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

The solution for y^{''}-3y^'+2y=(x+1)^2 is y=c_{1}e^{2x}+c_{2}e^x+(x^2)/2+(5x)/2+15/4

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

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

y = c_{1} e^{2 x} + c_{2} e^{x} + \frac{x ^{2}}{2} + \frac{5 x}{2} + \frac{15}{4}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{x} + \frac{x ^{2}}{2} + \frac{5 x}{2} + \frac{15}{4}

y = c_{1} e^{2 x} + c_{2} e^{x} + \frac{x ^{2}}{2} + \frac{5 x}{2} + \frac{15}{4}

(D^{2} - 4) y = cos^{2} (x)

y^{^{''}} - 2 y^{^{'}} - 3 y = 8 e^{2 t} + 10 t^{2}

2 y^{^{''}} - 3 y = x + 1

y^{^{''}} - 5 y^{^{'}} + 6 y = x

\frac{1}{6} y^{^{''}} + 6 y = 3 tan (6 t) - \frac{1}{3} e^{t}

What is the solution for y^{''}-3y^'+2y=(x+1)^2 ?
The solution for y^{''}-3y^'+2y=(x+1)^2 is y=c_{1}e^{2x}+c_{2}e^x+(x^2)/2+(5x)/2+15/4