What is the solution for y^{''}+y^'=2e^x+14 ?

The solution for y^{''}+y^'=2e^x+14 is y=c_{1}+c_{2}e^{-x}+e^x+14x

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y^{''}+y^'=2e^x+14

y^{^{''}} + y^{^{'}} = 2 e^{x} + 14

y = c_{1} + c_{2} e^{- x} + e^{x} + 14 x

Solution steps

y^{''} (x) + y^{'} (x) = 2 e^{x} + 14

Solve linear ODE:

y = c_{1} + c_{2} e^{- x} + e^{x} + 14 x

y = c_{1} + c_{2} e^{- x} + e^{x} + 14 x

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\frac{\partial}{\partial y} (cos (x y) - e^{z} + 2)

\frac{\partial}{\partial y} (r cos (y))

\frac{d}{d x} (4 x^{2} - 5 x + 4)

f (x) = x^{4 c o s (x)}

What is the solution for y^{''}+y^'=2e^x+14 ?
The solution for y^{''}+y^'=2e^x+14 is y=c_{1}+c_{2}e^{-x}+e^x+14x