What is the solution for f^{''}(x)=-(f^'(x))/(2(n+2)) ?

The solution for f^{''}(x)=-(f^'(x))/(2(n+2)) is f(x)=c_{1}+c_{2}e^{-x/(2n+4)}

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f^{''}(x)=-(f^'(x))/(2(n+2))

f^{^{''}} (x) = - \frac{f ^{^{'}} ( x )}{2 ( n + 2 )}

f (x) = c_{1} + c_{2} e^{- \frac{x}{2 n + 4}}

Solution steps

f^{''} (x) = - \frac{f ^{'} ( x )}{2 ( n + 2 )}

Solve linear ODE:

f (x) = c_{1} + c_{2} e^{- \frac{x}{2 n + 4}}

f (x) = c_{1} + c_{2} e^{- \frac{x}{2 n + 4}}

y^{^{''}} - y = 0, y (0) = 1

u^{^{''}} + 49 H (t) = 0, H (0) = 3, u^{^{'}} (0) = - 7

y^{^{'''}} - 6 y^{^{''}} + 11 y^{^{'}} - 6 y = 0, y (π) = 0

y^{^{''}} + y = 0, y^{^{''}} (0) = 0, y (0) = 1

4 y^{^{''}} - 4 y + 5 y = 0, y (0) = \frac{1}{2}, y^{^{'}} (0) = 1

What is the solution for f^{''}(x)=-(f^'(x))/(2(n+2)) ?
The solution for f^{''}(x)=-(f^'(x))/(2(n+2)) is f(x)=c_{1}+c_{2}e^{-x/(2n+4)}