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

The solution for y^{''}-3y^'+2y=x^2 is y=c_{1}e^{2x}+c_{2}e^x+(x^2)/2+(3x)/2+7/4

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

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

y = c_{1} e^{2 x} + c_{2} e^{x} + \frac{x ^{2}}{2} + \frac{3 x}{2} + \frac{7}{4}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{x} + \frac{x ^{2}}{2} + \frac{3 x}{2} + \frac{7}{4}

y = c_{1} e^{2 x} + c_{2} e^{x} + \frac{x ^{2}}{2} + \frac{3 x}{2} + \frac{7}{4}

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What is the solution for y^{''}-3y^'+2y=x^2 ?
The solution for y^{''}-3y^'+2y=x^2 is y=c_{1}e^{2x}+c_{2}e^x+(x^2)/2+(3x)/2+7/4