What is the solution for y^{''}-3y^{''}+3y^'-y=x-2e^x ?

The solution for y^{''}-3y^{''}+3y^'-y=x-2e^x is y=c_{1}e^{x/2}+c_{2}e^x-x-3+2xe^x

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y^{''}-3y^{''}+3y^'-y=x-2e^x

y^{^{''}} - 3 y^{^{''}} + 3 y^{^{'}} - y = x - 2 e^{x}

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{x} - x - 3 + 2 x e^{x}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{x} - x - 3 + 2 x e^{x}

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{x} - x - 3 + 2 x e^{x}

y^{^{''}} + 4 y^{^{'}} + 5 y = 1 - 5 x + 8 cos (x) - 8 sin (x)

y^{^{''}} + 16 y = 4 e^{2 t} + 4 sin (2 t)

x^{^{''}} + x = a cos (t)

y^{^{''}} + 36 y = 36 sec^{2} (6 x)

y^{^{'''}} + 25 y^{^{'}} = 10 sin (5 t) - 4 t^{2}

What is the solution for y^{''}-3y^{''}+3y^'-y=x-2e^x ?
The solution for y^{''}-3y^{''}+3y^'-y=x-2e^x is y=c_{1}e^{x/2}+c_{2}e^x-x-3+2xe^x