What is the solution for y^{''}-1/(a^2)y=0,y(0)=1,y^'(0)=2 ?

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

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y^{''}-1/(a^2)y=0,y(0)=1,y^'(0)=2

y^{^{''}} - \frac{1}{a ^{2}} y = 0, y (0) = 1, y^{^{'}} (0) = 2

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

Solution steps

y^{''} (t) - \frac{1}{a ^{2}} y = 0

Solve linear ODE:

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

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

y^{^{''}} + 4 y^{^{'}} + 20 y = 0, y (0) = 4, y^{^{'}} (0) = - 5

4 y^{^{''}} - 4 y^{^{'}} + 5 y = 0, y (0) = \frac{1}{2}, y^{^{'}} (0) = 1

y^{^{''}} + y = 0, y (\frac{π}{3}) = 2

9 y^{^{''}} + 24 y^{^{'}} - 33 y = 0

\frac{d ^{2} y}{d x ^{2}} = (4 y)

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