What is the solution for 2y^{''}+(a+1)y=0 ?

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

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2y^{''}+(a+1)y=0

2 y^{^{''}} + (a + 1) y = 0

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

Solution steps

2 y^{''} (t) + (a + 1) y = 0

Solve linear ODE:

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

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

\int {\frac{3 - x}{x ^{2} + 1} + x ln (x^{2})} d x

y = 1 + 3 x^{2} (ln (x^{2}))

f (x) = (x^{3} + 4) e^{x}

\int \frac{x ^{2} + 4}{x + 1} d x

n = 0 \sum \infty 3 (\frac{x}{4})^{n}

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