What is the solution for 4y^{''}-8y=2 ?

The solution for 4y^{''}-8y=2 is y=c_{1}e^{sqrt(2)t}+c_{2}e^{-sqrt(2)t}-1/4

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4y^{''}-8y=2

4 y^{^{''}} - 8 y = 2

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

Solution steps

4 y^{''} (t) - 8 y = 2

Solve linear ODE:

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

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

y^{^{''}} - y = 2.9 cos (t), y (0) = 0, y^{^{'}} (0) = 2.9

\frac{d ^{2} y}{d x ^{2}} + 3 \frac{d y}{d x} - 5 y = e^{x}

\frac{d ^{2} y}{d t ^{2}} - 4 \frac{d y}{d t} + 4 y = e^{- 4 t}

y^{^{''}} + 8 y^{^{'}} + 15 y = - 6 cos (5 x)

x^{^{''}} + b x^{^{'}} - c = 0

What is the solution for 4y^{''}-8y=2 ?
The solution for 4y^{''}-8y=2 is y=c_{1}e^{sqrt(2)t}+c_{2}e^{-sqrt(2)t}-1/4