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

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

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4y^{''}-y-1=0

4 y^{^{''}} - y - 1 = 0

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

Solution steps

4 y^{''} (t) - y - 1 = 0

Solve linear ODE:

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

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

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

\frac{d}{d x} (ln (cos (\frac{x - 1}{x})))

x \to 0 lim (x^{5} \cdot cos (\frac{1}{x}))

\int (\frac{1}{x ^{5}}) d x

\int_{1}^{3} 8^{x} - 2^{x} d x

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