What is the solution for y^{''}-3y^'+2y=x^2+8x+9 ?

The solution for y^{''}-3y^'+2y=x^2+8x+9 is y=c_{1}e^{2x}+c_{2}e^x+(x^2)/2+(11x)/2+49/4

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y^{''}-3y^'+2y=x^2+8x+9

y^{^{''}} - 3 y^{^{'}} + 2 y = x^{2} + 8 x + 9

y = c_{1} e^{2 x} + c_{2} e^{x} + \frac{x ^{2}}{2} + \frac{11 x}{2} + \frac{49}{4}

Solution steps

y^{''} (x) - 3 y^{'} (x) + 2 y = x^{2} + 8 x + 9

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{x} + \frac{x ^{2}}{2} + \frac{11 x}{2} + \frac{49}{4}

y = c_{1} e^{2 x} + c_{2} e^{x} + \frac{x ^{2}}{2} + \frac{11 x}{2} + \frac{49}{4}

(D^{4} - 8 D^{2} + 16) (y) = x e^{2 x}

\frac{d ^{2}}{d t ^{2}} (θ (t)) + 4 θ (t) = cos^{2} (2 t)

\frac{d ^{3} y}{d x ^{3}} - 12 (\frac{d y}{d x}) + 16 y = 32 x - 8

y^{^{''}} + y^{^{'}} - 6 y = 30 - 30 H (t - 4)

y^{^{''}} + 10 y^{^{'}} + 16 y = 3 cos (5 t) + 4 sin (5 t)

What is the solution for y^{''}-3y^'+2y=x^2+8x+9 ?
The solution for y^{''}-3y^'+2y=x^2+8x+9 is y=c_{1}e^{2x}+c_{2}e^x+(x^2)/2+(11x)/2+49/4