What is the solution for 2*y^{''}+y^'-y=0 ?

The solution for 2*y^{''}+y^'-y=0 is y=c_{1}e^{t/2}+c_{2}e^{-t}

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2*y^{''}+y^'-y=0

2 \cdot y^{^{''}} + y^{^{'}} - y = 0

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

Solution steps

2 \cdot y^{''} (t) + y^{'} (t) - y = 0

Solve linear ODE:

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

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

y^{^{''}} + 2 y^{^{'}} = 2 t + 5 - e^{- 2 t}

\frac{d}{d x} (x^{2 - 2 x})

y = \frac{ln ( x )}{e ^{x}}

\int csc (6 x) d x

\frac{2 s}{s ^{2} + 2 s + 1}

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