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^{'} (t) = y

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}}

y^{^{'''}} + 2 y^{^{''}} - 19 y^{^{'}} - 20 y = 0

\frac{d ^{2} y}{d x ^{2}} - 6 \frac{d y}{d x} + 14 y = 0

y^{^{''}} - 13 y^{^{'}} + 30 y = 0

y^{^{'''}} - 5 y^{^{''}} - y^{^{'}} + 5 y = 0

2 y^{^{''}} + 5 y^{^{'}} - 3 y = 0, y (0) = - 1, y^{^{'}} (0) = 17

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}