What is the solution for y^{''}-y^'-2y=0,y(0)=a,y^'(0)=2 ?

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

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y^{''}-y^'-2y=0,y(0)=a,y^'(0)=2

y^{^{''}} - y^{^{'}} - 2 y = 0, y (0) = a, y^{^{'}} (0) = 2

y = e^{2 t} (a + \frac{2 - 2 a}{3}) - \frac{e ^{- t} ( 2 - 2 a )}{3}

Solution steps

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

Solve linear ODE:

y = e^{2 t} (a + \frac{2 - 2 a}{3}) - \frac{e ^{- t} ( 2 - 2 a )}{3}

y = e^{2 t} (a + \frac{2 - 2 a}{3}) - \frac{e ^{- t} ( 2 - 2 a )}{3}

y^{^{''}} + 4 y^{^{'}} + 3 y = 0, y (0) = 1, y^{^{'}} (0) = 0

2 y^{^{''}} + 6 y^{^{'}} + 8 y = 0, y (0) = 5, y^{^{'}} (0) = 0

x^{^{''}} + x^{^{'}} + 2 x = 0

y^{^{''}} + y^{^{'}} - 20 y = 0, y (0) = 1, y (0) = 22

20 y^{^{''}} + 60 y = 0

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