What is the solution for y^{''}-10y^'+24y=48x^2+10 ?

The solution for y^{''}-10y^'+24y=48x^2+10 is y=c_{1}e^{6x}+c_{2}e^{4x}+2x^2+(5x)/3+17/18

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y^{''}-10y^'+24y=48x^2+10

y^{^{''}} - 10 y^{^{'}} + 24 y = 48 x^{2} + 10

y = c_{1} e^{6 x} + c_{2} e^{4 x} + 2 x^{2} + \frac{5 x}{3} + \frac{17}{18}

Solution steps

y^{''} (x) - 10 y^{'} (x) + 24 y = 48 x^{2} + 10

Solve linear ODE:

y = c_{1} e^{6 x} + c_{2} e^{4 x} + 2 x^{2} + \frac{5 x}{3} + \frac{17}{18}

y = c_{1} e^{6 x} + c_{2} e^{4 x} + 2 x^{2} + \frac{5 x}{3} + \frac{17}{18}

What is the solution for y^{''}-10y^'+24y=48x^2+10 ?
The solution for y^{''}-10y^'+24y=48x^2+10 is y=c_{1}e^{6x}+c_{2}e^{4x}+2x^2+(5x)/3+17/18