What is the solution for y^{''}-10y^'+25y=30x^2+3 ?

The solution for y^{''}-10y^'+25y=30x^2+3 is y=c_{1}e^{5x}+c_{2}xe^{5x}+(6x^2)/5+(24x)/(25)+51/125

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y^{''}-10y^'+25y=30x^2+3

y^{^{''}} - 10 y^{^{'}} + 25 y = 30 x^{2} + 3

y = c_{1} e^{5 x} + c_{2} x e^{5 x} + \frac{6 x ^{2}}{5} + \frac{24 x}{25} + \frac{51}{125}

Solution steps

y^{''} (x) - 10 y^{'} (x) + 25 y = 30 x^{2} + 3

Solve linear ODE:

y = c_{1} e^{5 x} + c_{2} x e^{5 x} + \frac{6 x ^{2}}{5} + \frac{24 x}{25} + \frac{51}{125}

y = c_{1} e^{5 x} + c_{2} x e^{5 x} + \frac{6 x ^{2}}{5} + \frac{24 x}{25} + \frac{51}{125}

What is the solution for y^{''}-10y^'+25y=30x^2+3 ?
The solution for y^{''}-10y^'+25y=30x^2+3 is y=c_{1}e^{5x}+c_{2}xe^{5x}+(6x^2)/5+(24x)/(25)+51/125