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

The solution for y^{''}+y=2*e^x is y=c_{1}cos(x)+c_{2}sin(x)+e^x

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

y^{^{''}} + y = 2 \cdot e^{x}

y = c_{1} cos (x) + c_{2} sin (x) + e^{x}

Solution steps

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

Solve linear ODE:

y = c_{1} cos (x) + c_{2} sin (x) + e^{x}

y = c_{1} cos (x) + c_{2} sin (x) + e^{x}

y^{^{''}} - y^{^{'}} - 6 y = 10 x e^{3 x} + 17 e^{3 x}

y^{^{''}} - 4 y = x + sin (x)

y^{^{''}} - 4 y^{^{'}} + 4 y = e^{x} sin (x)

y^{^{''}} - 4 y^{^{'}} + 5 y = 10 x^{2} + 4 x + 8

y^{^{''}} + 2 y^{^{'}} + y = (\frac{1}{6}) (e^{- x})

What is the solution for y^{''}+y=2*e^x ?
The solution for y^{''}+y=2*e^x is y=c_{1}cos(x)+c_{2}sin(x)+e^x