What is the solution for y^{''}-4y=5,+y(0)=2 ?

The solution for y^{''}-4y=5,+y(0)=2 is y=c_{1}e^{2t}+c_{2}e^{-2t}-5/4

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

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

y = c_{1} e^{2 t} + c_{2} e^{- 2 t} - \frac{5}{4}

Solution steps

y^{''} (t) - 4 y = 5

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{- 2 t} - \frac{5}{4}

y = c_{1} e^{2 t} + c_{2} e^{- 2 t} - \frac{5}{4}

(d^{2} - 8 d + 9) y = 8 sin (5 x)

y^{^{''}} - 2 y^{^{'}} + 5 y = 4 sin (2 t)

y^{^{'''}} - 3 y^{^{''}} = sin (x)

2 y^{^{''}} - 4 y^{^{'}} + 4 y = \frac{e ^{x}}{cos ( x )}

y^{^{''}} - 10 y^{^{'}} + 26 y = 4

What is the solution for y^{''}-4y=5,+y(0)=2 ?
The solution for y^{''}-4y=5,+y(0)=2 is y=c_{1}e^{2t}+c_{2}e^{-2t}-5/4