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

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

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y^{''}-2y^'=-2e^x

y^{^{''}} - 2 y^{^{'}} = - 2 e^{x}

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

Solution steps

y^{''} (x) - 2 y^{'} (x) = - 2 e^{x}

Solve linear ODE:

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

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

y^{^{''}} - 16 y = (x + 1) e^{x}

y^{^{''}} + 9 y = 13 e^{2 x} + 10 sin (2 x)

y^{^{''}} - 3 y^{^{'}} = sin (2 x)

y^{^{''}} + y = 4 e^{x} + x^{2} - x

y^{^{''}} + 4 y^{^{'}} + 3 y = 1

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