What is the solution for y^{''}-2y=5 ?

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

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y^{''}-2y=5

y^{^{''}} - 2 y = 5

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

Solution steps

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

Solve linear ODE:

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

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

\int \frac{( 3 in x - 5 ) ^{4}}{x} d x

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

y = x^{2} - 1 arcsec (x)

6 (t) e^{- 2 t}

\frac{\partial}{\partial x} ((1 + 2 x y^{2})^{3})

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