What is the solution for y^{''}-4y=1,y^'(0)=2 ?

The solution for y^{''}-4y=1,y^'(0)=2 is y=(1+c_{2})e^{2t}+c_{2}e^{-2t}-1/4

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y^{''}-4y=1,y^'(0)=2

y^{^{''}} - 4 y = 1, y^{^{'}} (0) = 2

y = (1 + c_{2}) e^{2 t} + c_{2} e^{- 2 t} - \frac{1}{4}

Solution steps

y^{''} (t) - 4 y = 1

Solve linear ODE:

y = (1 + c_{2}) e^{2 t} + c_{2} e^{- 2 t} - \frac{1}{4}

y = (1 + c_{2}) e^{2 t} + c_{2} e^{- 2 t} - \frac{1}{4}

What is the solution for y^{''}-4y=1,y^'(0)=2 ?
The solution for y^{''}-4y=1,y^'(0)=2 is y=(1+c_{2})e^{2t}+c_{2}e^{-2t}-1/4