What is the solution for y^{''}-3y^'+2y=4x^2 ?

The solution for y^{''}-3y^'+2y=4x^2 is y=c_{1}e^{2x}+c_{2}e^x+2x^2+6x+7

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y^{''}-3y^'+2y=4x^2

y^{^{''}} - 3 y^{^{'}} + 2 y = 4 x^{2}

y = c_{1} e^{2 x} + c_{2} e^{x} + 2 x^{2} + 6 x + 7

Solution steps

y^{''} (x) - 3 y^{'} (x) + 2 y = 4 x^{2}

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{x} + 2 x^{2} + 6 x + 7

y = c_{1} e^{2 x} + c_{2} e^{x} + 2 x^{2} + 6 x + 7

\int \frac{x}{2 x} d x

x \to \infty lim (\frac{x}{k})

\frac{d}{d x} (6 x^{2} - x)

\frac{\partial}{\partial y} (x sin (yz))

x^{2} - x

What is the solution for y^{''}-3y^'+2y=4x^2 ?
The solution for y^{''}-3y^'+2y=4x^2 is y=c_{1}e^{2x}+c_{2}e^x+2x^2+6x+7