What is the solution for (D^2-3D+2)y=x+1 ?

The solution for (D^2-3D+2)y=x+1 is y=c_{1}e^{2x}+c_{2}e^x+x/2+5/4

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(D^2-3D+2)y=x+1

(D^{2} - 3 D + 2) y = x + 1

y = c_{1} e^{2 x} + c_{2} e^{x} + \frac{x}{2} + \frac{5}{4}

Solution steps

(D^{2} - 3 D + 2) y = x + 1

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{x} + \frac{x}{2} + \frac{5}{4}

y = c_{1} e^{2 x} + c_{2} e^{x} + \frac{x}{2} + \frac{5}{4}

4 y^{^{''}} + 4 y + y = 3 x e^{x}

y^{^{''}} + 4 y + 4 = 0

x^{^{''}} + 8 x^{^{'}} + 16 x = e^{- t} sin (4 t)

D (t) = \frac{100 ln ( 24 + 2 )}{24 + 2}

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

What is the solution for (D^2-3D+2)y=x+1 ?
The solution for (D^2-3D+2)y=x+1 is y=c_{1}e^{2x}+c_{2}e^x+x/2+5/4