 10 y^{^{'}} + 60 y = 0, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + 8 y^{^{'}} + 9 y = 0, y (0) = 5, y^{^{'}} (0) = 0

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

\frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} + 6 y = 0

- y^{^{''}} + 9 y = 0

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