What is the solution for y^{''}-2y^'-3y=2e^x-10sin(x) ?

The solution for y^{''}-2y^'-3y=2e^x-10sin(x) is y=c_{1}e^{3x}+c_{2}e^{-x}-1/2 e^x+2sin(x)-cos(x)

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y^{''}-2y^'-3y=2e^x-10sin(x)

y^{^{''}} - 2 y^{^{'}} - 3 y = 2 e^{x} - 10 sin (x)

y = c_{1} e^{3 x} + c_{2} e^{- x} - \frac{1}{2} e^{x} + 2 sin (x) - cos (x)

Solution steps

y^{''} (x) - 2 y^{'} (x) - 3 y = 2 e^{x} - 10 sin (x)

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{- x} - \frac{1}{2} e^{x} + 2 sin (x) - cos (x)

y = c_{1} e^{3 x} + c_{2} e^{- x} - \frac{1}{2} e^{x} + 2 sin (x) - cos (x)

y^{^{''}} - y^{^{'}} - 6 y = 5 t

y^{^{'''}} - y^{^{'}} = 3 t

y^{^{'''}} - y^{^{'}} = 5 t

x^{^{''}} + 4 x^{^{'}} + 13 x = t e^{- t}

y^{^{''}} + y^{^{'}} = sec (x)

What is the solution for y^{''}-2y^'-3y=2e^x-10sin(x) ?
The solution for y^{''}-2y^'-3y=2e^x-10sin(x) is y=c_{1}e^{3x}+c_{2}e^{-x}-1/2 e^x+2sin(x)-cos(x)